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CHAPTER 12

PROBLEM 12.1 Astronauts who landed on the moon during the Apollo 15, 16 and 17 missions brought back a large collection of rocks to the earth. Knowing the rocks weighed 139 lb when they were on the moon, determine (a) the weight of the rocks on the earth, (b) the mass of the rocks in slugs. The acceleration due to gravity on the moon is 5.30 ft/s2.

SOLUTION Since the rocks weighed 139 lb on the moon, their mass is

= m (a)

Wmoon 139 lb = = 26.226 lb ⋅ s 2 /ft g moon 5.30 ft/s 2

On the earth, Wearth = mg earth = w (26.226 lb ⋅ s 2 /ft)(32.2 ft/s 2 )

(b)

Since 1 slug = 1 lb ⋅ s 2 /ft,

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w = 844 lb  m = 26.2 slugs 

PROBLEM 12.2 The value of g at any latitude φ may be obtained from the formula g 32.09(1 + 0.0053 sin 2φ ) ft/s 2 =

which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly spherical. Knowing that the weight of a silver bar has been officially designated as 5 lb, determine to four significant figures (a) the mass is slugs, (b) the weight in pounds at the latitudes of 0°, 45°, and 60°.

SOLUTION g 32.09(1 + 0.0053 sin 2φ ) ft/s 2 =

φ = 0° :

g = 32.09 ft/s 2

φ= 45°:

g = 32.175 ft/s 2

φ= 90°:

g = 32.26 ft/s 2

m=

(a) Mass at all latitudes: (b)

Weight:

5.00 lb 32.175 ft/s 2

= m 0.1554 lb ⋅ s 2 /ft 

W = mg

φ = 0° :

W= (0.1554 lb ⋅ s 2 /ft)(32.09 ft/s 2 ) = 4.987 lb



φ= 45°:

W= (0.1554 lb ⋅ s 2 /ft)(32.175 ft/s 2 ) = 5.000 lb



φ= 90°:

W= (0.1554 lb ⋅ s 2 /ft)(32.26 ft/s 2 ) = 5.013 lb



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PROBLEM 12.3 A 400-kg satellite has been placed in a circular orbit 1500 km above the surface of the earth. The acceleration of gravity at this elevation is 6.43 m/s2. Determine the linear momentum of the satellite, knowing that its orbital speed is 25.6 × 103 km/h.

SOLUTION Mass of satellite is independent of gravity:

m = 400 kg

= v 25.6 × 103 km/h  1h  = (25.6 × 106 m/h)  7.111 × 103 m/s =  3600 s  = L mv = (400 kg)(7.111 × 103 m/s)

L =2.84 × 106 kg ⋅ m/s 

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PROBLEM 12.4 A spring scale A and a lever scale B having equal lever arms are fastened to the roof of an elevator, and identical packages are attached to the scales as shown. Knowing that when the elevator moves downward with an acceleration of 1 m/s 2 the spring scale indicates a load of 60 N, determine (a) the weight of the packages, (b) the load indicated by the spring scale and the mass needed to balance the lever scale when the elevator moves upward with an acceleration of 1 m/s2.

SOLUTION Assume

g = 9.81 m/s 2

m=

W g

ΣF =ma : Fs − W =−

W a g

 a W 1 −  = Fs g 

or

= W

Fs 60 = 1 a 1− 1− 9.81 g W = 66.8 N 

(b) = ΣF ma : Fs= −W

W a g

 a Fs W 1 +  = g  1   = 66.811 +   9.81 

For the balance system B, ΣM = 0: bFw − bF = 0 0 p Fw = Fp

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Fs = 73.6 N 

PROBLEM 12.4 (Continued)

But

a  = Fw Ww 1 +  g 

and

 a = Fp W p 1 +  g 

so that

Ww = W p

and

m = w

W p 66.81 = g 9.81

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mw = 6.81 kg 

PROBLEM 12.5 In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3 ft/s 2 while still on a level section of the highway. Knowing that the speed of the bus is 60 mi/h as it begins to climb the grade and that the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus up the grade when its speed has decreased to 50 mi/h.

SOLUTION First consider when the bus is on the level section of the highway. alevel = 3 ft/s 2

Σ= Fx ma: = P

We have

W alevel g

Now consider when the bus is on the upgrade. We have

= ΣFx ma: P − W = sin 7°

Substituting for P

W a′ g

W W alevel − W sin 7° = a′ g g

a′ =alevel − g sin 7°

or

= (3 − 32.2 sin 7°) ft/s 2 = −0.92419 ft/s 2

For the uniformly decelerated motion 2 v 2 =(v0 ) upgrade + 2a′( xupgrade − 0)

 5  = mi/h  v0  , we have when v 50 Noting that 60 mi/h = 88 ft/s, then =  6  2

5  2 2 =  6 × 88 ft/s  (88 ft/s) + 2(−0.92419 ft/s ) xupgrade  

or or

xupgrade = 1280.16 ft xupgrade = 0.242 mi 

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PROBLEM 12.6 A 0.2-lb model rocket is launched vertically from rest at time t = 0 with a constant thrust of 2 lb for one second and no thrust for t > 1 s. Neglecting air resistance and the decrease in mass of the rocket, determine (a) the maximum height h reached by the rocket, (b) the time required to reach this maximum height.

SOLUTION W = 0.2 lbs For 0 ≤ t ≤ 1 s Ft = 2 lbs

Given:

Free Body Diagram:

For t > 1 s Ft = 0 lbs For the thrust phase, 0 ≤ t ≤ 1 s :

ΣF = ma : Ft − W = ma = F  a g  t − 1= =  W  

= v at =

At t = 1 s :

= y

For free flight phase, t > 1 s :

( 32.2 ) 

2  2 − 1=  289.8 ft/s 0.2  

( 289.8)(1= )

1 2 = at 2

W a g

289.8 ft/s

1 2 = 1) 144.9 ft ( 289.8)( 2

a =− g =− 32.2 ft/s v = v1 + a ( t − 1) = 289.8 + ( − 32.2 )( t − 1)

(a) At maximum height:

= v 0,

t −= 1

289.8 = 9.00 s, = t 10.00 s 32.2

v 2 − v12 = 2a ( y − y1 ) = −2 g ( y − y1 )

0 − ( 289.8 ) v 2 − v12 y − y1 = 1304.1 ft − = − = 2g ( 2 )( 32.2 ) 2

ymax= h= 1304.1 + 144.9

(b) Time to reach max height:

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h = 1449 ft  t = 10.00 s 

PROBLEM 12.7 A tugboat pulls a small barge through a harbor. The propeller thrust minus the drag produces a net thrust that varies linearly with speed. Knowing that the combined weight of the tug and barge is 3600 kN, determine (a) the time required to increase the speed from an initial value v1 = 1.0 m/s to a final value v2 = 2.5 m/s, (b) the distance traveled during this time interval.

SOLUTION W = 3600 kN

Given:

Free Body Diagram:

W ⇒ m = 366972.5 kg g m/s, v2 2.5 m/s = v1 1.0 = m=

81000 v = 81000 - 27000v N 3

From Graph:

Fnet = 81000 −

From FBD:

ΣFx = max ⇒ Fnet = max ⇒ ax = 81000 − 27000v m 27000 = ( 3 − v ) m/s2 m

Fnet m

ax =

(a) From Kinematics:

ax = t

dv dv ⇒ dt = dt ax

m v2 dv ∫ 27000 v1 3 − v m 2.5 t = ln ( 3 − v ) |1 − 27000

∫ dt = 0

t = 18.84 s 

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PROBLEM 12.7 (Continued) (b) From Kinematics:

ax dx = vdv ⇒ dx = x

m

v2

vdv ax

vdv

∫ dx = 27000 ∫ 3 − v v1 0 x =

m 2.5 3 − v − 3ln 3 − v ) |1 ( 27000

x = 36.14 m 

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PROBLEM 12.8 Determine the maximum theoretical speed that may be achieved over a distance of 60 m by a car starting from rest, knowing that the coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) four-wheel drive, (b) front-wheel drive, (c) rear-wheel drive.

SOLUTION (a)

Four-wheel drive Σ= Fy 0: N1 + N 2 −= W 0

F1 + F2 = µ N1 + µ N 2 = µ ( N1 + N 2 ) = µ w = ΣFx ma : F1 + = F2

ma

µ w = ma µW

mg = µ = µ= g 0.80(9.81) m m a = 7.848 m/s 2

= a

= v 2 2= ax 2(7.848 m/s 2 )(60 = m) 941.76 m 2 /s 2 v = 30.69 m/s

(b)

v = 110.5 km/h 

Front-wheel drive

F2 = ma

µ (0.6 W ) = ma 0.6µW 0.6µ mg = m m = 0.6 = µ g 0.6(0.80)(9.81)

a =

a = 4.709 m/s 2 = v 2 2= ax 2(4.709 m/s 2 )(60 = m) 565.1 m 2 /s 2 v = 23.77 m/s

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v = 85.6 km/h 

PROBLEM 12.8 (Continued) (c)

Rear-wheel drive

F1 = ma

µ (0.4 W ) = ma 0.4µW 0.4µ mg = m m = 0.4 = µ g 0.4(0.80)(9.81)

a =

a = 3.139 m/s 2 = v 2 2= ax 2(3.139 m/s 2 )(60 = m) 376.7 m 2 /s 2 v = 19.41 m/s

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v = 69.9 km/h 

PROBLEM 12.9 If an automobile’s braking distance from 90 km/h is 45 m on level pavement, determine the automobile’s braking distance from 90 km/h when it is (a) going up a 5° incline, (b) going down a 3-percent incline. Assume the braking force is independent of grade.

SOLUTION Assume uniformly decelerated motion in all cases. For braking on the level surface, v0 90 km/h 25 m/s, vf 0 = = = x f − x0 = 45 m v 2f =+ v02 2a ( x f − x0 ) a= =

v 2f − v02 2( x f − x0 ) 0 − (25) 2 (2)(45)

= −6.9444 m/s 2

Braking force. Fb = ma W = a g 6.944 = − W 9.81 = −0.70789W

(a)

Going up a 5° incline. ΣF = ma

W a g F + W sin 5° a= − b g W = −(0.70789 + sin 5°)(9.81)

− Fb − W sin 5° =

= −7.79944 m/s 2 v 2f − v02 x f − x0 = 2a 0 − (25)2 = (2)(−7.79944)

x f − x0 = 40.1 m 

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PROBLEM 12.9 (Continued)

(b)

Going down a 3 percent incline.

3 = b 1.71835° 100 W − Fb + W sin b = a g −(0.70789 − sin b )(9.81) a= = tan b

= −6.65028 m/s 0 − (25)2 x= x = 0 f (2)(−6.65028)

x f − x0 = 47.0 m 

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PROBLEM 12.10 A mother and her child are skiing together, and the mother is holding the end of a rope tied to the child’s waist. They are moving at a speed of 7.2 km/h on a gently sloping portion of the ski slope when the mother observes that they are approaching a steep descent. She pulls on the rope with an average force of 7 N. Knowing the coefficient of friction between the child and the ground is 0.1 and the angle of the rope does not change, determine (a) the time required for the child’s speed to be cut in half, (b) the distance traveled in this time.

SOLUTION Draw free body diagram of child. ΣF = ma :

x-direction:

mg sin 5° − µk N − T cos15° = ma

y-direction:

N − mg cos 5° + T sin15° = 0

From y-direction, = N mg cos 5° − T sin15 = ° (20 kg)(9.81 m/s 2 ) cos 5° − (7 N)sin15° = 193.64 N

From x-direction,

µk N

T cos15° m m (0.1)(193.64 N) (7 N) cos 15° 2 = (9.81 m/s )sin 5° − − 20 kg 20 kg = a g sin 5° −

−

= −0.45128 m/s 2

(in x-direction.)

= v0 7.2 = km/h 2 m/s = vf

1 = v0 1 m/s 2

vf = v0 + at (a) (b)

x0 = 0

v f − v0 −1 m/s 2.2159 s t= = = a −0.45128 m/s 2 t = 2.22 s 

Time elapsed. Corresponding distance. x = x0 + v0t +

1 2 at 2

=0 + (2 m/s)(2.2159 s) +

1 (−0.45128 m/s 2 )(2.2159 s) 2 2 x = 3.32 m 

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PROBLEM 12.11 The coefficients of friction between the load and the flat-bed trailer shown are µ s = 0.40 and µk = 0.30. Knowing that the speed of the rig is 72 km/h, determine the shortest distance in which the rig can be brought to a stop if the load is not to shift.

SOLUTION Load: We assume that sliding of load relative to trailer is impending:

F = Fm = µs N Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration a max . ΣF= 0: N − W = 0 N = W y

= Fm µ= 0.40 W sN = ΣFx ma := Fm mamax = 0.40 W

W = amax amax 3.924 m/s 2 g a max = 3.92 m/s 2

Uniformly accelerated motion.

= v0 72 = km/h 20 m/s v2 = v02 + 2ax with v = 0 a= −amax = 3.924 m/s 2 = 0 (20) 2 + 2(−3.924) x

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x = 51.0 m 

PROBLEM 12.12 A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. Knowing that car A has a mass of 25 Mg and car B a mass of 20 Mg, and that the braking force is 30 kN on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is showing down.

SOLUTION = v0 90 = km/h 90/3.6 = 25 m/s (a)

Both cars:

a = 1.333 m/s 2

ΣFx = Σma: 60 × 103 N = (45 × 103 kg)a

v 2 = v10 + 2ax: 0 = (25) 2 + 2(−1.333) x x = 234 m 

Stopping distance: (b)

Car A: ΣFx = ma: 30 × 103 + P = (25 × 103 )a

P = (25 × 103 )(1.333) − 30 × 103

Coupling force:

P = +3332 N

P = 3.33 kN (tension) 

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PROBLEM 12.13 The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the incline, determine (a) the acceleration of each block, (b) the tension in the cable.

SOLUTION (a)

We note that aB =

1 aA . 2

Block A

= ΣFx mA a A : T − (200 lb)sin = 30°

200 aA 32.2

(1)

Block B

= ΣFy mB aB : 350 = lb − 2T (a)

(2)

Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T:

−2(200)sin 30= ° + 350 2

(b)

350  1  aA  32.2  2 

From Eq. (1), T − (200)sin 30° =

200 350  1  aA + aA 32.2 32.2  2 

150 =

575 aA 32.2

= aB

1 1 = aA (8.40 ft/s 2 ), 2 2

a A = 8.40 ft/s 2

200 (8.40) 32.2

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30° 

a B = 4.20 ft/s 2  T = 152.2 lb 

PROBLEM 12.14 Solve Problem 12.13, assuming that the coefficients of friction between block A and the incline are µ s = 0.25 and µk = 0.20. PROBLEM 12.13 The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the incline, determine (a) the acceleration of each block, (b) the tension in the cable.

SOLUTION We first determine whether the blocks move by computing the friction force required to maintain block A in equilibrium. T = 175 lb. When B in equilibrium, = ΣFx 0: 175 − 200sin 30°= − Freq 0 Freq = 75.0 lb = ΣFy 0: N − 200 cos= 30° 0 = N 173.2 lb

= FM µ= 0.25(173.2= lb) 43.3 lb sN Since Freq > Fm , blocks will move (A up and B down). We note that aB =

1 aA . 2

Block A

= F µ= (0.20)(173.2) = 34.64 lb. kN = ΣFx mA 0 A : − 200sin 30° − 34.64 = +T

200 aA 32.2

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(1)

PROBLEM 12.14 (Continued)

Block B

= ΣFy mB aB : 350lb = − 2T (a)

(2)

Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T:

−2(200)sin 30° − 2(34.64) = + 350 2 81.32 = = aB

(b)

350  1  aA 32.2  2 

575 aA 32.2

200 350  1  aA + aA 32.2 32.2  2  a A = 4.55 ft/s 2

1 1 (4.52 ft/s 2 ), = aA 2 2

From Eq. (1), T − (200)sin 30° − 34.64 =

30° 

aB = 2.28 ft/s 2  200 (4.52) 32.2

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T = 162.9 lb 

PROBLEM 12.15 Each of the systems shown is initially at rest. Neglecting axle friction and the masses of the pulleys, determine for each system (a) the acceleration of block A, (b) the velocity of block A after it has moved through 10 ft, (c) the time required for block A to reach a velocity of 20 ft/s.

SOLUTION Let y be positive downward for both blocks.

constant Constraint of cable: y A + yB = a A + aB = 0

For blocks A and B, Block A: Block B:

aB = − a A

ΣF = ma : W WA − T = A a A g

or

= T WA −

WA aA g

W W P + WB − T = B aB = − B aA g g P + WB − WA +

Solving for aA,

or

WA W aA = − B aA g g aA =

WA − WB − P g WA + WB

(1)

v A2 − (v A )02 = 2a A [ y A − ( y A )0 ] with (v A )0 = 0

= vA

2a A [ y A − ( y A )0 ]

(2)

= v A − (v A )0 a At with = (v A )0 0 t=

vA aA

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(3)

PROBLEM 12.15 (Continued)

(a)

Acceleration of block A.

= WA 200= lb, WB 100 = lb, P 0

System (1):

(a A )1 =

System (2):

= WA 200 lb, = WB 0,= P 50 lb

By formula (1),

(a A )2 =

By formula (1),

(c)

200 − 100 (32.2) 200

(a A )1 = 10.73 ft/s 2 

(a A )2 = 16.10 ft/s 2 

= WA 2200 = lb, WB 2100 = lb, P 0

System (3):

(b)

200 − 100 (32.2) 200 + 100

By formula (1),

( a A )3 =

2200 − 2100 (32.2) 2200 + 2100

(a A )3 = 0.749 ft/s 2 

v A at y A − ( y A )0 = 10 ft. Use formula (2). System (1):

(v A )1 = (2)(10.73)(10)

(v A )1 = 14.65 ft/s 

System (2):

(v A ) 2 = (2)(16.10)(10)

(v A )2 = 17.94 ft/s 

System (3):

(v A )3 = (2)(0.749)(10)

(v A )3 = 3.87 ft/s 

Time at v A = 20 ft/s. Use formula (3). System (1):

t1 =

20 10.73

t1 = 1.864 s 

System (2):

t2 =

20 16.10

t2 = 1.242 s 

System (3):

t3 =

20 0.749

t3 = 26.7 s 

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PROBLEM 12.16 Boxes A and B are at rest on a conveyor belt that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Knowing that the coefficients of kinetic friction between the belt and the boxes are ( µk ) A = 0.30 and ( µk ) B = 0.32, determine the initial acceleration of each box.

SOLUTION Assume that aB > a A so that the normal force NAB between the boxes is zero. A:

= ΣFy 0: NA − WA cos = 15° 0

A:

= NA WA cos 15°

or

FA = ( µk ) A NA

Slipping:

= 0.3WA cos 15°

= ΣFx mA a A : FA − = WA sin 15° mA a A 0.3WA cos 15° − WA sin 15° =

or

WA aA g

= a A (32.2 ft/s 2 )(0.3 cos 15° − sin 15°)

or

= 0.997 ft/s 2

B:

B:

= ΣFy 0: N B − WB cos = 15° 0

= N B WB cos 15°

or Slipping:

FB = ( µk ) B N B = 0.32WB cos 15°

= ΣFx mB aB : FB − W = B sin 15° mB aB or or

0.32WB cos 15° − WB sin 15° =

WB aB g

= aB (32.2 ft/s 2 )(0.32 = cos 15° − sin 15°) 1.619 ft/s 2

aB > a A ⇒ assumption is correct a A = 0.997 ft/s 2

15° 

a B = 1.619 ft/s 2

15° 

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PROBLEM 12.16 (Continued) Note: If it is assumed that the boxes remain in contact ( NAB ≠ 0), then assuming NAB to be compression,

= a A aB

and= find (ΣFx ma) for each box.

A:

0.3WA cos 15° − WA sin 15° − N AB =

WA a g

B:

0.32WB cos 15° − WB sin 15° + N AB =

WB a g

Solving yields a = 1.273 ft/s 2 and NAB = −0.859 lb, which contradicts the assumption.

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PROBLEM 12.17 A 5000-lb truck is being used to lift a 1000 lb boulder B that is on a 200 lb pallet A. Knowing the acceleration of the truck is 1 ft/s2, determine (a) the horizontal force between the tires and the ground, (b) the force between the boulder and the pallet.

SOLUTION aT = 1 m/s 2

Kinematics:

a= a= 0.5 m/s 2 A B

5000 = 155.28 slugs 32.2 200 = mA = 6.211 slugs 32.2 1000 = mB = 31.056 slugs 32.2

= mT

Masses:

Let T be the tension in the cable. Apply Newton’s second law to the lower pulley, pallet and boulder.

Vertical components

: 2T − (mA + mB ) g = (mA + mB )a A 2T − (37.267)(32.2) = (37.267)(0.5) T = 609.32 lb

Apply Newton’s second law to the truck.

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PROBLEM 12.17 (Continued)

Horizontal components (a)

: F −T = mT aT

Horizontal friction force between tires and ground. F= T + mT aT = 609.32 + (155.28)(1.0)

F = 765 lb 

Apply Newton’s second law to the boulder.

Vertical components + :

FAB − mB g = mB aB

FAB= mB ( g + a= ) 31.056(32.2 + 0.5)

(b)

Contact force between boulder and pallet:

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FAB = 1016 lb 

PROBLEM 12.18 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are µ s = 0.20 and µk = 0.15. If P = 0, determine (a) the acceleration of block B, (b) the tension in the cord.

SOLUTION From the constraint of the cord:

2 x A + xB/A = constant Then

2v A + vB/A = 0

and

2a A + aB/A = 0

Now

a= a A + a B/A B

Then

aB= a A + (−2a A )

or

aB = − a A

(1)

First we determine if the blocks will move for the given value of θ . Thus, we seek the value of θ for which the blocks are in impending motion, with the impending motion of A down the incline. B:

= ΣFy 0: N AB − WB cos = θ 0

or

N AB = mB g cos θ

Now

FAB = µ s N AB

B:

= 0.2mB g cos θ

ΣFx= 0: − T + FAB + WB sin θ= 0 or

= T mB g (0.2cos θ + sin θ ) A: Σ= Fy 0: N A − N AB − WA cos = θ 0

or

N = (mA + mB ) g cos θ A

Now

FA = µ s N A

A:

= 0.2(mA + mB ) g cos θ

ΣFx= 0: − T − FA − FAB + WA sin θ= 0

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PROBLEM 12.18 (Continued) T= mA g sin θ − 0.2(mA + mB ) g cos θ − 0.2mB g cos θ

or

= g[mA sin θ − 0.2(mA + 2mB ) cos θ ] Equating the two expressions for T

mB g (0.2cos θ + sin= θ ) g[mA sin θ − 0.2(mA + 2mB ) cos θ ] 8(0.2 + tan = θ ) [40 tan θ − 0.2(40 + 2 × 8)]

or

tan θ = 0.4

or

or= θ 21.8° for impending motion. Since θ < 25°, the blocks will move. Now consider the motion of the blocks. (a)

= ΣFy 0: N AB − WB = cos 25° 0

B:

= N AB mB g cos 25°

or Sliding:

= FAB µ= 0.15mB g cos 25° k N AB = ΣFx mB aB : − T + FAB + WB sin = 25° mB aB

or

T mB [ g (0.15cos 25° + sin 25°) − aB ] = = 8[9.81(0.15cos 25° + sin 25°) − aB ] = 8(5.47952 − aB )

A:

(N)

= ΣFy 0: N A − N AB − WA cos = 25° 0

or

NA = (mA + mB ) g cos 25°

Sliding:

FA = µk N A = 0.15(mA + mB ) g cos 25° = ΣFx mA a A : − T − FA − FAB + WA sin = 25° mA a A

Substituting and using Eq. (1) T mA g sin 25° − 0.15(mA + mB ) g cos 25° = − 0.15mB g cos 25° − mA (−aB ) = g[mA sin 25° − 0.15(mA + 2mB ) cos 25°] + mA aB = 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40aB = 91.15202 + 40aB

(N)

Equating the two expressions for T

8(5.47952 − = aB ) 91.15202 + 40aB or

aB = −0.98575 m/s 2 a B = 0.986 m/s 2

(b)

We have or

25° 

= T 8[5.47952 − (−0.98575)] T = 51.7 N 

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PROBLEM 12.19 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are µ s = 0.20 and µk = 0.15. If P = 40 N , determine (a) the acceleration of block B, (b) the tension in the cord.

SOLUTION From the constraint of the cord.

2 x A + xB/A = constant Then

2v A + vB/A = 0

and

2a A + aB/A = 0

Now

a= a A + a B/A B

Then

aB= a A + (−2a A )

or

aB = − a A

(1)

First we determine if the blocks will move for the given value of P. Thus, we seek the value of P for which the blocks are in impending motion, with the impending motion of a down the incline. B:

= ΣFy 0: N AB − WB cos = 25° 0

B:

= N AB mB g cos 25°

or

FAB = µ s N AB

Now

= 0.2 mB g cos 25°

ΣF = 0: − T + FAB + WB sin 25 = ° 0 x or

A:

= T 0.2 mB g cos 25° + mB g sin 25° = (8 kg)(9.81 m/s 2 ) (0.2 cos 25° + sin 25°) = 47.39249 N

A: or

= ΣFy 0: N A − N AB − WA cos 25° + P sin = 25° 0

N= (mA + mB ) g cos 25° − P sin 25° A

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PROBLEM 12.19 (Continued)

FA = µ s N A

Now

= FA 0.2[(mA + mB ) g cos 25° − P sin 25°]

or

ΣF = 0: − T − FA − FAB + WA sin 25° + P cos 25 = ° 0 x −T − 0.2[(mA + mB ) g cos 25° − P sin 25°] − 0.2mB g cos 25° + mA g sin 25° + P cos 25° = 0

or or

P(0.2 sin 25° + cos 25°)= T + 0.2[(mA + 2mB ) g cos 25°] − mA g sin 25°

Then

P(0.2 sin 25°= + cos 25°) 47.39249 N + 9.81 m/s 2 {0.2[(40 + 2 × 8) cos 25° − 40 sin 25°] kg} P = −19.04 N for impending motion.

or

Since P, < 40 N, the blocks will move. Now consider the motion of the blocks. (a)

= ΣFy 0: N AB − WB cos = 25° 0

B:

= N AB mB g cos 25°

or

FAB = µk N AB

Sliding:

= 0.15 mB g cos 25°

= ΣFx mB aB : − T + FAB + WB sin = 25° mB aB or

T mB [ g (0.15 cos 25° + sin 25°) − aB ] = = 8[9.81(0.15 cos 25° + sin 25°) − aB ] = 8(5.47952 − aB )

(N)

A:

= ΣFy 0: N A − N AB − WA cos 25° + P sin = 25° 0

or

N= (mA + mB ) g cos 25° − P sin 25° A

Sliding:

FA = µk N A = 0.15[(mA + mB ) g cos 25° − P sin 25°]

= ΣFx mA a A : − T − FA − FAB + WA sin 25° + P cos = 25° mA a A

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PROBLEM 12.19 (Continued)

Substituting and using Eq. (1) = T mA g sin 25° − 0.15[(mA + mB ) g cos 25° − P sin 25°] − 0.15 mB g cos 25° + P cos 25° − mA (−aB ) = g[mA sin 25° − 0.15(mA + 2mB ) cos 25°] + P(0.15 sin 25° + cos 25°) + mA aB = 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40(0.15 sin 25° + cos 25°) + 40aB = 129.94004 + 40aB

Equating the two expressions for T

8(5.47952 −= aB ) 129.94004 + 40aB or

aB = −1.79383 m/s 2 a B = 1.794 m/s 2

(b)

We have

25° 

= T 8[5.47952 − (−1.79383)] T = 58.2 N 

or

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PROBLEM 12.20 The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction between the two beams and between the lower beam and the bed of the trailer are 0.25 and 0.30, respectively. Knowing that the load does not shift, determine (a) the maximum acceleration of the trailer and the corresponding tension in the cable, (b) the maximum deceleration of the trailer.

SOLUTION (a) Maximum acceleration. The cable secures the upper beam to the cab; the lower beam has impending slip.

Σ= Fy 0: N1 − = W 0

For the upper beam,

N= W = mg 1

Σ= Fy 0: N 2 − N1 −= W 0

For the lower beam,

ΣF= ma : 0.25 N1 + 0.30 N= x 2 = a 0.85 = g

( 0.85)( 9.81)

For the upper beam, T = 0.25mg + ma =

a = 8.34 m/s 2

or

= N 2 2= W 2mg

( 0.25 + 0.60 ) mg=

ma



Σ = Fx ma : T − 0.25= N1 ma

( 0.25)(1500 )( 9.81) + (1500 )(8.34 ) =

16.19 × 103 N

(b) Maximum deceleration of trailer. Case 1: Assume that only the top beam has impending motion. As in Part (a) N1 = mg.

= ΣF ma = : 0.25mg ma = a 0.25 = g 2.45 m/s 2

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T = 16.19 kN 

PROBLEM 12.20 (Continued)

Case 2: Assume that both beams have impending slip. As before N 2 = 2mg.

= ΣF

( 2m= ) a : ( 0.30 )( 2mg ) ( 2m ) a

= a 0.30 = g 2.94 m/s 2

The smaller of deceleration value of Case 1 governs.

= a 2.45 m/s 2 ← 

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PROBLEM 12.21 A baggage conveyor is used to unload luggage from an airplane. The 10-kg duffel bag A is sitting on top of the 20-kg suitcase B. The conveyor is moving the bags down at a constant speed of 0.5 m/s when the belt suddenly stops. Knowing that the coefficient of friction between the belt and B is 0.3 and that bag A does not slip on suitcase B, determine the smallest allowable coefficient of static friction between the bags.

SOLUTION Since bag A does not slide on suitcase B, both have the same acceleration.

a=a

20°

Apply Newton’s second law to the bag A – suitcase B combination treated as a single particle.

= ΣFy ma y : − (mB + mA ) g cos 20= °+ N 0

= N (mA + mB ) g cos 30 = ° (30)(9.81) cos 20 = ° 276.55 N

µ B N (0.3)(276.55) = = 82.965 N Σ= Fx max : µ B N + (mA + mB ) g sin 20 = ° (mA + mB )a

µB N 82.965 9.81sin 20° − a g sin 20° + = = 30 mA mB a = 0.58972 m/s 2

a = 0.58972 m/s 2

Apply Newton’s second law to bag A alone.

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20°

PROBLEM 12.21 (Continued)

= ΣFy ma y : N AB − m= A g cos 20° 0 = N AB ma = g sin 20° (10)(9.81)= cos 20° 92.184 N

= ZFx max : M= mA a A g sin 20° − FAB = FAB mA ( g sin = 20° − a) (10)(9.81sin 20° − 0.58972) = 27.655 N Since bag A does not slide on suitcase B,

µs >

FAB 27.655 = = 0.300 N AB 92.184

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µ s > 0.300 

PROBLEM 12.22 To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are µ s = 0.40 and µk = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, (b) the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s.

SOLUTION Let a P be the acceleration of the plywood, aT be the acceleration of the truck, and a P/T be the acceleration of the plywood relative to the truck. (a)

Find the value of aT so that the relative motion of the plywood with respect to the truck is impending.

F1 µ= 0.40 N1 aP = aT and= s N1 ΣFy = mP a y : N1 − WP cos 20° = −mP aT sin 20° = N1 mP ( g cos 20° − aT sin 20°)

= ΣFx max : F1 − WP= sin 20° mP aT cos 20° = F1 mP ( g sin 20° + aT cos 20°)

mP ( g sin 20= ° + aT cos 20°) 0.40 mP ( g cos 20° − aT sin 20°) (0.40 cos 20° − sin 20°) g cos 20° + 0.40sin 20° = (0.03145)(9.81) = 0.309

aT =

aT = 0.309 m/s 2

(b)

xP/T = ( xP/T )o + (vP /T )t +

= a P /T

1 1 aP / T t 2 = 0 + 0 + aP / T t 2 2 2

2 xP /T (2)(2) = = 4.94 m/s 2 (0.9) 2 t2

a P / T = 4.94 m/s 2

20°

a P= aT + a P / T = (aT →) + (4.94 m/s 2

20°)

Fy = mP a y : N 2 − WP cos 20° = −mP aT sin 20° = N 2 mP ( g cos 20° − aT sin 20°)

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

PROBLEM 12.22 (Continued)

ΣFx = Σmax : F2 − WP sin 20° = mP aT cos 20° − mP aP / T = F2 mP ( g sin 20° + aT cos 20° − aP / T ) For sliding with friction

= F2 µ= 0.30 N 2 k N2

= mP ( g sin 20° + aT cos 20° − aP /T ) 0.30mP ( g cos 20° + aT sin 20°) aT =

(0.30 cos 20° − sin 20°) g + aP /T cos 20° + 0.30sin 20°

= (−0.05767)(9.81) + (0.9594)(4.94) = 4.17

aT = 4.17 m/s 2

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

PROBLEM 12.23 To transport a series of bundles of shingles A to a roof, a contractor uses a motor-driven lift consisting of a horizontal platform BC which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration a1 as shown. The lift then decelerates at a constant rate a2 and comes to rest at D, near the top of the ladder. Knowing that the coefficient of static friction between a bundle of shingles and the horizontal platform is 0.30, determine the largest allowable acceleration a1 and the largest allowable deceleration a2 if the bundle is not to slide on the platform.

SOLUTION F1 µ= 0.30 N1 Acceleration a1: Impending slip. = s N1 N1= − WA mA a1 sin 65°

ΣFy = mA a y :

N1 = WA + mA a1 sin 65° = mA ( g + a1 sin 65°)

= ΣFx mA a= mA a1 cos 65° x : F1 F1 = µ s N or

mA a1= cos 65° 0.30mA ( g + a1 sin 65°) 0.30 g cos 65° − 0.30 sin 65° = (1.990)(9.81)

a1 =

= 19.53 m/s 2

a1 = 19.53 m/s 2

F2 µ= 0.30 N 2 Deceleration a 2 : Impending slip. = s N2 ΣFy = ma y : N1 − WA = −mA a2 sin 65°

N1 = WA − mA a2 sin 65° ΣFx = max := F2 mA a2 cos 65°

F2 = µ s N 2

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65° 

PROBLEM 12.23 (Continued)

or

mA a= 2 cos 65° 0.30 m A ( g − a2 cos 65°) a2 =

0.30 g cos 65° + 0.30 sin 65°

= (0.432)(9.81)

= 4.24 m/s 2

a 2 = 4.24 m/s 2

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65° 

PROBLEM 12.24 An airplane has a mass of 25 Mg and its engines develop a total thrust of 40 kN during take-off. If the drag D exerted on the plane has a magnitude D = 2.25v 2 , where v is expressed in meters per second and D in newtons, and if the plane becomes airborne at a speed of 240 km/h, determine the length of runway required for the plane to take off.

SOLUTION F= ma: 40 × 103 N − 2.25v 2 = (25 × 103 kg)a a=v

Substituting

∫

x1 0

dv dv : 40 × 103 − 2.25 v 2 = (25 × 103 ) v dx dx

(25 × 103 )vdv 0 40 × 103 − 2.25v 2 25 × 103 − x1 = [ln(40 × 103 − 2.25v 2 )]v01 2(2.25)

dx =

=

∫

v1

25 × 103 40 × 103 ln 4.5 40 × 103 − 2.25v12

= = km/h 66.67 m/s For v1 240 = x1

25 × 103 40 × 103 = ln 5.556 ln1.333 4.5 40 × 103 − 2.25(66.67) 2

= 1.5982 × 103 m

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x1 = 1.598 km 

PROBLEM 12.25 A 4-kg projectile is fired vertically with an initial velocity of 90 m/s, reaches a maximum height and falls to the ground. The aerodynamic drag D has a magnitude D = 0.0024 v 2 where D and v are expressed in newtons and m/s, respectively. Knowing that the direction of the drag is always opposite to the direction of the velocity, determine (a) the maximum height of the trajectory, (b) the speed of the projectile when it reaches the ground.

SOLUTION Let y be the position coordinate of the projectile measured upward from the ground. The velocity and acceleration a taken to positive upward. D = kv 2. ΣFy = ma :

(a) Upward motion. − D − mg = ma

 D kv 2   a= − g +  = −  g +  m m     dv kv 2  k mg  = −  g + −  v2 + v  =  dy m  m k  

v dv k = − dy mg m v2 + k

∫

0

v dv k = − mg v0 2 m v + k

∫

h

dy 0

0

1  2 mg  kh ln  v + −  = 2  k v m 0

mg  1 1  kv 2 kh k ln = − ln  0 + 1 = − 2 v 2 + mg 2  mg m  0 k

= h

 m  kv02 + 1 ln  = 2k  mg 

= 335.36 m

 ( 0.0024 )( 90 )2  4 + 1 ln  ( 2 )( 0.0024 )  ( 4 )( 9.81)  h = 335 m 

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PROBLEM 12.25 (Continued) ΣF = ma :

(b) Downward motion. D − mg = ma

D kv 2 −g = −g m m dv kv 2 k  mg  = − g =−  − v2  v dy m m k  a =

v dv k = − dy mg m − v2 k

∫

vf 0

v dv k = − mg m

∫

0

dy h

vf

1  mg kh  ln  − v2  = 2  k m 0  mg − v 2f 1  k − ln  2  mg  k 

  kh  = m  

 kv 2f  2 kh  = − ln 1 −   mg  m  1−

kv 2f mg

e− 2 kh/m =

(

mg vf = ± 1 − e −2 kh/m k

vf =

)

( 4 )( 9.81) 1 − e−( 2)( 0.0024)(335.36)/4  0.0024 

= 73.6 m/s

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

v f = 73.6 m/s 

PROBLEM 12.26 A constant force P is applied to a piston and rod of total mass m to make them move in a cylinder filled with oil. As the piston moves, the oil is forced through orifices in the piston and exerts on the piston a force of magnitude kv in a direction opposite to the motion of the piston. Knowing that the piston starts from rest at t = 0 and x = 0, show that the equation relating x, v, and t, where x is the distance traveled by the piston and v is the speed of the piston, is linear in each of these variables.

SOLUTION = ΣF ma : P = − kv ma dv P − kv = a= dt m t v m dv dt = 0 0 P − kv m v = − ln ( P − kv) 0 k m = − [ln ( P − kv) − ln P] k m P − kv P − kv kt − ln = − t= or ln k P m m P − kv P = e− kt/m v or = (1 − e− kt/m ) m k

∫

∫

∫

x= = = x

Pt v dt = 0 k t

t 0

t

P k  −  − e − kt/m  k m 0

Pt P − kt/m Pt P + (e − 1) = − (1 − e− kt/m ) k m k m Pt kv − , which is linear. k m

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

PROBLEM 12.27 A spring AB of constant k is attached to a support at A and to a collar of mass m. The unstretched length of the spring is . Knowing that the collar is released from rest at x = x0 and neglecting friction between the collar and the horizontal rod, determine the magnitude of the velocity of the collar as it passes through Point C.

SOLUTION Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider B and x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle = L

2 + x 2

The elongation of the spring is e= L − , and the magnitude of the force exerted by the spring is Fs = ke = k (  2 + x 2 − )

cos θ =

By geometry,

x

 + x2 2

Σ= Fx max : − Fs cos= θ ma − k (  2 + x 2 − ) k x a= − x− m  2 + x 2

∫

v 0

v dv =

x

 + x2 2

= ma

   

0

∫ 0 a dx x

v 1 2 x k 0 − v = x−  2 x 2 0 m 0  + x2

∫

0  k 1 2 2 2  −  x −  + x   dx =  m2 x  0

k 1 2 1  −  0 −  2 − x02 +   2 + x02  v = 2 2 m  k = 2 2 + x02 − 2  2 + x02 v2 m k  2 =  + x02 − 2 2 + x02 + 2   m 

)

(

(

)

answer: v =

k m

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(

)

2 + x02 −  

PROBLEM 12.28 Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each. Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s, determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction.

SOLUTION Let the position coordinate y be positive downward.

y A + yD = constant

Constraint of cord AD:

v A += vD 0,

( yB − yD ) + ( yC − yD ) = constant

Constraint of cord BC:

vB + vC − 2= vD 0, Eliminate aD .

a A += aD 0

aB + aC − 2= aD 0

2a A + aB + aC = 0

(1)

We have uniformly accelerated motion because all of the forces are constant. y B = ( y B ) 0 + ( vB ) 0 t +

= aB Pulley D:

1 a B t 2 , ( vB ) 0 = 0 2

2[ yB − ( yB )0 ] (2)(3) = = 1.5 m/s 2 t2 (2)2

= ΣFy 0: 2TBC −= TAD 0

TAD = 2TBC Block A: or

= ΣFy ma y : WA= − TAD mA a A

aA =

WA − TAD WA − 2TBC = mA mA

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(2)

PROBLEM 12.28 (Continued)

Block C:

= ΣFy ma y : WC = − TBC mC aC

aC =

or

WC − TBC mC

(3)

Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving for TBC ,

 WC − TBC   W − 2TBC  2 A 0 =  + aB +  mA    mC   m g − 2TBC 2 A mA 

 mC g − TBC   + aB +  mC  

 0 = 

 4 1  + 3 g + aB   TBC =  mA mC 

 4 1  10 + 5  TBC = 3(9.81) + 1.5 or TBC = 51.55 N   Block B: (a)

= ΣFy ma y : P + WB −= TBC mB aB

Magnitude of P.

P = TBC − WB + mB aB =51.55 − 5(9.81) + 5(1.5) (b)

P = 10.00 N 

Tension in cord AD.

= TAD 2= TBC (2)(51.55)

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TAD = 103.1 N 

PROBLEM 12.29 A 40-lb sliding panel is supported by rollers at B and C. A 25-lb counterweight A is attached to a cable as shown and, in cases a and c, is initially in contact with a vertical edge of the panel. Neglecting friction, determine in each case shown the acceleration of the panel and the tension in the cord immediately after the system is released from rest.

SOLUTION (a)

F = Force exerted by counterweight

Panel:

40 T −F = a g

ΣFx = ma :

(1)

Counterweight A: Its acceleration has two components

a A = a P + a A /P = a → + a = ΣFx max= : F

25 a g

= ΣFg mag : 25 = −T

(2) 25 a g

(3)

Adding (1), (2), and (3): 40 + 25 + 25 a g 25 25 = a = g (32.2) 90 90

T − F + F + 25 − T =

a = 8.94 ft/s 2



Substituting for a into (3):

25 − T =

25  25  g g  90 

T = 25 −

625 90

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T = 18.06 lb 

PROBLEM 12.29 (Continued)

(b)

Panel:

ΣFy = ma :

40 a g

(1)

25 25 − T = a g

(2)

T=

Counterweight A: ΣFy = ma :

Adding (1) and (2):

40 + 25 a g 25 a= g 65

T + 25 − T =

a = 12.38 ft/s 2



Substituting for a into (1):

T = (c)

40  25  1000 g = g  65  65

T = 15.38 lb 

Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice versa. Panel:

ΣFx = ma :

T=

40 a g

(1)

Counterweight A: Same free body as in Part (b): ΣFy = ma :

25 25 − T = a g

(2)

Since Eqs. (1) and (2) are the same as in (b), we get the same answers:

a = 12.38 ft/s 2

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; T = 15.38 lb 

PROBLEM 12.30 An athlete pulls handle A to the left with a constant force of P = 100 N. Knowing that after the handle A has been pulled 30 cm its velocity is 3 m/s, determine the mass of the weight stack B.

SOLUTION Given:

P = 100 N x A = 0.30 m v A = 3 m/s

Kinematics:

constant x A + 4 yB = v A + 4vB = 0 a A + 4 aB = 0

(1)

Uniform Acceleration of handle A:

( v A )o2 +2a A ( xA − ( xA )o ) 32 = 02 + 2a A ( 0.3 − 0 )

= v A2

a A = 15 m/s 2

From (1): From FBD:

= aB 3.75 m/s 2 ↑

∑F

y

Free Body Diagram:

= mB aB

mB aB 4 P − mB g = mB = =

4P g + aB

4 (100 ) kg 9.81 + 3.75 mB = 29.50 kg 

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PROBLEM 12.31 A 10-lb block B rests as shown on a 20-lb bracket A. The coefficients of friction are µ s = 0.30 and µk = 0.25 between block B and bracket A, and there is no friction in the pulley or between the bracket and the horizontal surface. (a) Determine the maximum weight of block C if block B is not to slide on bracket A. (b) If the weight of block C is 10% larger than the answer found in a determine the accelerations of A, B and C.

SOLUTION Kinematics. Let x A and xB be horizontal coordinates of A and B measured from a fixed vertical line to the left of A and B. Let yC be the distance that block C is below the pulley. Note that yC increases when C moves downward. See figure.

The cable length L is fixed.

L = ( xB − x A ) + ( xP − x A ) + yC + constant Differentiating and noting that xP = 0,

vB − 2v A + vC = 0 −2a A + aB + aC = 0 Here, a A and aB are positive to the right, and aC is positive downward. Kinetics. Let T be the tension in the cable and FAB be the friction force between blocks A and B. The free body diagrams are: Bracket A:

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(1)

PROBLEM 12.31 (Continued)

Block B:

Block C:

= ΣFx max : 2T = − FAB

Bracket A:

WB aB g

= ΣFx max : F = AB − T

Block B:

WA aA g

(2) (3)

+ ΣFy ma y : N AB = = − WB 0

N AB = WB

or

= ΣFy ma y : m = C −T

Block C:

WC aC g

(4)

Adding Eqs. (2), (3), and (4), and transposing, W WA W a A + B aB + C aC = WC g g g

(5)

Subtracting Eq. (4) from Eq. (3) and transposing, W WB aB − C aC = FAB − WC g g

(a)

No slip between A and B.

aB = a A

From Eq. (1),

a= a= a= a A B C

From Eq. (5), For impending slip,

a=

WC g WA + WB + WC

= FAB µ= µ sWB s N AB

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(6)

PROBLEM 12.31 (Continued) Substituting into Eq. (6),

(WB − WC )(WC g ) = µ sWB − WC WA + WB + WC Solving for WC ,

WC = =

µ sWB (WA + WB ) WA + 2WB − µ sWB (0.30)(10)(20 + 10) 20 + (2)(10) − (0.30)(10)

WC = 2.43 lbs  (b)

WC = 2.6757 lbs

WC increased by 10%.

= FAB µ= µkWB k N AB

Since slip is occurring,

W WB aB − C aC =µkWB − WC g g

Eq. (6) becomes or

10aB − 2.6757aC = [(0.25)(10) − 2.6757](32.2)

(7)

With numerical data, Eq. (5) becomes

20a A + 10aB + 2.6757aC = (2.6757)(32.2)

(8)

Solving Eqs. (1), (7), and (8) gives = a A 3.144 = ft/s 2 , aB 0.881 = ft/s 2 , aC 5.407 ft/s 2

a A = 3.14 ft/s 2



a B = 0.881 ft/s 2



aC = 5.41 ft/s 2

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

PROBLEM 12.32 Knowing that µ = 0.30, determine the acceleration of each block when m A = m B = m C.

SOLUTION Given:

µ = 0.30 m= m= m= m A B c

Kinematics:

y A + 2 yB + xC = constant v A + 2vB + vC = 0

(1) aC + a A + 2aB = 0 Free Body Diagram of Block A: Equation of Motion:

∑F

y

= ma A

mg − T = ma A a A= g − Free Body Diagram of Block B:

T m

(2)

Equation of Motion:

∑F

y

= maB

mg − 2T = maB aB= g −

Free Body Diagram of Block B:

2T m

(3)

Equations of Motion:

∑F

y

=0

N = mg

∑F

x

= maC

maC 0.3mg − T = = aC 0.3 g −

T m

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(4)

Problem 12.32 (Continued)

Substitute (2), (3) and (4) → (1)

3.3 g − T=

6T 0 = m

1.1 mg 2

(5)

Substitute (5) → (2):

a A= g −

1.1 g 2

= a A 0.45g ↓ 

Substitute (5) → (3):

aB= g −

2.2 g 2

= aB 0.10g ↑ 

Substitute (5) → (4):

= aC 0.3 g −

1.1 g 2

= aC 0.25g ← 

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PROBLEM 12.33 Knowing that µ = 0.30, determine the acceleration of each block when mA = 5 kg, mB = 30 kg, and mC = 15 kg.

SOLUTION Given:

Kinematics:

= µ 0.30, = g 9.81 m/s 2 = mA 5= kg, mB 30= kg, mC 15 kg y A + 2 yB + xC = constant v A + 2vB + vC = 0 a A + 2aB + aC = 0

Free Body Diagram of Block A:

(1) Equation of Motion:

∑F

y

= mA a A

mA g − T = mA a A a A= g − Free Body Diagram of Block B:

T mA

(2)

Equation of Motion:

∑F

y

= mB aB

mB g − 2T = mB aB aB= g − Free Body Diagram of Block B:

2T mB

(3)

Equations of Motion:

∑F

y

=0

N = mC g

∑F

x

= mC aC

0.3mC g − T = mC aC

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Problem 12.33 (Continued)

= aC 0.3 g − Substitute (2), (3) and (4) → (1)

3.3 g −

T mC

(4)

T 4T T − − = 0 mA mB mC

Substitute (5) → (2):

a A= g −

8.25 g 5

= a A 6.377 m/s 2 ↑ 

Substitute (5) → (3):

aB= g −

16.5 g 30

= aB 4.415 m/s 2 ↓ 

Substitute (5) → (4):

= aC 0.3 g −

8.25 g 15

aC = −2.453 m/s 2 ← 

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PROBLEM 12.34 A 25-kg block A rests on an inclined surface, and a 15-kg counterweight B is attached to a cable as shown. Neglecting friction, determine the acceleration of A and the tension in the cable immediately after the system is released from rest.

SOLUTION Let the positive direction of x and y be those shown in the sketch, and let the origin lie at the cable anchor. Constraint of cable: x A = + yB / A constant or a A = + aB / A 0, where the positive directions of a A and aB / A are respectively the x and the y directions. Then aB / A = −a A (1) 20° ) + ( aB / A

First note that a B =a A + a B / A =( a A Block B: = ΣFx

mB ( aB ) x : = mB g sin 20° − N AB mB a A + N= AB

20° )

mB a A

mB g sin 20°

15 a A + N AB = 50.328

= ΣFy

mB ( aB )= : mB g cos 20° − T y mB aB / A= +T

(2) mB aB / A

mB g cos 20°

15 aB / A + T = 138.276

Block A: = ΣFx

mAa A : mA g sin= 20° + N AB + T mAa A − N AB= +T

(3) m Aa A

mA g sin 20°

25 aB / A − N AB + T = 83.880

(4)

Eliminate aB / A using Eq. (1), then add Eq. (4) to Eq. (2) and subtract Eq. (3). 55 a A = −4.068 or a A = −0.0740 m/s 2 , a A = 0.0740 m/s 2

20° 

From Eq. (1), aB / A = 0.0740 m/s 2 From Eq. (3), T = 137.2 N

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T = 137.2 N 

PROBLEM 12.35 Block B of mass 10-kg rests as shown on the upper surface of a 22-kg wedge A. Knowing that the system is released from rest and neglecting friction, determine (a) the acceleration of B, (b) the velocity of B relative to A at t = 0.5 s.

SOLUTION A:

= ΣFx mA a A : WA sin 30= ° + N AB cos 40° mA a A (a) NAB =

or

22 ( a A − 12 g ) cos 40°

a A + a B /A , where a B/A is directed along the top surface of A. Now we note: a= B B:

ΣFy′ = mB a y′ : NAB − WB cos 20° = −mB a A sin 50°

= NAB 10 ( g cos 20° − a A sin 50°)

or

Equating the two expressions for NAB 1   22  a A − g  2   = 10( g cos 20° − a A sin 50°) cos 40°

aA or =

(9.81)(1.1 + cos 20° cos 40°) = 6.4061 m/s 2 2.2 + cos 40° sin 50°

= ΣFx′ mB ax′ : WB sin = 20° mB aB/A − mB a A cos50° or

= aB/A g sin 20° + a A cos 50° = (9.81sin 20° + 6.4061cos 50°) m/s 2 = 7.4730 m/s 2

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PROBLEM 12.35 (Continued)

Finally

a= a A + a B/ A B

We have

aB2 = 6.40612 + 7.47302 − 2(6.4061 × 7.4730) cos 50°

or

aB = 5.9447 m/s 2

and or

7.4730 5.9447 = sin α sin 50° = α 74.4°

a B = 5.94 m/s 2

(b)

75.6° 

Note: We have uniformly accelerated motion, so that v= 0 + at

Now At t = 0.5 s: or

v B/A = v B − v A = a B t − a At = a B/At = vB/A 7.4730 m/s 2 × 0.5 s

v B/A = 3.74 m/s

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20° 

PROBLEM 12.36 A 450-g tetherball A is moving along a horizontal circular path at a constant speed of 4 m/s. Determine (a) the angle θ that the cord forms with pole BC, (b) the tension in the cord.

SOLUTION a= a= A n

First we note

ρ

ρ = l AB sin θ

where (a)

v A2

= ΣFy 0: TAB cos = θ − WA 0 TAB =

or

mA g cos θ

= ΣFx mA a= mA A : TAB sin θ

v A2

ρ

Substituting for TAB and ρ mA g v A2 sin θ = mA l AB sin θ cos θ

sin 2 θ = 1 − cos 2 θ

(4 m/s)2 1 − cos 2 θ = cos θ 1.8 m × 9.81 m/s 2

or

cos 2 θ + 0.906105cos θ − 1 =0 cos θ = 0.64479

Solving

= θ 49.9° 

or (b)

From above

= TAB

mA g 0.450 kg × 9.81 m/s 2 = cos θ 0.64479

TAB = 6.85 N 

or

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PROBLEM 12.37 During a hammer thrower’s practice swings, the 7.1-kg head A of the hammer revolves at a constant speed v in a horizontal circle as shown. If ρ = 0.93 m and θ= 60°, determine (a) the tension in wire BC, (b) the speed of the hammer’s head.

SOLUTION First we note

a= a= A n

v A2

ρ

(a)= ΣFy 0: TBC = sin 60° − WA 0 or

7.1 kg × 9.81 m/s 2 sin 60° = 80.426 N

TBC =

TBC = 80.4 N  a A : TBC cos 60° mA = ΣFx mA= (b) or

v A2 =

v A2

ρ

(80.426 N) cos 60° × 0.93 m 7.1 kg

v A = 2.30 m/s 

or

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PROBLEM 12.38 Human centrifuges are often used to simulate different acceleration levels for pilots. When aerospace physiologists say that a pilot is pulling 9g’s, they mean that the resultant normal force on the pilot from the bottom of the seat is nine times their weight. Knowing that the centrifuge starts from rest and has a constant angular acceleration of 1.5 RPM per second until the pilot is pulling 9g’s and then continues with a constant angular velocity, determine (a) how long it will take for the pilot to reach 9g’s (b) the angle θ of the normal force once the pilot reaches 9 g’s. Assume that the force parallel to the seat is zero.

SOLUTION Given:

RPM/s = α 1.5 = ω0 = 0

0.157 rad/s 2

N = 9 mg R=7 m Free Body Diagram of Pilot:

Equations of Motion:

∑F

y

= ma y

n

N sin θ − mg = m ( 0) N sin θ = mg

∑F

(1)

= man

N cos θ = mRω 2

ω=

N cos θ mR

(2)

Substitute N=9mg into (1): 9mg sin θ = mg

1   θ 6.379° =

θ = sin −1   9

Substitute N=9mg and θ into (2):

9*9.81cos 6.379° 7 ω = 3.540 rad/s

ω=

For constant angular acceleration:

ω = ω0 + α t 3.540= 0 + 0.157 * t

t = 22.55 s 

(a) Solving for t:

= θ 6.379° 

From earlier:

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PROBLEM 12.39 A single wire ACB passes through a ring at C attached to a sphere which revolves at a constant speed v in the horizontal circle shown. Knowing that the tension is the same in both portions of the wire, determine the speed v.

SOLUTION = ΣFx ma: T (sin 30= ° + sin 45°)

mv 2

ρ

(1)

= ΣFy 0: T (cos 30° + cos= 45°) − mg 0

T (cos 30° + cos 45°) = mg

Divide Eq. (1) by Eq. (2):

(2)

sin 30° + sin 45° v 2 = cos 30° + cos 45° ρ g

v 2 0.76733 m/s 2 ) 12.044 m 2 /s 2 = = ρ g 0.76733 (1.6 m)(9.81 =

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v = 3.47 m/s 

PROBLEM 12.40* Two wires AC and BC are tied at C to a sphere which revolves at a constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed 60 N.

SOLUTION From the solution of Problem 12.39, we find that both wires remain taut for

3.01 m/s ≤ v ≤ 3.96 m/s 

To determine the values of v for which the tension in either wire will not exceed 60 N, we recall Eqs. (1) and (2) from Problem 12.39:

TAC sin 30° + TBC sin 45° =

mv 2

ρ

(1)

TAC cos30° + TBC cos 45° = mg

(2)

Subtract Eq. (1) from Eq. (2). Since sin 45° = cos 45°, we obtain

°) mg − TAC (cos 30° − sin 30=

mv 2 (3)

ρ

Multiply Eq. (1) by cos 30°, Eq. (2) by sin 30°, and subtract:

° sin 30°) TBC (sin 45° cos 30° − cos 45=

= sin15° TBC

mv 2

ρ

mv 2

ρ

cos 30° − mg sin 30°

cos 30° − mg sin 30°

(4)

Making TAC = 60 N, m = 5= kg, ρ 1.6 = m, g 9.81 m/s 2 in Eq. (3), we find the value v1 of v for which

TAC = 60 N:

60(cos 30° − sin= 30°) 5(9.81) − 21.962 = 49.05 −

5v12 1.6

v12 0.32

v12 = 8.668,

We have TAC ≤ 60 N for v ≥ v1 , that is, for

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v1 = 2.94 m/s v ≥ 2.94 m/s 

PROBLEM 12.40* (Continued)

Making = TBC 60= N, m 5= kg, ρ 1.6= m, g 9.81 m/s 2 in Eq. (4), we find the value v2 of v for which

TBC = 60 N:

= 60sin 15°

5v22 cos 30° − 5(9.81) sin 30° 1.6

15.529 = 2.7063v22 − 24.523

v22 = 14.80,

v2 = 3.85 m/s

We have TBC ≤ 60 N for v ≤ v2 , that is, for Combining the results obtained, we conclude that the range of allowable value is

v ≤ 3.85 m/s  3.01 m/s ≤ v ≤ 3.85 m/s 

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PROBLEM 12.41 A 1-kg sphere is at rest relative to a parabolic dish which rotates at a constant rate about a vertical axis. Neglecting friction and knowing that r = 1 m, determine (a) the speed v of the sphere, (b) the magnitude of the normal force exerted by the sphere on the inclined surface of the dish.

SOLUTION r 2 dy , = r= 1= dx 2

Given:

y=

Free Body Diagram of Sphere:

Equations of Motion:

tan θ

θ= 45°

or

= + ↑ ΣFy 0: N cos = θ − mg 0 mg cosθ

N = = ΣFn

man= : N sin θ

man

mg tan θ = m

v2 r

v 2 = gr tan θ

(a) v 2 = (b) N =

9.81)(1)(1.0000 ) (=

mg = cos 45°

9.81 m 2 /s 2

(1)( 9.81) cos 45°

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v = 3.13 m/s  N = 13.87 N 

PROBLEM 12.42* As part of an outdoor display, a 12-lb model C of the earth is attached to wires AC and BC and revolves at a constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed 26 lb.

SOLUTION a= a= C n

First note

vC2

ρ

ρ = 3 ft

where

= ΣFx mC aC : TCA sin = 40° + TCB sin15°

WC vC2 g ρ

= ΣFy 0: TCA cos 40° − TCB cos15 = ° − WC 0

Note that Eq. (2) implies that (a)

when

= TCB (T = (TCA )max CB ) max , TCA

(b)

when

= TCB (T = TCA (TCA )min CB ) min ,

Case 1: TCA is maximum.

Let Eq. (2) or

TCA = 26 lb (26 lb) cos 40° − TCB cos15° − (12 lb) = 0

TCB = 8.1964 lb (TCB )(TCA )max < 26 lb

OK

[(TCB ) max = 8.1964 lb]

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(1) (2)

PROBLEM 12.42* (Continued)

Eq. (1)

= (vC2 )(TCA )max

(32.2 ft/s 2 )(3 ft) (26sin 40° + 8.1964 sin15°) lb 12 lb

(vC )(TCA )max = 12.31 ft/s

or

(cos15°)(Eq. 1) + (sin15°)(Eq. 2)

Now we form

40° sin15° TCA sin 40° cos15° + TCA cos=

WC vC2 cos15° + WC sin15° g ρ

= TCA sin 55°

WC vC2 cos15° + WC sin15° g ρ

or

(3)

(vc )max occurs when TCA = (TCA )max

(vC )max = 12.31 ft/s Case 2: TCA is minimum.

Because (TCA )min occurs when TCB = (TCB )min , let TCB = 0 (note that wire BC will not be taut).

TCA cos 40° − (12 lb) = 0

Eq. (2)

TCA = 15.6649 lb, 26 lb OK

or

Note: Eq. (3) implies that when TCA = (TCA ) min , vC = (vC )min . Then Eq. (1)

= (vC2 )min

(32.2 ft/s 2 )(3 ft) (15.6649 lb)sin 40° 12 lb

(vC )min = 9.00 ft/s

or

0 < TCA ≤ TCB < 6 lb when

9.00 ft/s < vC < 12.31 ft/s 

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PROBLEM 12.43* The 1.2-lb flyballs of a centrifugal governor revolve at a constant speed v in the horizontal circle of 6-in. radius shown. Neglecting the weights of links AB, BC, AD, and DE and requiring that the links support only tensile forces, determine the range of the allowable values of v so that the magnitudes of the forces in the links do not exceed 17 lb.

SOLUTION v2

First note

= a a= n

where

ρ = 0.5 ft

ρ

W v2 g ρ

(1)

= ΣFy 0: TDA cos 20° − TDE= cos 30° − W 0

(2)

= ΣFx ma : TDA sin 20°= + TDE sin 30°

Note that Eq. (2) implies that (a)

when

= TDE (T = TDA (TDA )max DE ) max ,

(b)

when

= TDE (T = TDA (TDA )min DE ) min ,

Case 1: TDA is maximum.

Let Eq. (2) or Now let Eq. (2) or

TDA = 17 lb

(17 lb) cos 20° − TDE cos30° − (1.2 lb) = 0 = TDE 17.06 lb unacceptable (> 17 lb) TDE = 17 lb

TDA cos 20° − (17 lb) cos30° − (1.2 lb) = 0 = TDA 16.9443 lb OK ( ≤ 17 lb)

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PROBLEM 12.43* (Continued)

(TDA ) max = 16.9443 lb (TDE )max = 17 lb

(v 2 )(TDA )max Eq. (1) =

(32.2 ft/s 2 )(0.5 ft) (16.9443sin 20° + 17 sin 30°) lb 1.2 lb

v(TDA )max = 13.85 ft/s

or

(cos 30°) × [Eq. (1)] + (sin 30°) × [Eq. (2)]

Now form

20° sin 30° TDA sin 20° cos 30° + TDA cos =

W v2 cos 30° + W sin 30° g ρ

T= DA sin 50°

W v2 cos 30° + W sin 30° g ρ

or

(3)

vmax occurs when TDA = (TDA )max vmax = 13.85 ft/s Case 2: TDA is minimum.

Because (TDA ) min occurs when TDE = (TDE ) min , let TDE = 0. Eq. (2)

TDA cos 20° − (1.2 lb) = 0

or

TDA = 1.27701 lb, 17 lb

OK

Note: Eq. (3) implies that when TDA = (TDA ) min , v = vmin . Then

= (v 2 )min Eq. (1)

(32.2 ft/s 2 ) (0.5 ft) (1.27701 lb)sin 20° 1.2 lb

vmin = 2.42 ft/s

or

0 < TAB , TBC , TAD , TDE < 17 lb when

2.42 ft/s < v < 13.85 ft/s 

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PROBLEM 12.44 A 130-lb wrecking ball B is attached to a 45-ft-long steel cable AB and swings in the vertical arc shown. Determine the tension in the cable (a) at the top C of the swing, (b) at the bottom D of the swing, where the speed of B is 13.2 ft/s.

SOLUTION (a)

At C, the top of the swing, vB = 0; thus

= an

vB2 = 0 LAB

= ΣFn 0: TBA − WB cos = 20° 0 TBA= (130 lb) × cos 20°

or

TBA = 122.2 lb 

or (b)

= ΣFn man : TBA= − WB mB

or

(vB ) 2D LAB

 130 lb TBA (130 lb) +  = 2  32.2 ft/s

  (13.2 ft/s)   45 ft 

2

    

TBA = 145.6 lb 

or

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PROBLEM 12.45 During a high-speed chase, a 2400-lb sports car traveling at a speed of 100 mi/h just loses contact with the road as it reaches the crest A of a hill. (a) Determine the radius of curvature ρ of the vertical profile of the road at A. (b) Using the value of ρ found in part a, determine the force exerted on a 160-lb driver by the seat of his 3100-lb car as the car, traveling at a constant speed of 50 mi/h, passes through A.

SOLUTION (a)

Note:

100 mi/h = 146.667 ft/s

Wcar v A2 g ρ

= ΣFn man= : Wcar or

(146.667 ft/s)2 32.2 ft/s 2 = 668.05 ft

ρ=

ρ = 668 ft 

or (b)

⇒ at 0; 50= mi/h 73.333 ft/s Note: v is constant= = ΣFn man : W= −N

or

W v A2 g ρ

  (73.333 ft/s)2 = N (160 lb) 1 −  2  (32.2 ft/s )(668.05 ft)  N = 120.0 lb 

or

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PROBLEM 12.46 An airline pilot climbs to a new flight level along the path shown. Knowing that the speed of the airplane decreases at a constant rate from 180 m/s at point A to 160 m/s at point C, determine the magnitude of the abrupt change in the force exerted on a 90-kg passenger as the airplane passes point B.

SOLUTION

8° π = 0.13963 rad 180°

Angle change over arc AB. θ =

= ( 6000 )( 0.13963 )

ρθ =

Length of arc: s= AB

837.76 m

sBC= 800 m, s AC= 837.76 + 800= 1637.76 m 160

= ∫180 v dv

1637.76 at 0

∫

1602 1802 or = − 2 2

ds

at (1637.76 )

at = −2.076 m/s 2 837.76 vB ∫180 v dv = ∫0 at ds

vB2 =

vB2 1802 − = ( −2.076 )(837.76 ) 2 2

or

28922 m 2 /s 2 vB 170.06 m/s = 90 )( 9.81) (=

Weight of passenger: = mg

882.9 N

Just before point B. v 170.06 = = m/s, ρ 6000 m an =

Free Body Diagram of Pilot:

v2 =

ρ

(170.06= )2 6000

4.820 m/s 2

ΣFn =N1 − W = −m ( an )1 : N1 = 882.9 − ( 90 )( 4.820 ) = 449.1 N ΣFt =Ft =mat :

Just after point B.

( Ft )1

=( 90 )( −2.076 ) =−186.8 N

v= 170.06 m/s,

Σ= Fy 0 : N 2 −= W 0

ρ = ∞, an = 0 N= W = 882.9 N 2

ΣFx = mat : Ft = mat = ( 90 )( −2.076 ) =−186.8 N Ft does not change.

N increases by 433.8 N. magnitude of change of force = 434 N 

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PROBLEM 12.47 The roller-coaster track shown is contained in a vertical plane. The portion of track between A and B is straight and horizontal, while the portions to the left of A and to the right of B have radii of curvature as indicated. A car is traveling at a speed of 72 km/h when the brakes are suddenly applied, causing the wheels of the car to slide on the track ( µk = 0.20). Determine the initial deceleration of the car if the brakes are applied as the car (a) has almost reached A, (b) is traveling between A and B, (c) has just passed B.

SOLUTION = ΣFn man : N − = mg m

(a)

v2

ρ

 v2  = N m  g +  ρ   v2 µk m  g + F µ= = kN ρ 

  

= ΣFt mat= : F mat a= t Given data:

 F v2  = µk  g +  ρ m 

= µk 0.20, = v 72= km/h 20 m/s

= g 9.81 = m/s 2 , ρ 30 m

 (20) 2  = at 0.20 9.81 +  30  

(b)

an = 0

at = 4.63 m/s 2 

ΣFn= man= 0: N − mg= 0 N = mg = F µ= µk mg k N

= ΣFt mat= : F mat

a= t

F = µk = g 0.20(9.81) m

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at = 1.962 m/s 2 

PROBLEM 12.47 (Continued)

(c)

= ΣFn man : mg = −N

mv 2

ρ

 v2  = N m  g −  ρ   v2  µ = − F µ= N m g  k k   ρ  

= ΣFt mat= : F mat  F v2 at = =µk  g − m ρ 

  (20)2   =0.20 9.81 −  45    at = 0.1842 m/s 2 

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PROBLEM 12.48 A spherical-cap governor is fixed to a vertical shaft that rotates with angular velocity ω. When the string-supported clapper of mass m touches the cap, a cutoff switch is operated electrically to reduce the speed of the shaft. Knowing that the radius of the clapper is small relative to the cap, determine the minimum angular speed at which the cutoff switch operates.

SOLUTION Given:

ωmin is the angular velocity when the clapper hits the cap

Geometry From Figure:

Distance AB is 300 mm, the length of the Clapper Arm Distance BC and AC is 300 mm, the radius of the Spherical Cap. Therefore  ABC is equilateral, so θ = 60° and α = 30°

R = 0.3*cos α

Free Body Diagram of Clapper:

Equations of Motion:

∑F

y

= ma y

n

FA cos θ − mg = m ( 0) FA =

Substitute (1)into (2):

ωwin =

mg cos θ

∑F

(1)

= man

2 FA sin θ = mRωmin

ωwin =

FA sin θ mR

(2)

m g tan θ mR

9.81tan 60° 0.3cos 30° = 8.087 rad/s =

ωwin =77.23 rpm 

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PROBLEM 12.49 A series of small packages, each with a mass of 0.5 kg, are discharged from a conveyor belt as shown. Knowing that the coefficient of static friction between each package and the conveyor belt is 0.4, determine (a) the force exerted by the belt on a package just after it has passed Point A, (b) the angle θ defining the Point B where the packages first slip relative to the belt.

SOLUTION Assume package does not slip. = at 0, F f ≤ µ s N

On the curved portion of the belt

a= n

v 2 (1 m/s)2 = = 4 m/s 2 ρ 0.250 m

For any angle θ

mv 2 ΣFy = −man = − ma y : N − mg cos θ =

ρ

= N mg cos θ −

mv 2

ρ

(1)

ΣFx = max : −F f + mg sin θ = mat = 0

F f = mg sin θ

(a)

At Point A,

θ = 0° = N (0.5)(9.81)(1.000) − (0.5)(4)

(b)

At Point B,

(2)

Ff = µs N mg sin θ µ s (mg cos θ − man ) =  an  4   sin = θ µ s  cos θ − =  0.40 cos θ − 9.81  g   

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N = 2.905 N 

PROBLEM 12.49 (Continued)

Squaring and using trigonometic identities,

θ 0.16cos 2 θ − 0.130479cos θ + 0.026601 1 − cos 2= 1.16cos 2 θ − 0.130479cos θ − 0.97340 = 0 cos θ = 0.97402

= θ 13.09° 

Check that package does not separate from the belt. N =

F f mg sin θ =

µs

µs

N > 0.

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PROBLEM 12.50 A 54-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a constant rate. Knowing that the pilot’s apparent weights at Points A and C are 1680 N and 350 N, respectively, determine the force exerted on her by the seat of the trainer when the trainer is at Point B.

SOLUTION First we note that the pilot’s apparent weight is equal to the vertical force that she exerts on the seat of the jet trainer.

ΣFn man : N= m At A: = A −W or

v A2

ρ

 1680 N  = v A2 (1200 m)  − 9.81 m/s 2   54 kg  = 25,561.3 m 2 /s 2

ΣFn man : N= m At C: = C +W or

vC2

ρ

 350 N  = vC2 (1200 m)  + 9.81 m/s 2   54 kg  = 19,549.8 m 2 /s 2

Since at = constant, we have from A to C vC2 =v A2 + 2at ∆s AC

or or

19,549.8 = m 2/s 2 25,561.3 m 2 /s 2 + 2at (π × 1200 m)

at = −0.79730 m/s 2

Then from A to B vB2 =v A2 + 2at ∆s AB π  = 25,561.3 m 2/s 2 + 2(−0.79730 m/s 2 )  × 1200 m  2   = 22,555 m 2/s 2

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PROBLEM 12.50 (Continued)

At B:

: NB m = ΣFn man=

vB2

ρ

22,555 m 2 /s 2 1200 m

or

N B = 54 kg

or

N B = 1014.98 N = ΣFt mat : W = + PB m | at |

or or Finally,

= PB (54 kg)(0.79730 − 9.81) m/s 2

PB = 486.69 N ( Fpilot ) B =

N B2 + PB2 =

(1014.98) 2 + (486.69) 2

= 1126 N

or

(Fpilot ) B = 1126 N

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25.6° 

PROBLEM 12.51 A carnival ride is designed to allow the general public to experience high acceleration motion. The ride rotates about Point O in a horizontal circle such that the rider has a speed v0. The rider reclines on a platform A which rides on rollers such that friction is negligible. A mechanical stop prevents the platform from rolling down the incline. Determine (a) the speed v0 at which the platform A begins to roll upwards, (b) the normal force experienced by an 80-kg rider at this speed.

SOLUTION Radius of circle:

R = 5 + 1.5cos 70° = 5.513 m ΣF = ma:

Components up the incline,

70°:

−mA g cos 20° = −

mv02 sin 20° R

1

(a)

Components normal to the incline, N − mg = sin 20°

(b)

1

 g R  2  (9.81 m/s) (5.513 m  2 = = Speed v0 : v0 =    12.1898 m/s tan 20°  tan 20°   

v0 = 12.19 m/s 

20°. mv02 cos 20°. R

Normal= force: N (80)(9.81)sin 20° +

80(12.1898) 2 = cos 20° 2294 N 5.513

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N = 2290 N 

PROBLEM 12.52 A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (See Sample Problem 12.7 for the definition of rated speed). Knowing that a racing car starts skidding on the curve when traveling at a speed of 180 mi/h, determine (a) the banking angle θ , (b) the coefficient of static friction between the tires and the track under the prevailing conditions, (c) the minimum speed at which the same car could negotiate that curve.

SOLUTION W = mg

Weight

a=

Acceleration

v2

ρ

= ΣFx max : F + W sin = θ ma cos θ = F

mv 2

ρ

cos θ − mg sin θ

(1)

= ΣFy ma y : N − W cos = θ ma sin θ

= N (a)

mv 2

ρ

sin θ + mg cos θ

(2)

Banking angle. Rated = speed v 120 = mi/h 176 ft/s. F = 0 at rated speed. = 0

mv 2

ρ

cos θ − mg sin θ

v2 (176) 2 = = 0.96199 ρ g (1000) (32.2) = θ 43.89°

θ tan=

(b)

Slipping outward.

= θ 43.9° 

= v 180 = mi/h 264 ft/s

F = µN

µ=

µ =

F v 2 cos θ − ρ g sin θ = N v 2 sin θ + ρ g cos θ

(264) 2 cos 43.89° − (1000) (32.2)sin 43.89° (264) 2 sin 43.89° + (1000) (32.2) cos 43.89°

= 0.39009

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µ = 0.390 

PROBLEM 12.52 (Continued)

(c)

Minimum speed.

F = −µ N v 2 cos θ − ρ g sin θ −µ =2 v sin θ + ρ g cos θ ρ g (sin θ − µ cos θ ) v2 = cos θ + µ sin θ =

(1000) (32.2) (sin 43.89° − 0.39009 cos 43.89°) cos 43.89° + 0.39009 sin 43.89°

= 13,369 ft 2 /s 2 v = 115.62 ft/s

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v = 78.8 mi/h 

PROBLEM 12.53 Tilting trains, such as the American Flyer which will run from Washington to New York and Boston, are designed to travel safely at high speeds on curved sections of track which were built for slower, conventional trains. As it enters a curve, each car is tilted by hydraulic actuators mounted on its trucks. The tilting feature of the cars also increases passenger comfort by eliminating or greatly reducing the side force Fs (parallel to the floor of the car) to which passengers feel subjected. For a train traveling at 100 mi/h on a curved section of track banked through an angle θ = 6° and with a rated speed of 60 mi/h, determine (a) the magnitude of the side force felt by a passenger of weight W in a standard car with no tilt (φ = 0), (b) the required angle of tilt φ if the passenger is to feel no side force. (See Sample Problem 12.7 for the definition of rated speed.)

SOLUTION Rated speed:

= vR 60 = mi/h 88 ft/s, 100 = mi/h 146.67 ft/s

From Sample Problem 12.6, vR2 = g ρ tan θ

= ρ

or

vR2 (88) 2 = = 2288 ft g tan θ 32.2 tan 6°

Let the x-axis be parallel to the floor of the car.

= ΣFx max : Fs + W sin (θ= + φ ) man cos (θ + φ ) = (a)

mv 2

ρ

cos (θ + φ )

φ = 0.  v2  cos (θ + φ ) − sin (θ + φ )  Fs W  =  gρ   (146.67)2  cos 6° − sin 6° = W  (32.2)(2288)  = 0.1858W

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Fs = 0.1858W 

PROBLEM 12.53 (Continued)

(b)

For Fs = 0,

v2 cos (θ + φ ) − sin (θ + φ ) = 0 gρ v2 (146.67)2 = = 0.29199 g ρ (32.2)(2288) θ += φ 16.28° = φ 16.28° − 6°

tan (θ + φ ) =

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= φ 10.28° 

PROBLEM 12.54 Tests carried out with the tilting trains described in Problem 12.53 revealed that passengers feel queasy when they see through the car windows that the train is rounding a curve at high speed, yet do not feel any side force. Designers, therefore, prefer to reduce, but not eliminate, that force. For the train of Problem 12.53, determine the required angle of tilt φ if passengers are to feel side forces equal to 10% of their weights.

SOLUTION = vR 60 = mi/h 88 ft/s, 100 = mi/h 146.67 ft/s

Rated speed: From Sample Problem 12.6,

vR2 = g ρ tan θ

or

= ρ

vR2 (88) 2 = = 2288 ft g tan θ 32.2 tan 6°

Let the x-axis be parallel to the floor of the car.

= ΣFx max : Fs + W sin (θ= + φ ) man cos (θ + φ ) =

Solving for Fs,

Now So that Let Then

mv 2

ρ

cos (θ + φ )

 v2  cos (θ + φ ) − sin (θ + φ )  = Fs W   gρ 

v2 (146.67)2 = = 0.29199 = and Fs 0.10W g ρ (32.2)(2288) = 0.10W W [0.29199 cos (θ + φ ) − sin (θ + φ )] = u sin (θ + φ ) cos (θ + φ ) = 1 − u 2 0.10 = 0.29199 1 − u 2 − u or 0.29199 1 − u 2 = 0.10 + u

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PROBLEM 12.54 (Continued)

Squaring both sides,

0.08526(1 − u 2 ) = 0.01 + 0.2u + u 2

or

1.08526u 2 + 0.2u − 0.07526 = 0

The positive root of the quadratic equation is u = 0.18685 Then,

θ += φ sin −1 = u 10.77° = φ 10.77° − 6°

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= φ 4.77° 

PROBLEM 12.55 A 3-kg block is at rest relative to a parabolic dish which rotates at a constant rate about a vertical axis. Knowing that the coefficient of static friction is 0.5 and that r = 2 m, determine the maximum allowable velocity v of the block.

SOLUTION Let b be the slope angle of the dish. tan= b At = r 2 m, tan b= 1

or

dy = dr

1 r 2

b= 45°

Draw free body sketches of the sphere. = ΣFy 0: N cos b − µ S N sin = b − mg 0

N =

mg cos b − µ S sin b

= ΣFn man: N sin b + µ = S N cos b

mv 2

ρ

mg (sin b + µ S N cos b ) mv 2 = cos b − µ S sin b ρ

sin b + µ S cos b sin 45° + 0.5cos 45° = (2)(9.81) = 58.86 m 2 /s 2 v 2 ρ= g cos b − µ S sin b cos 45° − 0.5sin 45° v = 7.67 m/s 

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PROBLEM 12.56 A polisher is started so that the fleece along the circumference undergoes a constant tangential acceleration of 4 m/s 2 . Three seconds after a polisher is started from rest, small tufts of fleece from along the circumference of the 225-mm-diameter polishing pad are observed to fly free of the pad. At this instant, determine (a) the speed v of a tuft as it leaves the pad, (b) the magnitude of the force required to free a tuft if the average mass of a tuft is 1.6 mg.

SOLUTION (a) = at constant ⇒ uniformly acceleration motion Then

v= 0 + at t

At t = 3 s:

v = (4 m/s 2 )(3 s) v = 12.00 m/s 

or (b) = ΣFt mat= : Ft mat or

= Ft (1.6 × 10−6 kg)(4 m/s 2 ) = 6.4 × 10−6 N

: Fn m = ΣFn man= At t = 3 s:

Fn (1.6 × 10−6 kg) =

v2

ρ (12 m/s)2 m) ( 0.225 2

= 2.048 × 10−3 N

Finally,

F = tuft

Ft 2 + Fn2 = (6.4 × 10−6 N) 2 + (2.048 × 10−3 N) 2

or

F= 2.05 × 10−3 N  tuft

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PROBLEM 12.57 A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 10 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.24 m/s 2 , determine the coefficient of static friction between the trunk and the turntable.

SOLUTION First we note that (aB )t = constant implies uniformly accelerated motion.

vB= 0 + ( a B ) t t At t = 10 s:

= vB (0.24 = m/s 2 )(10 s) 2.4 m/s

In the plane of the turntable

= ΣF mB a B= : F mB (a B )t + mB (a B )n Then

= F mB (aB )t2 + (aB )n2 = mB (aB )t2 +

( ) vB2

2

ρ

+ ΣF= 0: N − W = 0 y or At t = 10 s: Then

N = mB g = F µ= µ s mB g sN  v2 = µ s mB g mB (aB )t2 +  ρB 

  

2

1/ 2

or

1 = µs 9.81 m/s 2

2   (2.4 m/s)2    2 2 + (0.24 m/s )      2.5 m   

µ s = 0.236 

or

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PROBLEM 12.58 The carnival ride from Prob 12.51 is modified so that the 80 kg riders can move up and down the inclined wall as the speed of the ride increases. Assuming that the friction between the wall and the carriage is negligible, determine the position h of the rider if the speed v0 = 13 m/s.

SOLUTION

= m 80 = kg, v0 13 m/s

Given: Geometry from figure:

R= 5 + h cos 70°

Free Body Diagram of Rider:

Equations of Motion:

∑F

y

= ma y

n

N cos 70° − mg = m ( 0 ) N=

Substitute (1)into (2):

v= = 13

mg cos 70°

∑F

(1)

= man

mv 2 R N sin 70° R v= m N sin 70° =

(2)

m g tan 70° R m 9.81tan 70° ( 5 + h cos 70° )

h = 3.714 m 

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PROBLEM 12.59 The carnival ride from Prob 12.51 is modified so that the 80 kg riders can move up and down the inclined wall as the speed of the ride increases. Knowing that the coefficient of static friction between the wall and the platform is 0.2, determine the range of values of the constant speed v0 for which the platform will remain in the position shown.

SOLUTION

= m 80 kg, = µ s 0.2, = h 1.5 m

Given: Geometry from figure:

5 + 1.5cos 70° R= = 5.513 m

Minimum and Maximum speeds occur if slipping is impending:

f = µs N Finding the minimum speed so that rider does not slide down the wall: Free Body Diagram of Rider:

Equations of Motion:

∑F

y

= ma y

N cos 70° + f sin 70° − mg = m ( 0 ) N cos 70° + µ s N sin 70° − mg = 0 mg cos 70° + µ s sin 70°

N=

N = 1480.9 N

∑F

n

= man

2 mvmin R 2 mvmin N sin 70° − µ s N cos 70° = R

N sin 70° − f cos 70° =

vmin =

N ( sin 70° − µ s cos 70° ) R m vmin = 9.430 m/s 

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PROBLEM 12.59 (Continued) Finding the maximum speed so that rider does not slide up the wall: Free Body Diagram of Rider:

Equations of Motion:

∑F

y

= ma y

N cos 70° − f sin 70° − mg = m ( 0 ) N cos 70° − µ s N sin 70° − mg = 0 mg cos 70° − µ s sin 70°

N=

N = 5093.4 N

∑F

n

= man

2 mvmin R 2 mvmin N sin 70° + µ s N cos 70° = R

N sin 70° + f cos 70° =

vmin =

N ( sin 70° + µ s cos 70° ) R m vmin = 18.81 m/s 

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PROBLEM 12.60 A semicircular slot of 10-in. radius is cut in a flat plate which rotates about the vertical AD at a constant rate of 14 rad/s. A small, 0.8-lb block E is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are µ s = 0.35 and µk = 0.25, determine whether the block will slide in the slot if it is released in the position corresponding to (a) θ= 80°, (b) θ= 40°. Also determine the magnitude and the direction of the friction force exerted on the block immediately after it is released.

SOLUTION First note

Then

1 (26 − 10 sin θ ) ft 12 vE = ρφABCD

= ρ

a= n

vE2 = ρ (φABCD ) 2

ρ

1  =  (26 − 10 sin θ ) ft  (14 rad/s) 2 12  98 = (13 − 5 sin θ ) ft/s 2 3

Assume that the block is at rest with respect to the plate.

= ΣFx max : N + W cos = θ m

or

vE2

ρ

  v2 N= W  − cos θ + E sin θ  gρ  

ΣFy =ma y : −F + W sin θ =−m

or

sin θ

vE2

ρ

cos θ

  v2 F W  sin θ + E cos θ  =   gρ  

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PROBLEM 12.60 (Continued)

(a)

We have

θ= 80°

Then 1 98   = N (0.8 lb)  − cos80° + × (13 − 5sin 80°) ft/s 2 × sin 80° 2 3 32.2 ft/s   = 6.3159 lb 1 98   = F (0.8 lb) sin 80° + × (13 − 5sin 80°) ft/s 2 × cos80° 2 3 32.2 ft/s   = 1.92601 lb

Now

F= µ= 0.35(6.3159= lb) 2.2106 lb max sN

The block does not slide in the slot, and F = 1.926 lb

(b)

We have

80° 

θ= 40°

Then 1 98   = N (0.8 lb)  − cos 40° + × (13 − 5sin 40°) ft/s 2 × sin 40° 2 3 32.2 ft/s   = 4.4924 lb 1 98   = F (0.8 lb) sin 40° + × (13 − 5sin 40°) ft/s 2 × cos 40° 2 3 32.2 ft/s   = 6.5984 lb

Now

Fmax = µ s N ,

from which it follows that

F > Fmax Block E will slide in the slot and

a= a n + a E/plate E = a n + (a E/plate )t + (a E/plate )n

At t = 0, the block is at rest relative to the plate, thus (a E/plate )n = 0 at t = 0, so that a E/plate must be directed tangentially to the slot.

= ΣFx max : N + W = cos 40° m or

vE2

ρ

sin 40°

  v2 N W  − cos 40° + E sin 40°  (as above) = gρ   = 4.4924 lb

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PROBLEM 12.60 (Continued)

F = µk N

Sliding:

= 0.25(4.4924 lb) = 1.123 lb Noting that F and a E/plane must be directed as shown (if their directions are reversed, then ΣFx is while ma x is ), we have the block slides downward in the slot and

F = 1.123 lb

40° 

Alternative solutions. (a)

Assume that the block is at rest with respect to the plate.

= ΣF ma : W= + R ma n

Then

W W g = = 2  man W vE ρ (φ ABCD )2 g ρ

tan (φ − 10= °)

=

32.2 ft/s 2 98 (13 − 5sin 80°) ft/s 2 3

or

φ −= 10° 6.9588°

and

= φ 16.9588°

Now

= tan φs µ= µ s 0.35 s

so that

(from above)

= φs 19.29°

0 < φ < φs ⇒ Block does not slide and R is directed as shown. Now Then

F R= = sin φ and R

W sin (φ − 10°)

sin16.9588° sin 6.9588° = 1.926 lb

F = (0.8 lb)

The block does not slide in the slot and

F = 1.926 lb

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80° 

PROBLEM 12.60 (Continued)

(b)

Assume that the block is at rest with respect to the plate.

= ΣF ma : W= + R ma n From Part a (above), it then follows that tan = (φ − 50°)

or

g =  ρ (φ ABCD )2

φ −= 50° 5.752°

and

= φ 55.752°

Now

= φs 19.29°

so that

32.2 ft/s 2 98 (13 − 5sin 40°) ft/s 2 3

φ > φs

The block will slide in the slot and then or

φ = φk , where = tan φk µ= µk 0.25 k = φk 14.0362°

To determine in which direction the block will slide, consider the free-body diagrams for the two possible cases.

Now

ΣF= ma : W + R= ma n + ma E/plate

From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding downward. Then

R cos φk m = ΣFx max : W cos 40° += Now Then or

vE2

ρ

sin 40°

F = R sin φk F W vE2 sin 40° W cos 40= °+ tan φk g ρ

  v2 F µkW  − cos 40° + E sin 40°  =   gρ   (see the first solution) = 1.123 lb

The block slides downward in the slot and

F = 1.123 lb

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40° 

PROBLEM 12.61 A small block B fits inside a lot cut in arm OA which rotates in a vertical plane at a constant rate. The block remains in contact with the end of the slot closest to A and its speed is 1.4 m/s for 0 ≤ θ ≤ 150°. Knowing that the block begins to slide when = θ 150°, determine the coefficient of static friction between the block and the slot.

SOLUTION Draw the free body diagrams of the block B when the arm is at = θ 150°. = v a= 0, t

= g 9.81 m/s 2

= ΣFt mat : − mg sin 30° = +N 0 = N mg sin 30°

= ΣFn man : mg = cos 30° − F m = F mg cos30° − Form the ratio

µ= s

v2

ρ

mv 2

ρ

F , and set it equal to µ s for impending slip. N

F = N

g cos 30° − v 2 /ρ 9.81 cos 30° − (1.4) 2 /0.3 = g sin 30° 9.81 sin 30°

µ s = 0.400 

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PROBLEM 12.62 The parallel-link mechanism ABCD is used to transport a component I between manufacturing processes at stations E, F, and G by picking it up at a station when θ = 0 and depositing it at the next station when = θ 180°. Knowing that member BC remains horizontal throughout its motion and that links AB and CD rotate at a constant rate in a vertical plane in such a way that vB = 2.2 ft/s, determine (a) the minimum value of the coefficient of static friction between the component and BC if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending.

SOLUTION = ΣFx max= : F

W vB2 cos θ g ρ

W vB2 − sin θ ma y : N − W = + ΣFy = g ρ or

  v2 N W 1 − B sin θ  =   gρ  

Now

  v2 µ= µ s W 1 − B sin θ  F= max sN gρ  

and for the component not to slide

F < Fmax or or

  v2 W vB2 cos θ < µ s W 1 − B sin θ    g ρ gρ  

µs >

cos θ gρ vB2

− sin θ

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PROBLEM 12.62 (Continued) We must determine the values of θ which maximize the above expression. Thus

(

)

gρ − sin θ − (cos θ )(− cos θ ) d  cos θ  − sin θ vB2  gρ  = 0 = 2 gρ dθ  vB2 − sin θ  − sin θ v2

or

Now

B

)

= sin θ

vB2 = for µ s ( µ s )min gρ

= sin θ

(2.2 ft/s) 2 = 0.180373 (32.2 ft/s 2 ) ( 10 ft ) 12

θ= 10.3915° and θ = 169.609°

or (a)

(

From above, ( µ s ) min =

vB2 cos θ where sin θ = gρ gρ − sin θ v2 B

( µ s )min =

cos θ cos θ sin θ tan θ = = 1 − sin θ 1 − sin 2 θ sin θ

= tan10.3915°

( µ s )min = 0.1834 

or (b)

We have impending motion to the left for

= θ 10.39° 

to the right for

= θ 169.6° 

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PROBLEM 12.63 Knowing that the coefficients of friction between the component I and member BC of the mechanism of Problem 12.62 are µ s = 0.35 and µk = 0.25, determine (a) the maximum allowable constant speed vB if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending.

SOLUTION : F = ΣFx max=

W vB2 cos θ g ρ

W vB2 − sin θ ma y : N − W = + ΣFy = g ρ   v2 N W 1 − B sin θ  = gρ  

or

Fmax = µ s N

Now

  v2 = µ s W 1 − B sin θ  gρ   and for the component not to slide

F < Fmax or

or

  v2 W vB2 cos θ < µ s W 1 − B sin θ    g ρ gρ  

vB2 < µ s

gρ cos θ + µ s sin θ

( )

To ensure that this inequality is satisfied, vB2 must be less than or equal to the minimum value max of µ s g ρ /(cos θ + µ s sin θ ), which occurs when (cos θ + µ s sin θ ) is maximum. Thus d (cos θ + µ s sin θ ) = − sin θ + µ s cos θ = 0 dθ

or

tan θ = µ s

µ s = 0.35 or

= θ 19.2900°

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(1)

PROBLEM 12.63 (Continued)

(a)

The maximum allowed value of vB is then

gρ vB2 = where tan θ µ s µ= s max cos θ + µ s sin θ

( )

tan θ g ρ sin θ = g= ρ cos θ + (tan θ ) sin θ  10  = (32.2 ft/s 2 )  ft  sin 19.2900°  12 

(vB )max = 2.98 ft/s 

or (b)

First note that for 90° < θ < 180°, Eq. (1) becomes

vB2 < µ s

gρ cos α + µ s sin α

where α = 180° − θ . It then follows that the second value of θ for which motion is impending is = θ 180° − 19.2900° = 160.7100°

we have impending motion to the left for

= θ 19.29° 

to the right for

= θ 160.7° 

Alternative solution.

= ΣF ma : W= + R ma n

For impending motion, φ = φs . Also, as shown above, the values of θ for which motion is impending

(

minimize the value of vB, and thus the value of an is an =

vB2

ρ

). From the above diagram, it can be concluded

that an is minimum when ma n and R are perpendicular.

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PROBLEM 12.63 (Continued)

Therefore, from the diagram

θ= φ= tan −1 µ s s and or

(as above)

man = W sin φs m

vB2

or

ρ

= mg sin θ

vB2 = g ρ sin θ

(as above)

α = 180° − θ

(as above)

For 90° ≤ θ ≤ 180°, we have from the diagram

α = φs and or

man = W sin φs vB2 = g ρ sin θ

(as above)

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PROBLEM 12.64 A small 250-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at a constant rate of 7.5 rad/s. Determine the three values of θ for which the collar will not slide on the rod, assuming no friction between the collar and the rod.

SOLUTION If the collar is not sliding, it moves at constant speed on a circle of radius ρ = r sin θ . v = ρω

an Normal acceleration. =

v 2 ρ 2ω 2 = = (r sin θ ) ω 2

ρ

ρ

∑ Fy = ma y :

N cosθ − mg = 0 N =

mg cosθ

ΣFx = max :

N sin θ = ma mg sin θ = (m r sin θ )ω 2 cos θ

(1)

To satisfy (1), either: sin θ = 0

or

cos θ =

g rω 2

θ= 0° or 180° or = cosθ

9.81 = 0.3488 (0.5) (7.5) 2

θ = 0°, 180°, and 69.6° 

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PROBLEM 12.65 A small 250-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at a constant rate of 7.5 rad/s. Knowing that the coefficients of friction are µs = 0.25 and µk = 0.20, indicate whether the collar will slide on the rod if it is released in the position corresponding to (a) θ = 75°, (b) θ = 40°. Also, determine the magnitude and direction of the friction force exerted on the collar immediately after release.

SOLUTION If the collar is not sliding, it moves at constant speed on a circle of radius ρ = r sin θ . v = ρω = r 500 = mm 0.500 = m, ω 7.5 rad/s,

Given:

= m 250 = g 0.250 kg.

Normal acceleration:

= an

v 2 ρ 2ω 2 = = (r sin θ )ω 2

ρ

ρ

ΣF = ma: F − mg sin θ = − m (r sin θ ) ω 2 cos θ

F m ( g − rω 2 cos θ )sin θ =

ΣF = ma : N − mg cos θ = m(r sin θ )ω 2 sin θ = N m( g cos θ + r ω 2 sin 2 θ )

(a) θ= 75°.

F =(0.25) 9.81 − (0.500 cos 75°)(7.5) 2  sin 75°

= 0.61112 N = N (0.25) 9.81cos 75° + (0.500sin 2 75°)(7.5) 2 

= 7.1950 N = µ s N (0.25)(7.1950) = 1.7987 N

Since F < µ s N , the collar does not slide. F = 0.611 N

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75° 

PROBLEM 12.65 (Continued)

F = (0.25) 9.81 − (0.500 cos 40°) (7.5) 2  sin 40°

(b) θ= 40°.

= − 1.8858 N = N (0.25) 9.81cos 40° + (0.500sin 2 40°)(7.5) 2  = 4.7839 N = µ s N (0.25) = (4.7839) 1.1960 N

Since F > µ s N , the collar slides. Since the collar is sliding, F = µ k N .

∑ Fn = +

ma :

N − mg cosθ = man sin θ

= N mg cos θ + m (r sin θ )ω 2 sin θ = m  g cos θ + (r sin 2 θ )ω 2  = (0.25) 9.81cos 40° + (0.500sin 2 40°)(7.5) 2 

= 4.7839 N = F µ= (0.20) (4.7839) = 0.957 N kN

F = 0.957 N

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40° 

PROBLEM 12.66 An advanced spatial disorientation trainer allows the cab to rotate around multiple axes as well as extend inwards and outwards. It can be used to simulate driving, fixed wing aircraft flying, and helicopter maneuvering. In one training scenario, the trainer rotates and translates in the horizontal plane where the location of the pilot is defined by the relationships

π  and θ 0.1( 2t 2 − t ) , where r, θ, and t are expressed = r 10 + 2 cos  t  = 3 

in feet, radians, and seconds, respectively. Knowing that the pilot has a mass of 175 lbs, (a) find the magnitude of the resulting force acting on the pilot at t= 5 s (b) plot the magnitudes of the radial and transverse components of the force exerted on the pilot from 0 to 10 seconds.

SOLUTION

π  = r 10 + 2 cos  t  ft 3 

Given:

= θ 0.1( 2t 2 − t ) rad = m

175 lbs = 5.435 slugs 32.2 ft/s 2

Using Radial and Transverse Coordinates:

2π π  sin  t  ft/s 3 3  2π 2 π   r= − cos  t  ft/s 2 9 3  r = −

= θ 0.1( 4t − 1) rad/s θ = 0.4 rad/s 2

Free Body Diagram of pilot (top and side view) showing net forces in the radial and transverse directions: Equations of Motion:

(

F ∑=

ma = m  r − rθ 2 r

r

)

 2π 2 2 π    π  Fr = m  − cos  t  − 10 + 2 cos  t   0.12 ( 4t − 1)  3    3   9  Fθ ∑=

(

= ma m rθ + 2rθ θ

)

  0.4π  π  π  Fθ = m 10 + 2 cos  t   ( 0.4 ) − sin  t  ( 4t − 1)  3  3  3   

∑F

z

=0 ⇒ Fz =mg

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PROBLEM 12.66 (Continued) (a) Evaluate forces at t=5 s:

Fr = −221.78 lbs Fθ = 61.37 lbs

Fz = 175 lbs

F=

F = 289.1 lb 

Fr2 + Fθ2 + Fz2

(b) Plot of Fr and Fθ from t=0 to t=10 s: Radial and Transverse Components of Force for t=0 to 10 seconds 100 0 -100

Force, (lbs)

-200 -300 -400 -500 -600 -700 -800

Radial Component Transverse Component 0

1

2

3

4

5 6 Time, (sec)

7

8

9

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10



PROBLEM 12.67 An advanced spatial disorientation trainer is programmed to only rotate and translate in the horizontal plane. The pilot’s location is defined by the

(

relationships = r 8 1 − e−t

)

 

and θ = 2 / π  sin

π 

t  , where r, θ, and t 2 

are expressed in feet, radians, and seconds, respectively. Determine the radial and transverse components of the force exerted on the 175 lb pilot at t = 3 s.

SOLUTION

= r 8 (1 − e − t ) ft

Given:

 π  t  rad 2   175 lbs = m = 5.435 slugs 32.2 ft/s 2

θ = 2 / π  sin

Using Radial and Transverse Coordinates: −t

r = 8e ft/s  r = −8e − t ft/s 2

π θ = cos t rad/s 2 π π θ = − sin t rad/s 2 2 2

Free Body Diagram of pilot (top view) showing net forces in the radial and transverse directions: Equations of Motion:

∑F

r

= mar

(

= Fr m  r − rθ 2

)

π   Fr = m  −8e − t − 8 (1 − e − t ) cos 2 t  2  

∑ Fθ = maθ

(

Fθ m rθ + 2rθ =

)

π    π π Fθ = m 8 (1 − e − t )  −  sin t − 16e − t cos t  2 2   2  Evaluate Fr and Fθ at t=3 s:

Fr = −2.165 lb  Fθ = 64.90 lb 

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PROBLEM 12.68 The 3-kg collar B slides on the frictionless arm AA′. The arm is attached to drum D and rotates about O in a horizontal plane at the rate θ = 0.75t , where θ and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of 0.5 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA′.

SOLUTION Kinematics dr = r= 0.5 m/s dt

We have At= t 0,= r 0:

∫

r 0

dr =

∫

t 0

0.5 dt

or

r = (0.5t ) m

Also,

 r =0

θ = (0.75t ) rad/s θ = 0.75 rad/s 2

Now

 0 − [(0.5t ) m][(0.75t ) rad/s]2 = ar = r − rθ 2 = −(0.28125t 3 ) m/s 2

and

a= rθ + 2rθ θ = [(0.5t ) m][0.75 rad/s 2 ] + 2(0.5 m/s)[(0.75t ) rad/s] = (1.125t ) m/s 2

Kinetics Σ= Fr mar : −= T (3 kg)(−0.28125t 3 ) m/s 2

or

T = (0.84375t 3 ) N 2 = ΣFθ mB a= θ : Q (3 kg)(1.125t ) m/s

or

Q = (3.375t ) N

Now require that

T =Q

or or

(0.84375t 3 ) N = (3.375t ) N

t 2 = 4.000 t = 2.00 s 

or

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PROBLEM 12.69 A 0.5-kg block B slides without friction inside a slot cut in arm OA which rotates in a vertical plane. The rod has a constant angular acceleration θ = 10 rad/s2. Knowing that when θ = 45º and r = 0.8 m the velocity of the block is zero, determine at this instant, (a) the force exerted on the block by the arm, (b) the relative acceleration of the block with respect to the arm.

SOLUTION Given:

r =0.8 m, θ = 10 rad/s 2 = m 0.5 = kg, W mg

θ =45°,

Using Radial and Transverse Components:

vr= r= 0,

FBD of Block B

vθ= rθ= 0

θ = 0 (a)

(

= ΣFθ maθ : N − W cos= 45° m rθ + 2rθ

(

= N mg cos 45° + m rθ + 2rθ

)

)

= (0.5)(9.81) cos 45° + 0.5 ( 0.8 )(10 ) + 0  = 7.468

(b)

N = 7.47 N

(

Σ= Fr mar : mg sin 45 = ° m  r − rθ 2

45° 

)

mg sin 45° + rθ 2 m = g sin 45° + rθ 2 = ( 9.81) sin 45° + 0 = 6.937 m/s 2  = r

a B / rod = 6.94 m/s 2

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45° 

PROBLEM 12.70 Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that θ = 15 rad/s and θ = 250 rad/s 2 for the position θ= 20°, determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE.

SOLUTION Kinematics. From the geometry of the system, we have

r = − (2b sin θ )θ

r = 2b cos θ

Then

 r= −2b(θsin θ + θ 2 cos θ )

and

 ar = r − rθ 2 = −2b(θsin θ + θ 2 cos θ ) − (2b cos θ )θ 2 = − 2b(θsin θ + 2θ 2 cos θ )

Now

 20  = − 2 ft  [(250 rad/s 2 )sin 20° + 2(15 rad/s) 2 cos 20°] = −1694.56 ft/s 2 12   aθ = rθ + 2rθ = (2b cos θ )θ + 2(−2bθ sin θ )θ = 2b(θ cos θ − 2θ 2 sin θ )

and

 20  ft  [(250 rad/s 2 ) cos= 20° − 2(15 rad/s) 2 sin 20°] 270.05 ft/s 2 = 2  12  Kinetics. (a)

1 4

lb

We have

Fr =mar =

and

1 lb 4 × (270.05 ft/s 2 ) = Fθ = maθ = 2.0967 lb 32.2 ft/s 2

32.2 ft/s 2

Fr = −13.16 lb 

× (−1694.56 ft/s 2 ) =−13.1565 lb

Fθ = 2.10 lb 

ΣFr : −Fr = −Q cos 20°

(b) or

= Q

1 = (13.1565 lb) 14.0009 lb cos 20°

ΣFθ : Fθ = P − Q sin 20° or

= P (2.0967 + 14.0009sin 20°)= lb 6.89 lb P = 6.89 lb

70° 

Q = 14.00 lb

40° 

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PROBLEM 12.71 The two blocks are released from rest when r = 0.8 m and θ= 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine (a) the initial tension in the cable, (b) the initial acceleration of block A, (c) the initial acceleration of block B.

SOLUTION Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Constraint of cable:

r + yB = constant, r + vB = 0,  r + aB = 0

 r = −aB

(1)

ΣFx mAa A : = T cos θ mAa= mAa A sec θ For block A, = A or T

(2)

For block B,

or

= ΣFy mB aB : = mB g − T mB aB

Adding Eq. (1) to Eq. (2) to eliminate= T, mB g mAa A secθ + mB aB

(3) (4)

Radial and transverse components of a A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.  r − rθ 2 = ar = a A ⋅ er = −a A cos θ

(5)

Noting that initially θ = 0, using Eq. (1) to eliminate r, and changing signs gives

aB = a A cosθ

(6)

Substituting Eq. (6) into Eq. (4) and solving for a A , = aA

mB g = mA sec θ + mB cos θ

(25) (9.81) = 5.48 m/s 2 20sec30° + 25cos 30°

From Eq. = (6), aB 5.48cos30 = ° 4.75 m/s 2 (a)

From Eq. (2), T (20)(5.48)sec30 = = ° 126.6

(b)

Acceleration of block A.

(c)

Acceleration of block B.

T = 126.6 N 

a A = 5.48 m/s 2



a B = 4.75 m/s 2 

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PROBLEM 12.72 The velocity of block A is 2 m/s to the right at the instant when r = 0.8 m and θ= 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine, at this instant, (a) the tension in the cable, (b) the acceleration of block A, (c) the acceleration of block B.

SOLUTION Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Radial and transverse components of v A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.

r = vr = v A ⋅ er = −v A cos30° = −2cos30° = −1.73205 m/s rθ = vθ = v A ⋅ eθ = −v A sin 30° = 2sin= 30° 1.000 m/s 2 = θ

vθ 1.000 = = 1.25 rad/s r 0.8

constant, Constraint of cable: r + yB = r + vB = 0,  r + aB = 0 or

 r = −aB

(1)

ΣFx mAa A := T cosθ mAa= mAa A secθ For block A, = A or T

(2)

= ΣFy mB aB: m= mB aB Bg − T

(3)

For block B,

Adding Eq. (1) to Eq. (2) to eliminate= T, mB g mAa A secθ + mB aB

(4)

Radial and transverse components of a A. Use a method similar to that used for the components of velocity.  r − rθ 2 = ar = a A ⋅ er = −a A cos θ

(5)

Using Eq. (1) to eliminate r and changing signs gives = aB a A cosθ − rθ 2

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(6)

PROBLEM 12.72 (Continued)

Substituting Eq. (6) into Eq. (4) and solving for a A ,

= aA

(

)

mB g + rθ 2 = mA sec θ + mB cos θ

(25)[9.81 + (0.8)(1.25) 2 ] = 6.18 m/s 2 20sec30° + 25cos 30°

2 From Eq. = (6), aB 6.18cos 30° − (0.8)(1.25) = 4.10 m/s 2

(a)

From Eq. (2), T (20)(6.18)sec30 = = ° 142.7

(b)

Acceleration of block A.

(c)

Acceleration of block B.

T = 142.7 N 

a A = 6.18 m/s 2

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

a B = 4.10 m/s 2 

PROBLEM 12.73* Slider C has a weight of 0.5 lb and may move in a slot cut in arm AB, which rotates at the constant rate θ0 = 10 rad/s in a horizontal plane. The slider is attached to a spring of constant k = 2.5 lb/ft, which is unstretched when r = 0. Knowing that the slider is released from rest with no radial velocity in the position r = 18 in. and neglecting friction, determine for the position r = 12 in. (a) the radial and transverse components of the velocity of the slider, (b) the radial and transverse components of its acceleration, (c) the horizontal force exerted on the slider by arm AB.

SOLUTION Let l0 be the radial coordinate when the spring is unstretched. Force exerted by the spring. Fr = −k (r − l0 ) Σ Fr = mar : − k (r − l0 ) = m( r − rθ 2 ) kl k   r = θ 2 −  r + 0 m m 

(1)

But

d dr dr dr = (r) = r dt dr dt dr  kl  k   = rdr rdr = θ 2 −  r + 0  dr m m 

 = r

Integrate using the condition r = r0 when r = r0 . 1 2 r  1   2 k  2 kl0 =  θ − r + r 2 r0  2  m m

(

 r r  r0

)

kl 1 2 1 2 1  2 k  2 θ −  r − r02 + 0 (r − r0 ) r − r0= 2 2 2  m m 2kl0 k  (r − r0 ) r 2 =r02 +  θ 2 −  r 2 − r02 + m m 

(

Data:

= m

)

W 0.5 lb = = 0.01553 lb ⋅ s 2 /ft g 32.2 ft/s 2

= θ 10 = rad/s, k 2.5 = lb/ft, l0 0 = r0 (= vr )0 0,= r0 18= in. 1.5 ft,= r 12= in. 1.0 ft

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PROBLEM 12.73* (Continued)

(a)

Components of velocity when r = 12 in. 2.5   r 2 = 0 + 102 − (1.02 − 1.52 ) + 0  0.01553   = 76.223 ft 2 /s 2 vr = r = ±8.7306 ft/s

Since r is decreasing, vr is negative

(b)

r = −8.7306 ft/s

vr = −8.73 ft/s 

v= r= θ (1.0)(10) θ

vθ = 10.00 ft/s 

Components of acceleration.

Fr =−kr + kl0 =−(2.5)(1.0) + 0 =−2.5 lb ar =

Fr 2.5 = − m 0.01553

ar = 161.0 ft/s 2 

aθ = rθ + 2rθ =+ 0 (2)(−8.7306)(10) aθ = −174.6 ft/s 2 

(c)

Transverse component of force.

= Fθ ma = (0.01553)(−174.6) θ

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Fθ = −2.71 lb 

PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.25), express the radial and transverse components of the velocity v of the particle as functions of θ .

SOLUTION Since the particle moves under a central force, h = constant.  h= r v = h r 2θ= 0 0 0

Using Eq. (12.27),

r0 v0 r0 v0 cos 2θ v0 = = θ = cos 2θ r0 r2 r0 2

or Radial component of velocity.

vr= r=

= r0

r0 dr  d  θ=  dθ dθ  cos 2θ

 sin 2θ θ θ= r0 3/ 2 (cos 2 ) θ 

sin 2θ v0 cos 2θ (cos 2θ )3/ 2 r

vr = v0

sin 2θ cos 2θ



Transverse component of velocity.

vθ=

h r0 v0 cos 2θ = r r0

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vθ = v0 cos 2θ 

PROBLEM 12.75 For the particle of Problem 12.74, show (a) that the velocity of the particle and the central force F are proportional to the distance r from the particle to the center of force O, (b) that the radius of curvature of the path is proportional to r3. PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.25), express the radial and transverse components of the velocity v of the particle as functions of θ.

SOLUTION Since the particle moves under a central force, h = constant. Using Eq. (12.27),  h= r v = h r 2θ= 0 0 0

r0 v0 r0 v0 cos 2θ v0 cos 2θ = θ = or = r0 r2 r0 2

Differentiating the expression for r with respect to time, = r

 r0 dr  d  sin 2θ sin 2θ v0 sin 2θ = = = = θ θ r0 cos 2θ v0  θ r0 3/ 2 3/ 2 dθ dθ  cos 2θ  r (cos 2θ ) (cos 2θ ) cos 2θ 0

Differentiating again,  r =

(a)

sin 2θ   2 cos 2 2θ + sin 2 2θ  v0 2 2 cos 2 2θ + sin 2 2θ dr  d  θ θ = = =  v0  θ v0 dθ dθ  r0 (cos 2θ )3/ 2 cos 2θ  cos 2θ

vr= r= v0

sin 2θ cos 2θ

=

v0 r sin 2θ r0 v=

θ v= r= θ

(vr )2 + (vθ )2 =

v0 r cos 2θ r0

v0 r sin 2 2θ + cos 2 2θ r0

v=

v0 r  r0

v 2 2 cos 2 2θ + sin 2 2θ r0 v0 2  − cos 2 2θ ar =− r rθ 2 =0 r0 cos 2θ cos 2θ r0 2 =

v0 2 cos 2 2θ + sin 2 2θ v0 v0 2 r = = r0 r0 2 r0 cos 2θ cos 2θ

= Fr ma = r

mv02 r r02

:

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Fr =

mv02 r r02



PROBLEM 12.75 (Continued)

Since the particle moves under a central force, aθ = 0. Magnitude of acceleration. a=

ar 2 + aθ 2 =

v0 2 r r0 2

Tangential component of acceleration. = at

v0 2 r dv d  v0 r  v0  = = = sin 2θ r   dt dt  r0  r0 r0 2

Normal component of acceleration. at =

r  cos 2θ =  0  r

But

an =

Hence, (b)

But a= n

a 2 − at 2 =

v2

ρ

or

v0 2 r r0 2

1 − sin 2 2θ =

v0 2 r cos 2θ r0 2

2

v0 2 r

= ρ

v 2 v0 2 r 2 r = ⋅ an r0 2 v0 2

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ρ=

r3  r02

PROBLEM 12.76 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F along a semicircular path of diameter OA. Observing that r = r0 cos θ and using Eq. (12.25), show that the speed of the particle is v = v0 /cos 2 θ .

SOLUTION Since the particle moves under a central force, h = constant.  h= r v = h r 2θ= 0 0 0

Using Eq. (12.27),

r0 v0 r0 v0 v0 = θ = = 2 2 2 r r0 cos θ r0 cos 2 θ

or Radial component of velocity.

vr = r =

d (r0 cos θ ) = −(r0 sin θ )θ dt

Transverse component of velocity.

θ (r0 cos θ )θ v= r= θ Speed.

v = vr 2 + vθ 2 = r0θ =

r0 v0 r0 cos θ 2

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v=

v0 cos 2 θ



PROBLEM 12.77 For the particle of Problem 12.76, determine the tangential component Ft of the central force F along the tangent to the path of the particle for (a) θ = 0, (b) θ= 45°. PROBLEM 12.76 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F along a semicircular path of diameter OA. Observing that r = r0 cos θ and using Eq. (12.25), show that the speed of the particle is v = v0 /cos 2 θ .

SOLUTION Since the particle moves under a central force, h = constant Using Eq. (12.27),  h= r v = h r 2θ= 0 0 0 r0 v0 r0 v0 v0 = = θ = 2 2 2 r r0 cos θ r0 cos 2 θ

Radial component of velocity. vr = r =

d (r0 cos θ ) = −(r0 sin θ )θ dt

Transverse component of velocity.

θ (r0 cos θ )θ v= r= θ Speed.

v = vr 2 + vθ 2 = r0θ =

r0 v0 r0 cos θ 2

=

v0 cos 2 θ

Tangential component of acceleration.

= at

v0 dv (−2)(− sin θ )θ 2v0 sin θ = v0 = ⋅ 3 3 dt cos θ cos θ r0 cos 2 θ =

2v0 2 sin θ r0 cos5 θ

Tangential component of force. = = Ft ma t : Ft

2mv0 2 sin θ r0 cos5 θ

θ 0,= Ft 0 (a)= (b)

θ =° 45 ,

2mv sin 45° Ft = 0 5 cos 45°

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Ft = 0  Ft =

8mv0 2  r0

PROBLEM 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and that the moon requires 27.32 days to complete one full revolution about the earth.

SOLUTION Mm [Eq. (12.28)] r2

We have

F =G

and

F = F= ma= m n n

Then

or

Now

G

Mm v2 = m r r2 M= v=

r 2 v G 2π r

τ 2

so that Noting that and

= M

2

r  2π r  1  2π  3 = r   G τ  G  τ 

= τ 27.32 = days 2.3604 × 106 s = r 238,910 = mi 1.26144 × 109 ft

we have M =

or

v2 r

2

  2π (1.26144 × 109 ft)3  4  6 −9 4   34.4 × 10 ft /lb ⋅ s  2.3604 × 10 s  1

M= 413 × 1021 lb ⋅ s 2 /ft 

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PROBLEM 12.79 Show that the radius r of the moon’s orbit can be determined from the radius R of the earth, the acceleration of gravity g at the surface of the earth, and the time τ required for the moon to complete one full revolution about the earth. Compute r knowing that τ = 27.3 days, giving the answer in both SI and U.S. customary units.

SOLUTION Mm r2

We have

F =G

and

F = F= ma= m n n

Then

G

GM r

GM = gR 2

Now

v2 =

so that

= τ

For one orbit,

[Eq. (12.30)]

g gR 2 or v = R r r 2π r 2π r = v R gr

 gτ 2 R 2 r =  2  4π

or

v2 r

Mm v2 = m r r2 v2 =

or

[Eq. (12.28)]

1/ 3

  



Q.E.D.

= τ 27.3 = days 2.35872 × 106 s

Now

= = R 3960 mi 20.9088 × 106 ft 1/3

SI:

 9.81 m/s 2 × (2.35872 × 106 s) 2 × (6.37 × 106 m) 2  = r  = 382.81 × 106 m  2 4π   = r 383 × 103 km 

or U.S. customary units: 1/3

 32.2 ft/s 2 × (2.35872 × 106 s)2 × (20.9088 × 106 ft) 2  r  1256.52 × 106 ft = =  2 4π  

or

= r 238 × 103 mi 

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PROBLEM 12.80 Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the ground. Determine (a) the altitude of these satellites above the surface of the earth, (b) the velocity with which they describe their orbit. Give the answers in both SI and U.S. customary units.

SOLUTION For gravitational force and a circular orbit,

= Fr

GMm mv 2 = r r2

or = v

GM r

Let τ be the period time to complete one orbit. = vτ 2π r

But

Then

= v 2τ 2

GM τ 2 r r or = = 4π 2 3

GM τ 2 = 4π 2 r 2 r  GM τ 2  2  4π

1/3

  

= = τ 23.934 h 86.1624 × 103 s

Data: (a)

or

= g 9.81 m/s 2 ,= R 6.37 × 106 m

In SI units:

GM =gR 2 =(9.81)(6.37 × 106 ) 2 =398.06 × 1012 m3 /s 2 1/3

 (398.06 × 1012 )(86.1624 × 103 ) 2  r  42.145 × 106 m = =  2 4π  

h = 35,800 km 

altitude h =r − R =35.775 × 106 m In U.S. units:

2 = g 32.2 ft/s= , R 3960 = mi 20.909 × 106 ft

GM =gR 2 =(32.2)(20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2 1/3

 (14.077 × 1015 )(86.1624 × 103 ) 2  6 r  = =  138.334 × 10 ft 2 4π  

altitude h =r − R =117.425 × 106 ft

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h = 22, 200 mi 

PROBLEM 12.80 (Continued)

(b)

In SI units: = v

GM = r

398.06 × 1012 = 3.07 × 103 m/s 42.145 × 106

= v

GM = r

14.077 × 1015 = 10.09 × 103 ft/s 6 138.334 × 10

v = 3.07 km/s 

In U.S. units: = v 10.09 × 103 ft/s 

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PROBLEM 12.81 Show that the radius r of the orbit of a moon of a given planet can be determined from the radius R of the planet, the acceleration of gravity at the surface of the planet, and the time τ required by the moon to complete one full revolution about the planet. Determine the acceleration of gravity at the surface of the planet Jupiter knowing that R = 71,492 km and that τ = 3.551 days and r = 670.9 × 103 km for its moon Europa.

SOLUTION Mm r2

We have

F =G

and

F = F= ma= m n n

Then or Now so that For one orbit,

G

[Eq. (12.28)] v2 r

Mm v2 = m r r2 v2 =

GM r

GM = gR 2

= v2

τ =

[Eq. (12.30)]

gR 2 g = or v R r r 2π r 2π r = v R gr

or

 gτ 2 R 2 r =  2  4π

Solving for g,

g = 4π 2

1/3

  



Q.E.D.

r3 τ 2 R2

and noting that τ = 3.551 days = 306,806 s, then g Jupiter = 4π 2 = 4π 2

3 rEur 2 τ Eur RJup

(670.9 × 106 m)3 (306,806 s) 2 (71.492 × 106 m) 2 g Jupiter = 24.8 m/s 2 

or Note:

g Jupiter ≈ 2.53g Earth

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PROBLEM 12.82 The orbit of the planet Venus is nearly circular with an orbital velocity of 126.5 × 103 km/h. Knowing that the mean distance from the center of the sun to the center of Venus is 108 × 106 km and that the radius of the sun is 695 × 103 km, determine (a) the mass of the sun, (b) the acceleration of gravity at the surface of the sun.

SOLUTION Let M be the mass of the sun and m the mass of Venus. For the circular orbit of Venus, GMm mv 2 = ma = n r r2

GM = rv 2

where r is radius of the orbit. r = 108 × 106 km = 108 × 109 m

Data:

v= 126.5 × 103 km/hr = 35.139 × 103 m/s GM =(108 × 109 )(35.139 × 103 ) 2 =1.3335 × 1020 m3 /s 2

= M

(a) Mass of sun. (b) At the surface of the sun,

GM 1.3335 × 1020 m3 /s 2 = G 66.73 × 10−12

= M 1.998 × 1030 kg 

R =695.5 × 103 km =695.5 × 106 m GMm = mg R2

g =

GM = R2

1.3335 × 1020 (695.5 × 106 ) 2

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g = 276 m/s 2 

PROBLEM 12.83 A satellite is placed into a circular orbit about the planet Saturn at an altitude of 2100 mi. The satellite describes its orbit with a velocity of 54.7 × 103 mi/h. Knowing that the radius of the orbit about Saturn and the periodic time of Atlas, one of Saturn’s moons, are 85.54 × 103 mi and 0.6017 days, respectively, determine (a) the radius of Saturn, (b) the mass of Saturn. (The periodic time of a satellite is the time it requires to complete one full revolution about the planet.)

SOLUTION 2π rA

Velocity of Atlas.

vA =

where

v A =85.54 × 103 mi =451.651 × 106 ft

τA

= τ A 0.6017 = days 51,987s

and

Gravitational force.

vA =

(2π ) (451.651 × 106 ) = 54.587 × 103 ft/s 51,987

= F

GMm mv 2 = r r2

from which

2 GM = rv= constant

For the satellite,

rs vs2 = rA v A2

rs =

rA v A2 vs 2

vs = 54.7 × 103 mi/h = 80.227 × 103 ft/s

where

(451.651 × 106 )(54.587 × 103 ) 2 = 209.09 × 106 ft (80.227 × 103 ) 2 rs = 39, 600 mi

= rs

(a)

Radius of Saturn.

R= rs − (altitude) = 39,600 − 2100 (b)

R = 37,500 mi 

Mass of Saturn.

= M

rA v A2 (451.651 × 106 )(54.587 × 103 ) 2 = G 34.4 × 10−9 M= 39.1 × 1024 lb ⋅ s 2 /ft 

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PROBLEM 12.84 The periodic time (see Prob. 12.83) of an earth satellite in a circular polar orbit is 120 minutes. Determine (a) the altitude h of the satellite, (b) the time during which the satellite is above the horizon for an observer located at the north pole.

SOLUTION For gravitational force and a circular orbit, = Fr

mv 2 r

GMm = r2

v or =

GM r

Let τ be the periodic time to complete one orbit. GM = vτ 2= πr or τ 2π r r 1/ 3

Solving for r,

 GM τ 2  r =  2    4π 

For earth, R =3960 mi =20.909 × 106 ft, g =32.2 ft/s 2

(

GM =gR 2 =( 32.2 ) 20.909 × 106

)

2

=14.077 × 1015 ft 3/s 2

Data: = τ 120 = min 7200 s

(

)

1/ 3

 14.077 × 1015 ( 7200 )2   = = r  26.441 × 106 ft   4π 2  

(a) altitude h = r − R = 5.532 × 106 ft (b) cos= θ

R = r

h = 1048 mi 

20.909 × 106 = 0.79078 26.441 × 106

= θ 37.74°

= t AB

2θ τ = 360°

7200 ) ( 75.48)( = 360

1509.6 s t AB = 25.2 min 

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PROBLEM 12.85 A 500 kg spacecraft first is placed into a circular orbit about the earth at an altitude of 4500 km and then is transferred to a circular orbit about the moon. Knowing that the mass of the moon is 0.01230 times the mass of the earth and that the radius of the moon is 1737 km, determine (a) the gravitational force exerted on the spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if the periodic times (see Problem 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the surface of the moon.

SOLUTION = RE 6.37 × 106 m

First note that

rE = RE + hE = (6.37 × 106 + 4.5 × 106 ) m

Then

(a)

= 10.87 × 106 m F=

We have

GMm r2

[Eq. (12.28)]

GM = gR 2

and

[Eq. (12.29)]

R 2 m = = F gR W  r2 r

Then

For the earth orbit,

2

 6.37 × 106 m  F = (500 kg)(9.81 m/s )   6  10.87 × 10 m 

2

2

F = 1684 N 

or (b)

From the solution to Problem 12.78, we have 2

M=

Then

Now

τ=

1  2π  3 r G  τ 

2π r 3/ 2 GM

τE = τM ⇒

2π rE3/ 2 GM E

2π r 3/ 2 =M GM M

(1)

1/3

or or

 MM  1/3 6 = rM =  rE (0.01230) (10.87 × 10 m) M  E  = rM 2.509 × 106 m

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

rM = 2510 km 

PROBLEM 12.85 (Continued)

(c)

GM = gR 2

We have

[Eq.(12.29)]

Substituting into Eq. (1) 2π rE3/ 2 RE g E

=

2π rM3/ 2 RM g M 2

or

3

2

 RE   rM   RE   M M = g M =    gE     RM   rE   RM   M E

  gE 

using the results of Part (b). Then 2

 6370 km  2 gM =   (0.01230)(9.81 m/s )  1737 km  g moon = 1.62 m/s 2 

or Note:

g moon ≈

1 g earth 6

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PROBLEM 12.86 A space vehicle is in a circular orbit of 2200-km radius around the moon. To transfer it to a smaller circular orbit of 2080-km radius, the vehicle is first placed on an elliptic path AB by reducing its speed by 26.3 m/s as it passes through A. Knowing that the mass of the moon is 73.49 × 1021 kg, determine (a) the speed of the vehicle as it approaches B on the elliptic path, (b) the amount by which its speed should be reduced as it approaches B to insert it into the smaller circular orbit.

SOLUTION For a circular orbit,

: F m = Σ Fn man = F =G

Eq. (12.28): Then

G

or

v2 r Mm r2

Mm v2 =m 2 r r GM v2 = r

66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg 2200 × 103 m

Then

2 (v A )circ =

or

(v A )circ = 1493.0 m/s

and

2 (vB )circ =

or

(vB )circ = 1535.5 m/s

(a)

We have

66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg 2080 × 103 m

(v= (v A )circ + ∆v A A )TR = (1493.0 − 26.3) m/s = 1466.7 m/s

Conservation of angular momentum requires that

rA m(v A )TR = rB m(vB )TR or

2200 km × 1466.7 m/s 2080 km = 1551.3 m/s

= (vB )TR

(vB )TR = 1551 m/s 

or (b)

Now or

(v B= )circ (vB )TR + ∆vB ∆= vB (1535.5 − 1551.3) m/s ∆vB = −15.8 m/s 

or

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PROBLEM 12.87 As a first approximation to the analysis of a space flight from the earth to Mars, assume the orbits of the earth and Mars are circular and coplanar. The mean distances from the sun to the earth and to Mars are 149.6 × 106 km and 227.8 × 106 km, respectively. To place the spacecraft into an elliptical transfer orbit at point A, its speed is increased over a short interval of time to v A which is 2.94 km/s faster than the earth’s orbital speed. When the spacecraft reaches point B on the elliptical transfer orbit, its speed vB is increased to the orbital speed of Mars. Knowing that the mass of the sun is 332.8 × 103 times the mass of the earth, determine the increase in speed required at B.

SOLUTION

(

For earth, ( GM )earth =gR 2 =( 9.81) 6.37 × 106

(

)

2

=398.06 × 1012 m3/s 2

)

For sun, ( GM )sun =× 332.8 103 ( GM )earth = 132.474 × 1018 m3 /s 2 For circular orbit of earth, rE = 149.6 × 106 km = 149.6 × 109 m/s

( GM )sun

= vE For transfer orbit AB,

= rE

= rA rE ,

132.474 × 1018 = 29.758 × 103 m/s 149.6 × 109

= rB r= 227.8 × 109 m M

v A= vE + ( ∆v ) A= 29.758 × 103 + 2.94 × 103= 32.698 × 103 m/s mrAv A = mrBvB

= vB

rAv A = rB

(149.6 × 10 )(32.698 × 10=) 9

3

227.8 × 10

9

21.473 × 103 m/s

For circular orbit of Mars, = vM

( GM )sun

= rM

132.474 × 1018 24.115 × 103 m/s = 9 227.8 × 10

Speed increase at B.

( ∆v )B

= vM − vB = 24.115 × 103 − 21.473 × 103 = 2.643 × 103 m/s

( ∆v )B

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= 2.64 km/s 

PROBLEM 12.88 To place a communications satellite into a geosynchronous orbit (see Problem 12.80) at an altitude of 22,240 mi above the surface of the earth, the satellite first is released from a space shuttle, which is in a circular orbit at an altitude of 185 mi, and then is propelled by an upperstage booster to its final altitude. As the satellite passes through A, the booster’s motor is fired to insert the satellite into an elliptic transfer orbit. The booster is again fired at B to insert the satellite into a geosynchronous orbit. Knowing that the second firing increases the speed of the satellite by 4810 ft/s, determine (a) the speed of the satellite as it approaches B on the elliptic transfer orbit, (b) the increase in speed resulting from the first firing at A.

SOLUTION For earth,

= R 3960 = mi 20.909 × 106 ft GM =gR 2 =(32.2)(20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2

rA = 3960 + 185 = 4145 mi = 21.8856 × 106 ft rB = 3960 + 22, 240 = 26, 200 mi = 138.336 × 106 ft

Speed on circular orbit through A. (v A )circ = =

GM rA 14.077 × 1015 21.8856 × 106

= 25.362 × 103 ft/s

Speed on circular orbit through B. (vB )circ = =

GM rB 14.077 × 1015 138.336 × 106

= 10.088 × 103 ft/s

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PROBLEM 12.88 (Continued)

(a)

Speed on transfer trajectory at B. (vB ) tr= 10.088 × 103 − 4810 = 5.278 × 103 Conservation of angular momentum for transfer trajectory.

5280 ft/s 

rA (v A ) tr = rB (vB ) tr (v A ) tr =

rB (vB ) tr rA

(138.336 × 106 )(5278) 21.8856 × 106 = 33.362 × 103 ft/s =

(b)

Change in speed at A. ∆v= (v A ) tr − (v A )circ A = 33.362 × 103 − 25.362 × 103 = 8.000 × 103

∆v A = 8000 ft/s 

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PROBLEM 12.89 A space vehicle is in a circular orbit of 1400-mi radius around the moon. To transfer to a smaller orbit of 1300-mi radius, the vehicle is first placed in an elliptic path AB by reducing its speed by 86 ft/s as it passes through A. Knowing that the mass of the moon is 5.03 × 1021 lb ⋅ s 2 /ft, determine (a) the speed of the vehicle as it approaches B on the elliptic path, (b) the amount by which its speed should be reduced as it approaches B to insert it into the smaller circular orbit.

SOLUTION Circular orbits: v =

GM r

= rA 1400 = mi 7.392 × 106 ft

= ( vA )1

10 ) (34.4 × 10 )(5.03 ×= −9

21

7.392 × 106

4.8382 × 103 ft/s

= rB 1300 = mi 6.864 × 106 ft

= ( vB ) 2

10 ) (34.4 × 10 )(5.03 × = −9

21

6.864 × 106

5.0208 × 103 ft/s

(a) Transfer orbit AB.

v)A ( v= ( vA )1 + ( ∆= A )2

4.8382 × 103 −= 86 4.7522 × 103 ft/s

mrA ( v A )2 = mrB ( vB )1

= ( vB )1

rA ( v A )2 = rB

10 ) ( 7.392 × 10 )( 4.7522 ×= 6

6.864 × 103

3

5.1178 × 103 ft/s

( v= B )1

5.12 × 103 ft/s 

(b) Speed change at B.

5.0208 × 103 ft/s − 5.1178 × 103 ft/s = −97.0 ft/s ( ∆vB ) = ( vB )2 − ( vB )1 = Speed reduction at B.

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∆vB = 97.0 ft/s 

PROBLEM 12.90 A 1 kg collar can slide on a horizontal rod, which is free to rotate about a vertical shaft. The collar is initially held at A by a cord attached to the shaft. A spring of constant 30 N/m is attached to the collar and to the shaft and is undeformed when the collar is at A. As the rod rotates at the rate θ = 16 rad/s, the cord is cut and the collar moves out along the rod. Neglecting friction and the mass of the rod, determine (a) the radial and transverse components of the acceleration of the collar at A, (b) the acceleration of the collar relative to the rod at A, (c) the transverse component of the velocity of the collar at B.

SOLUTION F= k (r − rA ) sp

First note (a)

Fθ = 0 and at A,

Fr = − Fsp = 0

(a A )r = 0 

(a A )θ = 0  ΣFr = mar :

(b)

Noting that

−Fsp = m( r − rθ 2 )

acollar/rod =  r , we have at A = 0 m[acollar/rod − (150 mm)(16 rad/s) 2 ] acollar/rod = 38400 mm/s 2 (acollar/rod ) A = 38.4 m/s 2 

or (c)

After the cord is cut, the only horizontal force acting on the collar is due to the spring. Thus, angular momentum about the shaft is conserved.

= rA m (v A )θ rB= m (vB )θ where (v A )θ rA θ0 Then or

= (vB )θ

150 mm = [(150 mm)(16 rad/s)] 800 mm/s 450 mm

(vB )θ = 0.800 m/s 

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PROBLEM 12.91 A 1-lb ball A and a 2-lb ball B are mounted on a horizontal rod which rotates freely about a vertical shaft. The balls are held in the positions shown by pins. The pin holding B is suddenly removed and the ball moves to position C as the rod rotates. Neglecting friction and the mass of the rod and knowing that the initial speed of A is v A = 8 ft/s, determine (a) the radial and transverse components of the acceleration of ball B immediately after the pin is removed, (b) the acceleration of ball B relative to the rod at that instant, (c) the speed of ball A after ball B has reached the stop at C.

SOLUTION Let r and θ be polar coordinates with the origin lying at the shaft. Constraint of rod: θ B = θ A + π radians; θB = θA = θ; θB = θA = θ. (a) Components of acceleration Sketch the free body diagrams of the balls showing the radial and transverse components of the forces acting on them. Owing to frictionless sliding of B along the rod, ( FB ) r = 0. Radial component of acceleration of B.

Fr = mB (aB )r :

(aB ) r = 0 

Transverse components of acceleration.

(a A )θ =rAθ + 2rAθ =raθ

(aB= )θ rBθ + 2rBθ

(1)

Since the rod is massless, it must be in equilibrium. Draw its free body diagram, applying Newton’s 3rd Law.

= ΣM 0 0: rA ( FA )θ + rB (= FB )θ rAmA (a A )θ + rB mB = (aB )θ 0 rAmArAθ + rB mB (rBθ + 2rBθ) = 0

θ =

−2mB rBθ mArA2 + mB rB 2

t = 0, rB 0= At = so that θ 0. From Eq. (1),

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(aB )θ = 0 

PROBLEM 12.91 (Continued)

(b)

Acceleration of B relative to the rod.

At= = 96 in./s, = θ t 0, (v A= )θ 8 ft/s

(v A )θ 96 = = 9.6 rad/s rA 10

 rB − rBθ 2 = (aB ) r = 0 2  2 (8) (9.6)  = rB r= = 737.28 in./s 2 Bθ

 rB = 61.4 ft/s 2 

(c)

Speed of A.

Substituting

d (mr 2θ) for rFθ in each term of the moment equation dt

gives

(

)

(

)

d d 0 mArA2θ + mB rB 2θ = dt dt

Integrating with respect to time,

mArA2θ + mB rB 2θ=

( m r θ ) + ( m r θ ) 2

A A

2

0

B B

0

Applying to the final state with ball B moved to the stop at C,  WA 2 WB 2    WA 2 WB  rA + rC  θ f = rA + (rB )02  θ0   g g  g   g  WArA2 + WB (rB )02  θ f θ0 = = WArA2 + WB rC 2

(1)(10) 2 + (2)(8) 2 = (9.6) 3.5765 rad/s (1)(10) 2 + (2)(16) 2

 (v= r= (10)(3.5765) = 35.765 in./s A) f Aθ f

(v A ) f = 2.98 ft/s 

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PROBLEM 12.92 Two 2.6-lb collars A and B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about CD. The two collars are connected by a cord running over a pulley that is attached to the frame at O and a stop prevents collar B from moving. The frame is rotating at the rate θ = 12 rad/s and r = 0.6 ft when the stop is removed allowing collar A to move out along rod OE. Neglecting friction and the mass of the frame, determine, for the position r = 1.2 ft, (a) the transverse component of the velocity of collar A, (b) the tension in the cord and the acceleration of collar A relative to the rod OE.

SOLUTION Masses: m= m= A B

2.6 = 0.08075 lb ⋅ s 2 /ft 32.2

(a) Conservation of angular momentum of collar A: ( H 0 ) 2 = ( H 0 )1

mAr1(vθ )1 = mAr2 (vθ )2 r1(vθ )1 r12θ1 (0.6) 2 (12) = = = 3.6 1.2 r2 r2

(= vθ ) 2

(vθ )2 = 3.60 ft/s  = θ2

(vθ ) 2 3.6 = = 3.00 rad/s 1.2 rA

(b) Let y be the position coordinate of B, positive upward with origin at O.  Constraint of the cord: r − y constant = = or y  r

Kinematics: (aB ) y=

Collar B: Collar A:

 y=  r

and

(a A ) r=  r − rθ 2

ΣFy= mB aB : T − WB= mB  y= mB  r

(1)

ΣF= mA (a A ) r : − T= mA ( r − rθ 2 ) r

(2)

Adding (1) and (2) to eliminate T, −WB = (mA + mB ) r + mArθ 2

a A/rod =  r=

mArθ 2 − WB (0.08075)(1.2)(3.00) 2 − (2.6) = = −10.70 ft/s 2 0.08075 + 0.08075 mA + mB

T= mB ( r + g )= (0.08075)(−10.70 + 32.2)

T = 1.736 lb 

a A / rod = 10.70 ft/s 2 radially inward. 

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PROBLEM 12.93 A small ball swings in a horizontal circle at the end of a cord of length l1 , which forms an angle θ1 with the vertical. The cord is then slowly drawn through the support at O until the length of the free end is l2 . (a) Derive a relation among l1 , l2 , θ1 , and θ 2 . (b) If the ball is set in motion so that initially l1 = 0.8 m and θ= 1 35°, determine the angle θ 2 when l2 = 0.6 m.

SOLUTION (a)

For state 1 or 2, neglecting the vertical component of acceleration, = ΣFy 0: T cos= θ −W 0 T = W cos θ

= ΣFx man := = θ cos θ T sin θ W sin But ρ =  sin θ

mv 2

ρ

so that

= v2

ρW

= sin 2 θ cos θ  g sin θ tan θ m

v1 = 1 g sin θ1 tan θ1

v2 =  2 g sin θ 2 tan θ 2

and

0: H= constant ΣM= y y

r1mv1 = r2 mv2

or

v11 sin θ1 = v2  2 sin θ 2

3/2 3/2 1 g sin θ1 sin θ1 tan θ1 =  2 sin θ 2 sin θ 2 tan θ 2

31 sin 3 θ1 tan θ1 = 32 sin 3 θ 2 tan θ 2  (b)

35°, With θ1 =

1 =0.8 m, and  2 =0.6 m

(0.8)3 sin 3 35° tan 35° =(0.6)3 sin 3 θ 2 tan θ 2

sin 3 θ 2 tan θ 2 − 0.31320 = 0

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= θ 2 43.6° 

PROBLEM 12.94 A particle of mass m is projected from Point A with an initial velocity v 0 perpendicular to OA and moves under a central force F along an r r0 /(2 − cos θ ). Using Eq. elliptic path defined by the equation= (12.35), show that F is inversely proportional to the square of the distance r from the particle to the center of force O.

SOLUTION u=

1 2 − cos θ , = r r0

d 2u 2 F +u = = 2 r0 mh 2u 2 dθ

Solving for F,

F =

du sin θ , = dθ r0

d 2u cos θ = r0 dθ 2

by Eq. (12.37).

2mh 2u 2 2mh 2 = r0 r0 r 2

Since m, h, and r0 are constants, F is proportional to

1 r2

2 , or inversely proportional to r .

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PROBLEM 12.95 A particle of mass m describes the logarithmic spiral r = r0 ebθ under a central force F directed toward the center of force O. Using Eq. (12.35) show that F is inversely proportional to the cube of the distance r from the particle to O.

SOLUTION u=

1 1 −bθ = e r r0

du b = − e−bθ dθ r0 d 2u b 2 −bθ e = r0 dθ 2

d 2u b 2 + 1 −bθ F = + u = e r0 dθ 2 mh 2u 2 F =

=

(b 2 + 1)mh 2u 2 −bθ e r0

(b 2 + 1)mh 2u 2 (b 2 + 1)mh 2 = r r3

Since b, m, and h are constants, F is proportional to

1 r3

, or inversely proportional to r 3.

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PROBLEM 12.96 A particle of mass m describes the path defined by the equation = r r0 (6 cosθ − 5) under a central force F directed away from the center of force O. Using Eq. (12.35), show that F is inversely proportional to the square of the distance r from the particle to O.

SOLUTION 6 cos θ − 5 r0 6sin θ du = − dθ r0 2 6 cos θ d u = − dθ r0

u=

1 = r

d 2u 5 F + u =− = 2 2 r0 dθ 2 mh u F = −

Solving for F,

Substitute u =

by Eq. (12.35).

5mh 2u 2 r0

1 r

F = −

5mh 2  r0r 2

1 , or inversely proportional to r 2. The minus sign r2 indicates that the force is repulsive, as shown in Fig. P12.96.

Since m, h, and r0 are constants, F is proportional to

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PROBLEM 12.97 A particle of mass m describes the parabola y = x 2 4r0 under a central force F directed toward the center of force C. Using Eq. (12.35) and Eq. (12.37 ') with ε = 1, show that F is inversely proportional to the square of the distance r from the particle to the center of force and that the angular momentum per unit mass h = 2GMr0 .

SOLUTION From Fig. P12.97, But

y=

x2 4r0

= x r sin θ ,

y= r0 − r cosθ

or r0 − r cosθ =

(

)

r 2 sin 2 θ = 4r0

r 2 1 − cos 2 θ

 1 − cos 2 θ  4r0 

 2 0  r + ( cosθ ) r − r0 = 

4r0

Solving the quadratic equation for r, r =

2r0 1 − cos 2 θ

 2   − cosθ ± cos 2 θ − 4  1 − cos θ  4r0   

2r0 θ ± 1) ( − cos= 1 − cos 2 θ since r > 0. Simplifying gives =

r =

2r0 1 + cosθ

   ( −r0 )    

2r0 (1 − cosθ ) 1 − cos 2 θ

1 1 + cosθ or = = u r 2r0

(1)

sin θ cosθ du d 2u = − = − and dθ 2r0 2r0 dθ 2

d 2u + u= dθ 2 Solving for F,

1 = 2r0 mh 2u 2 2r0

F =

Since m, h, and r0 are constants, F is proportional to

F mh 2u 2 F =

mh 2  2r0r 2

1 , or inversely proportional to r 2. r2

1 GM = 2 (1 + ε cosθ ) =u r h GM 1 Comparing with (1) shows that ε = 1 and 2 = 2r0 h By Eq. (12.37 ') ,

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h=

2GMr0 

PROBLEM 12.98 It was observed that during its second flyby of the earth, the Galileo spacecraft had a velocity of 14.1 km/s as it reached its minimum altitude of 303 km above the surface of the earth. Determine the eccentricity of the trajectory of the spacecraft during this portion of its flight.

SOLUTION For earth,= R 6.37 × 106 m r0 = 6.37 × 106 + 303. × 103 = 6.673 × 106 m h = r0v0 = (6.673 × 106 )(14.1 × 103 ) = 94.09 × 109 m 2 /s GM = gR 2 =(9.81)(6.37 × 106 ) 2 =398.06 × 1012 m3 /s 2

1 GM (1 + ε ) = r0 h2

ε 1 +=

h2 = r0GM

(94.09 × 109 ) 2 = 3.33 (6.673 × 106 )(398.06 × 1012 ) = ε 3.33 − 1

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ε = 2.33 

PROBLEM 12.99 It was observed that during the Galileo spacecraft’s first flyby of the earth, its maximum altitude was 600 mi above the surface of the earth. Assuming that the trajectory of the spacecraft was parabolic, determine the maximum velocity of Galileo during its first flyby of the earth.

SOLUTION = R 3960 = mi 20.909 × 106 ft For the earth: GM = gR 2 =(32.2) (20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2

For a parabolic trajectory, ε = 1. 1 GM Eq. (12.39′) := (1 + cos θ ) r h2

1 At θ 0, = = r0

2GM = h2

2GM r0 2v0 2

or = v0

2GM r0

At r0 = 3960 + 600 = 4560 mi = 24.077 × 106 ft, = v0

(2)(14.077 × 1015 ) = 34.196 × 103 ft/s 24.077 × 106

v0 = 6.48 mi/s 

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PROBLEM 12.100 As a space probe approaching the planet Venus on a parabolic trajectory reaches Point A closest to the planet, its velocity is decreased to insert it into a circular orbit. Knowing that the mass and the radius of Venus are 4.87 × 1024 kg and 6052 km, respectively, determine (a) the velocity of the probe as it approaches A, (b) the decrease in velocity required to insert it into the circular orbit.

SOLUTION First note (a)

rA = (6052 + 280) km = 6332 km

From the textbook, the velocity at the point of closest approach on a parabolic trajectory is given by v0 =

2GM r0 1/ 2

Thus,

(v A ) par

 2 × 66.73 × 10−12 m3 /kg ⋅ s 2 × 4.87 × 1024 kg  =  6332 × 103 m   = 10,131.4 m/s

(v A ) par = 10.13 km/s 

or (b)

We have

(v= (v A ) par + ∆v A A )circ

Now

(v A )circ = =

Then

= ∆v A

GM r0 1 2

1 2

Eq. (12.44)

(v A ) par

(v A ) par − (v A ) par

 1  =  − 1 (10.1314 km/s)  2  = −2.97 km/s

| ∆v A | = 2.97 km/s 

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PROBLEM 12.101 It was observed that as the Voyager I spacecraft reached the point of its trajectory closest to the planet Saturn, it was at a distance of 185 × 103 km from the center of the planet and had a velocity of 21.0 km/s. Knowing that Tethys, one of Saturn’s moons, describes a circular orbit of radius 295 × 103 km at a speed of 11.35 km/s, determine the eccentricity of the trajectory of Voyager I on its approach to Saturn.

SOLUTION For a circular orbit,

v=

Eq. (12.44)

GM r

For the orbit of Tethys, GM = rT vT2

For Voyager’s trajectory, we have

1 GM (1 + ε cos θ ) = r h2 where h = r0 v0 At O,

= r r0= ,θ 0

Then

1 GM (1 + ε ) = r0 (r0 v0 )2

or

r0 v02 r v2 −= 1 0 02 − 1 GM rT vT

= ε

2

185 × 103 km  21.0 km/s  = ×  −1 295 × 103 km  11.35 km/s 

ε = 1.147 

or

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PROBLEM 12.102 A satellite describes an elliptic orbit about a planet of mass M. Denoting by r0 and r1 , respectively, the minimum and maximum values of the distance r from the satellite to the center of the planet, derive the relation

1 1 2GM + =2 r0 r1 h where h is the angular momentum per unit mass of the satellite.

SOLUTION Using Eq. (12.39),

1 GM = + C cos θ A rA h2

and

1 GM = + C cos θ B . rB h2

But so that Adding,

θ B =θ A + 180°, cos θ A = − cos θ B . 1 1 1 1 2GM + = + = rA rB r0 r1 h2

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PROBLEM 12.103 A space probe is describing a circular orbit about a planet of radius R. The altitude of the probe above the surface of the planet is α R and its speed is v0. To place the probe in an elliptic orbit which will bring it closer to the planet, its speed is reduced from v0 to b v0 , where b < 1, by firing its engine for a short interval of time. Determine the smallest permissible value of b if the probe is not to crash on the surface of the planet.

SOLUTION GM rA

For the circular orbit,

v0 =

where

rA =R + α R =R(1 + α )

Eq. (12.44),

= GM v02 R(1 + α )

Then

From the solution to Problem 12.102, we have for the elliptic orbit, 1 1 2GM + =2 rA rB h

= h h= rA (v A ) AB A

Now

= [ R(1 + α )]( b v0 ) Then

2v 2 R(1 + α ) 1 1 + =0 R(1 + α ) rB [ R(1 + α ) b v0 ]2 =

2 b R(1 + α ) 2

Now b min corresponds to rB → R. Then

1 1 2 + =2 R(1 + α ) R b min R(1 + α )

b min =

or

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2  2 +α

PROBLEM 12.104 A satellite describes a circular orbit at an altitude of 19 110 km above the surface of the earth. Determine (a) the increase in speed required at point A for the satellite to achieve the escape velocity and enter a parabolic orbit, (b) the decrease in speed required at point A for the satellite to enter an elliptic orbit of minimum altitude 6370 km, (c) the eccentricity ε of the elliptic orbit.

SOLUTION

(

)

For earth, GM = gR 2 = ( 9.81) 6.37 × 106 = 398.06 × 1012 m3/s 2 rA =6370 + 19110 =25480 km =25.48 × 106 m

= vcirc

GM = rA

= vesc

398.06 × 1012 = 3.9525 × 103 m/s 6 25.48 × 10

2GM = rA

= 2 vcirc 5.5897 × 103 m/s

(a) Increase in speed at A. ∆v= vesc − vcirc= 1.637 × 103 m/s

∆ = v 1.637 × 103 m/s 

Elliptical orbit with rB = 6370 + 6370 = 12740 km = 12.74 × 106 m. Using Eq. (12.39), But

and

GM 1 = + C cosθ B . rB h2

θB = θ A + 180°, so that cosθ A = − cosθ B

Adding,

h =

GM 1 = + C cosθ A rA h2

1 1 += rA rB

2GMrArB = rA + rB

rA + rB = rArB

2GM h2

( 2 ) ( 398.06 × 1012 )( 25.48 × 106 )(12.74 × 106 ) 38.22 × 106

= 82.230 × 109 m 2 /s

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PROBLEM 12.104 (Continued) v= A

h 82.230 × 109 = = 3.2272 × 103 m/s 6 rA 25.48 × 10 ∆v =725 m/s 

(b) Decrease in speed. ∆v= vcirc − v A= 725 m/s (c)

1 1 −= rB rA

rA − rB = C cosθ B − C cos = 2C A rArB = C

rA − rB = 2rArB

Ch 2 By Eq. (12.40), = ε = GM

12.74 × 106

( 2 ) ( 25.48 × 10

6

= 19.623 × 10−9 m −1 6 12.74 × 10

)(

)

(19.623 × 10 )(82.230 × 10 ) −9

9

2

398.06 × 1012

ε = 0.333 

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PROBLEM 12.105 A space probe is to be placed in a circular orbit of 5600 mi radius about the planet Venus in a specified plane. As the probe reaches A, the point of its original trajectory closest to Venus, it is inserted in a first elliptic transfer orbit by reducing its speed by Δv A . This orbit brings it to Point B with a much reduced velocity. There the probe is inserted in a second transfer orbit located in the specified plane by changing the direction of its velocity and further reducing its speed by ΔvB . Finally, as the probe reaches Point C, it is inserted in the desired circular orbit by reducing its speed by ΔvC . Knowing that the mass of Venus is 0.82 times the mass of the earth, that = rA 9.3 × 103 mi and = rB 190 × 103 mi, and that the probe approaches A on a parabolic trajectory, determine by how much the velocity of the probe should be reduced (a) at A, (b) at B, (c) at C.

SOLUTION R =3690 mi =20.9088 × 106 ft, g =32.2 ft/s 2

For Earth,

GM earth =gR 2 =(32.2)(20.9088 × 106 ) 2 =14.077 × 1015 ft 3 /s 2 = GM 0.82GM = 11.543 × 1015 ft 3 /s 2 earth

For Venus, For a parabolic trajectory with

rA = 9.3 × 103 mi = 49.104 × 106 ft

(v A= )1 v= esc

First transfer orbit AB.

2GM = rA

(2)(11.543 × 1015 ) = 21.683 × 103 ft/s 49.104 × 106

rB = 190 × 103 mi = 1003.2 × 106 ft

At Point A, where = θ 180°

1 GM GM = + C cos 180 = ° −C 2 2 rA hAB hAB

(1)

1 GM GM =2 + C cos 0 =2 + C rB hAB hAB

(2)

At Point B, where θ = 0°

Adding,

r + rA 2GM 1 1 += B = 2 rA rB rA rB hAB

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PROBLEM 12.105 (Continued)

Solving for hAB , (2)(11.543 × 1015 )(49.104 × 106 )(1003.2 × 106 ) = 1.039575 × 1012 ft 2 /s 6 1052.3 × 10

2GMrA rB = rB + rA

hAB = (v A= )2

hAB 1.039575 × 1012 = = 21.174 × 103 ft/s rA 49.104 × 106

(v= B )1

hAB 1.039575 × 1012 1.03626 × 103 ft/s = = 6 rB 1003.2 × 10 = rC 5600 = mi 29.568 × 106 ft

Second transfer orbit BC. At Point B, where θ = 0

1 GM GM =2 + C cos 0 =2 + C rB hBC hBC At Point C, where = θ 180°

1 GM GM = + C cos 180 = ° −C 2 2 rC hBC hBC Adding,

1 1 r + rC 2GM += B = 2 rB rC rB rC hBC = hBC

2GMrB rC = rB + rC

(2)(11.543 × 1015 )(1003.2 × 106 )(29.568 × 106 ) = 814.278 × 109 ft 2 /s 1032.768 × 106

(vB= )2

hBC 814.278 × 109 = = 811.69 ft/s rB 1003.2 × 106

(v= C )1

hBC 814.278 × 109 = = 27.539 × 103 ft/s rC 29.568 × 106

Final circular orbit.

= rC 29.568 × 106 ft

(vC ) 2 =

GM = rC

11.543 × 1015 = 19.758 × 103 ft/s 29.568 × 106

Speed reductions. (a)

At A:

(v A )1 − (v A )= 21.683 × 103 − 21.174 × 103 2

∆v A = 509 ft/s 

(b)

At B:

(vB )1 − (vB ) 2 = 1.036 × 103 − 811.69

∆vB = 224 ft/s 

(c)

At C:

(vC )1 − (vC )= 27.539 × 103 − 19.758 × 103 2

∆vC = 7.78 × 103 ft/s 

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PROBLEM 12.106 For the space probe of Problem 12.105, it is known that r= 9.3 × 103 mi and that the velocity of the probe is A reduced to 20,000 ft/s as it passes through A. Determine (a) the distance from the center of Venus to Point B, (b) the amounts by which the velocity of the probe should be reduced at B and C, respectively.

SOLUTION Data from Problem 12.105:

rC = 29.568 × 106 ft, M = 0.82 M earth

For Earth,

R =3960 mi =20.9088 × 106 ft, g =32.2 ft/s 2

GM earth =gR 2 =(32.2)(20.9088 × 106 ) 2 =14.077 × 1015 m3 /s 2 = GM 0.82GM = 11.543 × 1015 ft 3 /s 2 earth

For Venus,

vA = 20, 000 ft/s, rA = 9.3 × 103 mi = 49.104 × 106 ft

Transfer orbit AB:

hAB = rA v A =× (49.104 106 )(20, 000) = 982.08 × 109 ft 2 /s

At Point A, where = θ 180°

1 GM GM = + C cos 180 = ° −C 2 2 rA hAB hAB At Point B, where θ = 0°

1 GM GM =2 + C cos 0 =2 + C rB hAB hAB Adding,

1 1 2GM + =2 rA rB hAB 1 2GM 1 = − 2 rB rA hAB =

(2)(11.543 × 1015 ) 1 − 9 2 (982.08 × 10 ) 49.104 × 106

= 3.57125 × 10−9 ft −1

(a)

Radial coordinate rB . = rB 280.01 × 106 ft

(v= B )1

= rB 53.0 × 103 mi 

hAB 982.08 × 109 = = 3.5073 × 103 ft/s 6 rB 280.01 × 10

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PROBLEM 12.106 (Continued)

Second transfer orbit BC. = rC 5600 = mi 29.568 × 106 ft At Point B, where θ = 0

1 GM GM =2 + C cos 0 =2 + C rB hBC hBC At Point C, where = θ 180°

1 GM GM = + C cos 180 = ° −C 2 2 rC hBC hBC r + rC 2GM 1 1 += B = 2 rB rC rB rC hBC

Adding,

hBC =

2GMrB rC = rB + rC

(2)(11.543 × 1015 )(280.01 × 106 )(29.568 × 106 ) 309.578 × 106

= 785.755 × 109 ft 2 /s

Circular orbit with

(vB= )2

hBC 785.755 × 109 = = 2.8062 × 103 ft/s rB 280.01 × 106

(v= C )1

hBC 785.755 × 109 = = 26.575 × 103 ft/s rC 29.568 × 106

= rC 29.568 × 106 ft = (vC ) 2

(b)

GM = rC

11.543 × 1015 = 19.758 × 103 ft/s 29.568 × 106

Speed reductions at B and C. At B:

(vB )1 − (vB )= 3.5073 × 103 − 2.8062 × 103 2

∆vB = 701 ft/s  At C:

(vC )1 − (vC )= 26.575 × 103 − 19.758 × 103 2 ∆vC = 6.82 × 103 ft/s 

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PROBLEM 12.107 As it describes an elliptic orbit about the sun, a spacecraft reaches a maximum distance of 202 × 106 mi from the center of the sun at Point A (called the aphelion) and a minimum distance of 92 × 106 mi at Point B (called the perihelion). To place the spacecraft in a smaller elliptic orbit with aphelion at A′ and perihelion at B′, where A′ and B′ are located 164.5 × 106 mi and 85.5 × 106 mi, respectively, from the center of the sun, the speed of the spacecraft is first reduced as it passes through A and then is further reduced as it passes through B′. Knowing that the mass of the sun is 332.8 × 103 times the mass of the earth, determine (a) the speed of the spacecraft at A, (b) the amounts by which the speed of the spacecraft should be reduced at A and B′ to insert it into the desired elliptic orbit.

SOLUTION First note

Rearth 3960 mi 20.9088 × 106 ft = = 202 × 106 mi = 1066.56 × 109 ft rA = rB = 92 × 106 mi = 485.76 × 109 ft

From the solution to Problem 12.102, we have for any elliptic orbit about the sun 1 1 2GM sun + = 2 r1 r2 h

(a)

For the elliptic orbit AB, we have

= r1 rA , r= rB , = h h= rA v A 2 A Also,

= GM sun G[(332.8 × 103 ) M earth ] 2 = gRearth (332.8 × 103 )

Then

using Eq. (12.30).

1 1 2 gR 2 (332.8 × 103 ) + = earth rA rB (rA v A ) 2 1/ 2

or

R v A = earth rA

 665.6 g × 103    1 1   + rA rB  

3960 mi  665.6 × 103 × 32.2 ft/s 2 = 202 × 106 mi  1066.561× 109 ft + 485.761× 109 ft  = 52, 431 ft/s or

1/ 2

   

= v A 52.4 × 103 ft/s 

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PROBLEM 12.107 (Continued)

(b)

From Part (a), we have 1 1 = 2GM sun (rA v A ) 2  +   rA rB 

Then, for any other elliptic orbit about the sun, we have

(

2 1 1 1 (rA v A ) rA + + = r1 r2 h2

1 rB

)

For the elliptic transfer orbit AB′, we have

= r1 rA , r= rB′ , = h h= rA (v A ) tr 2 tr Then

(

2 1 1 1 1 (rA v A ) rA + rB + = rA rB′ [rA (v A ) tr ]2

)

1/ 2

or

 1 + rA  r1 + r1  rB A B    (v A ) tr v= v = A A  r1 + r1   1 + rA rB′  A B′  

1/ 2

   

1/ 2

 1 + 202  92 = (52, 431 ft/s)   1 + 202  85.5   = 51,113 ft/s

Now Then

= htr (= hA ) tr (hB′ ) tr : rA (= v A ) tr rB′ (vB′ ) tr (vB′ ) tr=

202 × 106 mi × 51,113 ft/s= 120,758 ft/s 85.5 × 106 mi

For the elliptic orbit A′B′, we have

= r1 rA= rB= ′ , r2 ′ , h rB′ vB′ Then

or

(

2 1 1 1 (rA v A ) rA + + = rA′ rB′ (rB′ vB′ ) 2

r  vB ′ = v A A  rB′  

1 rB

1 rA

+

1 rB

1 rA′

+

1 rB′

) 1/ 2

   

1 1 202 × 106 mi  202 × 106 + 92 × 106 = (52, 431 ft/s) 85.5 × 106 mi  164.51× 106 + 85.51× 106  = 116,862 ft/s

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1/ 2

   

PROBLEM 12.107 (Continued)

Finally, or

(v A ) tr= v A + ∆v A ∆= v A (51,113 − 52, 431) ft/s |∆v A | = 1318 ft/s 

or and

= vB′ (vB′ ) tr + ∆vB

or

= ∆vB′ (116,862 − 120, 758) ft/s = −3896 ft/s

|∆vB | = 3900 ft/s 

or

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PROBLEM 12.108 Halley’s comet travels in an elongated elliptic orbit for which the minimum distance from the sun is approximately 12 rE , where = rE 150 × 106 km is the mean distance from the sun to the earth. Knowing that the periodic time of Halley’s comet is about 76 years, determine the maximum distance from the sun reached by the comet.

SOLUTION We apply Kepler’s Third Law to the orbits and periodic times of earth and Halley’s comet: 2

τH   aH    =   τE   aE  Thus

3

τ  aH = aE  H   τE 

2/3

 76 years  = rE    1 year  = 17.94rE

But

2/3

1 (rmin + rmax ) 2 11  = rE rE + rmax  17.94  2 2  = aH

1 rE 2 = (35.88 − 0.5)rE

= rmax 2(17.94 rE ) − = 35.38 rE

= rmax (35.38)(150 × 106 km) rmax = 5.31 × 109 km 

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PROBLEM 12.109 Based on observations made during the 1996 sighting of comet Hyakutake, it was concluded that the trajectory of the comet is a highly elongated ellipse for which the eccentricity is approximately ε = 0.999887. Knowing that for the 1996 sighting the minimum distance between the comet and the sun was 0.230 RE , where RE is the mean distance from the sun to the earth, determine the periodic time of the comet.

SOLUTION For Earth’s orbit about the sun, 2π RE 2π RE 3/ 2 GM = , τ0 = RE v0 GM

= v0

or

= GM

2π RE3/ 2

τ0

(1)

For the comet Hyakutake, 1 GM = = (1 + ε ), r0 h2 a= = h = τ =

r0 1 (r0 + r1 )= , b= 2 1− ε

r0 r1 =

1+ ε r0 1− ε

GMr0 (1 + ε ) 2π r02 (1 + ε )1/ 2 2π ab = h (1 − ε )3/ 2 GMr0 (1 + ε ) 2π r03/ 2 2π r03/ 2τ 0 = 3 3/ 2 GM (1 − ε )3/ 2 2π RE (1 − ε )

 r  = 0   RE 

3/ 2

1 (1 − ε )3/ 2

= (0.230)3/ 2

Since

1 GM 1+ ε = (1 + ε ), r1 = r0 2 1− ε r1 h

τ0

1 τ 0 91.8 × 103τ 0 = 3/ 2 (1 − 0.999887)

= τ 0 1 yr, = τ (91.8 × 103 )(1.000)

= τ 91.8 × 103 yr 

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PROBLEM 12.110 A space probe is to be placed in a circular orbit of radius 4000 km about the planet Mars. As the probe reaches A, the point of its original trajectory closest to Mars, it is inserted into a first elliptic transfer orbit by reducing its speed. This orbit brings it to Point B with a much reduced velocity. There the probe is inserted into a second transfer orbit by further reducing its speed. Knowing that the mass of Mars is 0.1074 times the mass of the earth, that rA = 9000 km and rB = 180,000 km, and that the probe approaches A on a parabolic trajectory, determine the time needed for the space probe to travel from A to B on its first transfer orbit.

SOLUTION For earth,= R 6373= km 6.373 × 106 m GM = gR 2 =(9.81) (6.373 × 106 ) 2 =398.43 × 1012 m3 /s 2

For Mars, GM = (0.1074)(398.43 × 1012 ) = 42.792 × 1012 m3 /s 2 = rA 9000 km = 9.0 × 106 m

= rB 180000 km = 180 × 106 m

For the parabolic approach trajectory at A,

2GM = rA

= (v A )1

(2)(42.792 × 1012 ) = 3.0837 × 103 m/s 9.0 × 106

First elliptic transfer orbit AB.

1 GM = + C cosθ A 2 rA hAB

Using Eq. (12.39), But

θB = θ A + 180°,

Adding,

1 1 += rA rB

= hAB

1 GM = + C cosθ B . 2 rB hAB

cosθ A = − cosθ B .

so that

rA + rB = rArB

and

2GM 2 hAB

2GMrArB = rA + rB

(2)(42.792 × 1012 )(9.0 × 106 )(180 × 106 ) 189.0 × 106

= hAB 27.085 × 109 m 2 /s a= b=

1 1 (rA + rB ) = (9.0 × 106 + 180 × 106 ) = 94.5 × 106 m 2 2 rArB =

(9.0 × 106 )(180 × 106 ) = 40.249 × 106 m

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PROBLEM 12.110 (Continued)

Periodic time for full ellipse: For half ellipse AB,

τ= AB = τ AB

τ =

2π ab h

1 π ab τ = 2 h

π (94.5 × 106 )(40.249 × 106 ) 27.085 × 109

= 444.81 × 103 s

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τ AB = 122.6 h 

PROBLEM 12.111 A spacecraft and a satellite are at diametrically opposite positions in the same circular orbit of altitude 500 km above the earth. As it passes through point A, the spacecraft fires its engine for a short interval of time to increase its speed and enter an elliptic orbit. Knowing that the spacecraft returns to A at the same time the satellite reaches A after completing one and a half orbits, determine (a) the increase in speed required, (b) the periodic time for the elliptic orbit.

SOLUTION g = 9.81 m/s 2

For earth,= R 6.37 × 106 m ,

(

GM =gR 2 =( 9.81) 6.37 × 106

)

2

=398.06 × 1012 m3/s 2

For circular orbit of satellite, r0 = 6370 + 500 = 6870 km = 6.87 × 106 m/s

= v0

= τ0

398.06 × 1012 = 7.6119 × 103 m/s 6.87 × 106

GM = r0

2π r0 = v0

( 2π ) ( 6.87 × 106 )

= 5.6708 × 103 s 7.6119 × 103

For elliptic orbit of spacecraft it is given that 3 = τ 0 8.5062 × 103 s 2

= τ

1 a= ( rA + rB ) , 2

Using Eq. (12.39), But

θB = θ A + 180°,

Adding,

so that

GM τ 2 = a = 4π 2

2π ab = h

2π ab a = GMb 2

or

4π 2

h=

GMb 2 a

2π a3 / 2 GM

10 ) (398.06 × 10 )(8.5062 ×= 12

1 GM = + C cosθ B . rB h2

cosθ A = − cosθ B .

1 1 r + rB 2a 2GM + = A = 2 = rA rB rArB b h2

Periodic time: τ =

3

1 GM and = + C cosθ A rA h2

b= rArB

3

2

729.558 × 1018 m3

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PROBLEM 12.111 (Continued)

a= 9.0023 × 106 m,

rA = r0 = 6.87 × 106 m

rB = 2a − rA = 11.1346 × 106 m,

h =

2π ab =

τ

(

)(

b=

rArB = 8.7461 × 106 m

)

2π 9.0023 × 106 8.7461 × 106 = 58.159 × 109 m 2 /s 3 8.5062 × 10

v= A

h 58.159 × 109 = = 8.4656 × 103 m/s 6 rA 6.87 × 10

(a) Increase in speed at A. ∆v A = v A − v0 = 8.4656 × 103 − 7.6119 × 103

∆v A = 854 m/s 

(b) Periodic time for elliptic orbit. As calculated above τ = 8.5062 s

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τ = 141.8 min 

PROBLEM 12.112 The Clementine spacecraft described an elliptic orbit of minimum altitude hA = 400 km and a maximum altitude of hB = 2940 km above the surface of the moon. Knowing that the radius of the moon is 1737 km and that the mass of the moon is 0.01230 times the mass of the earth, determine the periodic time of the spacecraft.

SOLUTION For earth,

= R 6370= km 6.370 × 106 m GM =gR 2 =(9.81)(6.370 × 106 ) 2 =398.06 × 1012 m3 /s 2

For moon,

GM = (0.01230)(398.06 × 1012 ) = 4.896 × 1012 m3 /s 2 rA = 1737 + 400 = 2137 km = 2.137 × 106 m

rB =1737 + 2940 = 4677 km = 4.677 × 106 m

Using Eq. (12.39),

1 GM = + C cos θ A rA h2

But

θ B =+ θ A 180°, so that cos θ A = − cos θ B .

Adding,

and

1 GM = + C cos θ B . rB h2

1 1 rA + rB 2GM += = 2 rA rB rA rB hAB = hAB

2GMrA rB = rA + rB

(2)(4.896 × 1012 )(2.137 × 106 )(4.677 × 106 ) 6.814 × 106

= 3.78983 × 109 m 2 /s 1 (rA + rB )= 3.402 × 106 m 2 = b = rA rB 3.16145 × 106 m a=

Periodic time.

τ =

2π ab 2π (3.402 × 106 )(3.16145 × 106 ) = = 17.831 × 103 s 9 hAB 3.78983 × 10

τ = 4.95 h 

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PROBLEM 12.113 Determine the time needed for the space probe of Problem 12.100 to travel from B to C.

SOLUTION From the solution to Problem 12.100, we have (v A ) par = 10,131.4 m/s

= (v A )circ

and

1 = (v A ) par 7164.0 m/s 2

rA = (6052 + 280) km = 6332 km

Also, For the parabolic trajectory BA, we have

1 GM v (1 + ε cos θ ) = 2 r hBA

[Eq. (12.39′)]

where ε = 1. Now

1 GM v (1 + 1) = 2 rA hBA

at A, θ = 0:

rA =

or at B, θ = −90°:

2 hBA 2GM v

1 GM v = (1 + 0) 2 rB hBA rB =

or

2 hBA GM v

rB = 2rA

As the probe travels from B to A, the area swept out is the semiparabolic area defined by Vertex A and Point B. Thus, (Area swept out)= A= BA BA

Now

2 4 2 rA= rB rA 3 3

dA 1 = h dt 2

where h = constant

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PROBLEM 12.113 (Continued)

Then

= A

2 ABA 1 = ht or t BA = hBA rA v A 2 hBA = t BA

2 × 43 rA2 8 rA = 3 vA rA v A

8 6332 × 103 m 3 10,131.4 m/s = 1666.63 s =

For the circular trajectory AC, t AC =

Finally,

π r π 6332 × 103 m 2 A = = 1388.37 s (v A )circ 2 7164.0 m/s

t BC = t BA + t AC = (1666.63 + 1388.37) s =3055.0 s

or

t BC = 50 min 55 s 

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PROBLEM 12.114 A space probe is describing a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O. As the probe passes through Point A, its velocity is reduced from v0 to b v0, where b < 1, to place the probe on a crash trajectory. Express in terms of n and b the angle AOB, where B denotes the point of impact of the probe on the planet.

SOLUTION r= r= nR 0 A

For the circular orbit,

v0 =

GM = r0

GM nR

The crash trajectory is elliptic. = v A b= v0

b 2GM nR

= = h r= nRv AvA A

b 2 nGMR

GM 1 = 2 2 h b nR 1 GM 1 + ε cos θ = 2 (1 + ε cos θ ) = 2 r b nR h At Point A, = θ 180°

1 1 1− ε = = or b 2 = 1 − ε or ε = 1− b 2 2 rA nR b nR At impact Point B, θ= π − φ

1 1 = rB R 1 1 + ε cos (π − φ ) 1 − ε cos φ = = R b 2 nR b 2 nR 1 − nb 2 1 − nb 2 ε cos φ = = 1 − nb 2 or cos φ = ε 1− b 2

φ = cos −1[(1 − nb 2 ) /(1 − b 2 )] 

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PROBLEM 12.115 A long-range ballistic trajectory between Points A and B on the earth’s surface consists of a portion of an ellipse with the apogee at Point C. Knowing that Point C is 1500 km above the surface of the earth and the range Rφ of the trajectory is 6000 km, determine (a) the velocity of the projectile at C, (b) the eccentricity ε of the trajectory.

SOLUTION For earth,= R 6370 = km 6.37 × 106 m GM = gR 2 =(9.81)(6.37 × 106 ) 2 =398.06 × 1012 m3 /s 2

For the trajectory, rC = 6370 + 1500 = 7870 km = 7.87 × 106 m rA = rB = R= 6.37 × 106 m,

rC 7870 = = 1.23548 rA 6370

Range A to B: = s AB 6000 = km 6.00 × 106 m

ϕ =

s AB = R

6.00 × 106 rad 53.968° = 0.94192 = 6.37 × 106

1 GM For an elliptic trajectory,= (1 + ε cos θ ) r h2

At A,

ϕ θ= 180° − = 153.016°,

At C,

1 = θ 180° , = rC

2

1 GM = (1 + ε cos153.016°) rA h2

GM (1 − ε ) h2

Dividing Eq. (1) by Eq. (2), rC 1 + ε cos153.016° = = 1.23548 1−ε rA

ε =

1.23548 − 1 = 0.68384 1.23548 + cos153.016°

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(1) (2)

PROBLEM 12.115 (Continued)

From Eq. = (2), h h=

GM (1 − ε )rC

(398.06 × 1012 )(0.31616)(7.87 × 106 ) = 31.471 × 109 m 2 /s

v= C

(a) Velocity at C. (b)

h 31.471 × 109 = = 4.00 × 103 m/s 6 rC 7.87 × 10

Eccentricity of trajectory.

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vC = 4 km/s 

ε = 0.684 

PROBLEM 12.116 A space shuttle is describing a circular orbit at an altitude of 563 km above the surface of the earth. As it passes through Point A, it fires its engine for a short interval of time to reduce its speed by 152 m/s and begin its descent toward the earth. Determine the angle AOB so that the altitude of the shuttle at Point B is 121 km. (Hint: Point A is the apogee of the elliptic descent trajectory.)

SOLUTION GM = gR 2 =(9.81)(6.37 × 106 ) 2 =398.06 × 1012 m3 /s 2 rA = 6370 + 563 = 6933 km = 6.933 × 106 m rB = 6370 + 121 = 6491 km = 6.491 × 106 m For the circular orbit through Point A, 398.06 × 1012 = 7.5773 × 103 m/s 6.933 × 106

GM = rA

= vcirc

For the descent trajectory,

= v A vcirc = + ∆v 7.5773 × 103 −= 152 7.4253 × 103 m/s h = rAv A = (6.933 × 106 )(7.4253 × 103 ) = 51.4795 × 109 m 2 /s 1 = r

At Point A,

= θ 180°,

GM (1 + ε cos θ ) h2

r = rA 1 GM = (1 − ε ) rA h2

1= −ε

h2 = GM rA

(51.4795 × 109 ) 2 = 0.96028 (398.06 × 1012 )(6.933 × 106 )

ε = 0.03972 1 GM = (1 + ε cos θ B ) rB h2

1 + ε cos θ B =

h2 (51.4795 × 109 ) 2 = = 1.02567 GM rB (398.06 × 1012 )(6.491 × 106 )

= cosθ B

= θ B 49.7°

1.02567 − 1 = 0.6463

ε

AOB = 180° − θ = 130.3° B

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AOB = 130.3° 

PROBLEM 12.117 As a spacecraft approaches the planet Jupiter, it releases a probe which is to enter the planet’s atmosphere at Point B at an altitude of 280 mi above the surface of the planet. The trajectory of the probe is a hyperbola of eccentricity ε = 1.031. Knowing that the radius and the mass of Jupiter are 44423 mi and 1.30 × 1026 slug, respectively, and that the velocity vB of the probe at B forms an angle of 82.9° with the direction of OA, determine (a) the angle AOB, (b) the speed vB of the probe at B.

SOLUTION r= (44.423 × 103 + 280) mi = 44.703 × 103 mi B

First we note (a)

We have

1 GM j = (1 + ε cos θ ) r h2

At A, θ = 0:

1 GM j (1 + ε ) = rA h2 h2 = rA (1 + ε ) GM j

or

θ θ=  AOB : At B, = B or Then or

[Eq. (12.39′)]

1 GM j = (1 + ε cos θ B ) rB h2 h2 = rB (1 + ε cos θ B ) GM j

rA (1 + ε ) = rB (1 + ε cos θ B ) cos= θB

 1  rA  (1 + ε ) − 1 ε  rB   1  44.0 × 103 mi (1 + 1.031) − 1  3 1.031  44.703 × 10 mi 

=

= 0.96902

or

= θ B 14.2988°

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 AOB = 14.30° 

PROBLEM 12.117 (Continued)

(b)

From above where

Then

= h 2 GM j rB (1 + ε cos θ B )

1 | rB × mv B | = rB vB sin φ m ) 97.1988° φ= (θ B + 82.9°= h=

(rB vB= sin φ ) 2 GM j rB (1 + ε cos θ B )

or 1/2

 1  GM j (1 + ε cos θ B )  vB =  sin φ  rB 

1/2

 34.4 × 10−9 ft 4 /lb ⋅ s 4 × (1.30 × 1026 slug)  1 × [1 + (1.031)(0.96902)]  6 sin 97.1988°  236.03 × 10 ft 

vB = 196.2 ft/s 

or

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PROBLEM 12.118 A satellite describes an elliptic orbit about a planet. Denoting by r0 and r1 the distances corresponding, respectively, to the perigee and apogee of the orbit, show that the curvature of the orbit at each of these two points can be expressed as 1 1 1 1  =  +  ρ 2  r0 r1 

SOLUTION Using Eq. (12.39),

1 GM = + C cos θ A rA h2

and

1 GM = + C cos θ B . rB h2

But so that Adding,

θ B =θ A + 180°, cos θ A = − cos θ B 1 1 2GM + =2 rA rB h

At Points A and B, the radial direction is normal to the path.

a= n

But

= Fn

v2 =

ρ

h2 r 2ρ

GMm mh 2 = = ma n r2 r 2ρ

1 GM 1  1 1 = =  +  2 ρ 2  rA rB  h

1 1 1 1  =  +   ρ 2  r0 r1 

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PROBLEM 12.119 (a) Express the eccentricity ε of the elliptic orbit described by a satellite about a planet in terms of the distances r0 and r1 corresponding, respectively, to the perigee and apogee of the orbit. (b) Use the result obtained in Part a and the data given in Problem RE 149.6 × 106 km, to determine the approximate 12.109, where = maximum distance from the sun reached by comet Hyakutake.

SOLUTION (a)

We have

1 GM = (1 + ε cos θ ) r h2

At A, θ = 0:

1 GM = (1 + ε ) r0 h2 h2 = r0 (1 + ε ) GM

or At B, = θ 180°:

or Then

Eq. (12.39′)

1 GM (1 − ε ) = r1 h2 h2 = r1 (1 − ε ) GM

r0 (1 + ε ) = r1 (1 − ε )

ε=

or (b)

r1 − r0  r1 + r0

1+ ε r0 1− ε

From above,

r1 =

where

r0 = 0.230 RE

Then

r1 =

1 + 0.999887 × 0.230(149.6 × 109 m) 1 − 0.999887

= r1 609 × 1012 m 

or

= = RE or r1 0.064 lightyears. Note: r1 4070

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PROBLEM 12.120 Derive Kepler’s third law of planetary motion from Eqs. (12.37) and (12.43).

SOLUTION For an ellipse,

2a = rA + rB and b = rA rB

Using Eq. (12.37),

1 GM = + C cos θ A rA h2

and

1 GM = + C cos θ B . rB h2

But so that Adding,

θ B =θ A + 180°, cos θ A = − cos θ B . 1 1 r +r 2a 2GM + = A B = 2 = rA rB rA rB b h2 h=b

GM a

By Eq. (12.43), = τ

τ2 =

2π ab 2π ab a 2π a 3/ 2 = = h b GM GM 4π 2 a3 GM

For Orbits 1 and 2 about the same large mass,

and

τ12 =

4π 2 a13 GM

τ 22 =

4π 2 a23 GM 2

3

 τ1   a1    =    τ 2   a2 

Forming the ratio,

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PROBLEM 12.121 Show that the angular momentum per unit mass h of a satellite describing an elliptic orbit of semimajor axis a and eccentricity ε about a planet of mass M can be expressed as = h

GMa (1 − ε 2 )

SOLUTION By Eq. (12.39′),

1 GM = (1 + ε cos θ ) r h2

At A, θ = 0°:

1 GM = = (1 + ε ) or rA h2

h2 rA = GM (1 + ε )

θ 180°: At B, =

1 GM (1 − ε ) or = = rB h2

h2 rB = GM (1 − ε )

h2 1  1 =  + GM  1 + ε 1 − ε

Adding,

rA + rB=

But for an ellipse,

rA + rB = 2a 2a =

2h 2 GM (1 − ε 2 )

2

2h  = 2  GM (1 − ε )

= h

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GMa (1 − ε 2 ) 

PROBLEM 12.122 In the braking test of a sports car its velocity is reduced from 70 mi/h to zero in a distance of 170 ft with slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static friction, determine (a) the coefficient of static friction, (b) the stopping distance for the same initial velocity if the car skids. Ignore air resistance and rolling resistance.

SOLUTION (a)

Coefficient of static friction.

= ΣFy 0:

N= −W 0

N =W

= v0 70 = mi/h 102.667 ft/s v 2 v02 − = at (s − s0 ) 2 2

at =

v 2 − v02 0 − (102.667)2 = = − 31.001 ft/s 2 2(s − s0 ) (2)(170)

= m | at | For braking without skidding µ µ= s , so that µ s N Σ= Ft mat : − µ s= N mat

µs = − (b)

mat a 31.001 = − t = W g 32.2

µ s = 0.963 

Stopping distance with skidding.

µ µ= (0.80)(0.963) = 0.770 Use = k ΣF =mat : µk N =−mat

µ N − k = −µk g = − 24.801 ft/s 2 at = m Since acceleration is constant,

(s − = s0 )

v 2 − v02 = 2at

0 − (102.667)2 (2)(− 24.801)

s − s0 = 212 ft 

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PROBLEM 12.123 A bucket is attached to a rope of length L = 1.2 m and is made to revolve in a horizontal circle. Drops of water leaking from the bucket fall and strike the floor along the perimeter of a circle of radius a. Determine the radius a when θ= 30°.

SOLUTION Initial velocity of drop = velocity of bucket

= ΣFy 0: = T cos 30° mg

(1)

= ΣFx max= : T sin 30° man

(2)

Divide (2) by (1):

tan 30= °

an = g

v2 pg

v 2 ρ g tan 30° Thus = But

= ρ L sin = 30° (1.2 m) sin= 30° 0.6 m

Thus = v 2 0.6(9.81) = tan 30° 3.398 m 2 /s 2

v = 1.843 m/s

Assuming the bucket to rotate clockwise (when viewed from above), and using the axes shown, we find that the components of the initial velocity of the drop are

= (v0 ) x 0,= (v0 ) y 0, = (v0 )2 1.843 m/s Free fall of drop y = y0 + (v0 ) y t −

1 2 gt 2

y = y0 −

1 2 gt 2

When drop strikes floor: y= 0

y0 −

1 2 gt = 0 2

But

y= 2L − L cos30= ° 2(1.2) − 1.2cos30= ° 1.361 m 0

Thus

1.361 −

1 (9.81) t 2 =0 2

t =0.5275

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PROBLEM 12.123 (Continued)

Projection on horizontal floor (uniform motion)

x= x0 + (v0 ) z = t L sin 30° + 0,

x= 0.6 m

z = z0 + (v0 ) z t = 0 + 1.843(0.527) = 0.971 m Radius of circle:= a = a

x2 + z 2

(0.6) 2 + (0.971) 2

a = 1.141 m 

Note: The drop travels in a vertical plane parallel to the yz plane.

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PROBLEM 12.124 A 12-lb block B rests as shown on the upper surface of a 30-lb wedge A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of A, (b) the acceleration of B relative to A.

SOLUTION Acceleration vectors:

a A = aA

30°, a B/A = aB/A

a= a A + a B/ A B Block B:

= ΣFx max : mB aB /A − m = B a A cos30° 0 = aB/A a A cos30°

(1)

ΣFy = ma y : N AB − WB = −mB a A sin 30° N AB = WB − (WB sin 30°)

Block A:

aA g

(2)

= ΣF ma : WA sin 30° = + N AB sin 30° WA WA sin 30° + WB sin 30° − (WB sin 2 30°)

= aA

aA g

aA a =WA A g g

(WA + WB )sin 30° (30 + 12)sin 30° = = (32.2) 20.49 ft/s 2 g 2 30 + 12sin 2 30° WA + WB sin 30°

a A = 20.49 ft/s 2

(a)

30° 

= aB/ A (20.49) = cos 30° 17.75 ft/s 2

(b)

a B/A = 17.75 ft/s 2

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

PROBLEM 12.125 A 500-lb crate B is suspended from a cable attached to a 40-lb trolley A which rides on an inclined I-beam as shown. Knowing that at the instant shown the trolley has an acceleration of 1.2 ft/s2 up and to the right, determine (a) the acceleration of B relative to A, (b) the tension in cable CD.

SOLUTION (a)

a A + a B/A , where a B/A is directed perpendicular to cable AB. First we note: a= B ΣFx = mB ax : 0 = −mB ax + mB a A cos 25°

B:

= aB/A (1.2 ft/s 2 ) cos 25°

or

a B/A = 1.088 ft/s 2

or (b)



For crate B = ΣFy mB a y : T= AB − WB

WB a A sin 25° g

A:

 (1.2 ft/s 2 )sin 25°  = TAB (500 lb) 1 +  32.2 ft/s 2   = 507.87 lb

or

For trolley A = ΣFx mA a A : TCD − TAB sin 25° −= WA sin 25°

or

WA aA g

 1.2 ft/s 2 = TCD (507.87 lb)sin 25° + (40 lb)  sin 25° + 32.2 ft/s 2 

  

TCD = 233 lb 

or

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PROBLEM 12.126 The roller-coaster track shown is contained in a vertical plane. The portion of track between A and B is straight and horizontal, while the portions to the left of A and to the right of B have radii of curvature as indicated. A car is traveling at a speed of 72 km/h when the brakes are suddenly applied, causing the wheels of the car to slide on the track (µk = 0.25). Determine the initial deceleration of the car if the brakes are applied as the car (a) has almost reached A, (b) is traveling between A and B, (c) has just passed B.

SOLUTION = v 72 = km/h 20 m/s

(a)

Almost reached Point A.

ρ = 30 m a= n

v 2 (20)2 = = 13.333 m/s 2 30 ρ

= ΣFy ma y : N R + N F= − mg man N R + N F = m( g + an ) F= µk ( N R + N F )= µk m( g + an )

Σ= Fx max : −= F mat F − = − µk ( g + an ) at = m

| at |= µk ( g + an )= 0.25(9.81 + 13.33) (b)

Between A and B.

| at | = 5.79 m/s 2 

ρ= ∞ an = 0

= | at | µ= (0.25)(9.81) kg (c)

Just passed Point B.

ρ = 45 m a= n

v 2 (20) 2 = = 8.8889 m/s 2 45 ρ

ΣFy = ma y : N R + N F − mg = −man

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| at | = 2.45 m/s 2 

PROBLEM 12.126 (Continued)

or

N R + N F = m( g − an ) F= µk ( N R + N F )= µk m( g − an )

Σ= Fx max : −= F mat F − = − µk ( g − an ) at = m

| at |= µk ( g − an )= (0.25)(9.81 − 8.8889) | at | = 0.230 m/s 2 

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PROBLEM 12.127 The parasailing system shown uses a winch to pull the rider in towards the boat, which is travelling with a constant velocity. During the interval when θ is between 20º and 40º, (where t = 0 at θ = 20º) the angle increases at the constant rate of 2 º/s. During this time, the length of the rope is defined by the

1 3

relationship = r 125 − t 3/2 , where r and t are expressed in meters and seconds, respectively. At the instant when the rope makes a 30 degree angle with the water, the tension in the rope is 18 kN. At this instant, what is the magnitude and direction of the force of the parasail on the 75 kg parasailor?

SOLUTION 1 3 = r 125 − t 2 m 3 30°, θ =° 2 /s= 0.0349 rad/s, θ=0 θ0 =20°, θ =

Given:

= = T 18 kN, m 75 kg

θ= θ 0 + θt 30 = 20 + 2t ⇒ t = 5 s

Kinematics:

a= aB + aP/ B P aP = 0 + ar e r + aθ eθ

Using Radial and Transverse Coordinates:

Free Body Diagram of parasailor (side view):

1 12 r = − t m/s 2 1 − 12  r = − t m/s 2 4 Equations of Motion:

∑F

r

= mar

(

FP cos b − T − mg sin θ= m  r − rθ 2 FP cos b = T + m  g sin θ +  r − rθ 2 

∑ Fθ = maθ

)

(

FP sin b − mg cos θ = m rθ + 2rθ

(1)

FP= sin b m  g cos θ + rθ + 2rθ 

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) (2)

PROBLEM 12.127 (Continued)

r= 121.27 m, r = −1.118 m/s,  r= −0.1118 m/s 2

Evaluate at t=5 s: (1)

2 FP cos b 18000 + 75 9.81sin 30° − 0.1118 − 121.27 ( 0.0349 )  ⇒ FP cos b 18348.4 N = =  

(2) = FP sin b 75 9.81cos 30° + 0 + 2 ( −1.118 )( 0.0349 )  = ⇒ FP sin b 631.33 N

FP sin b FP cos b

FP cos1.97° =18348.4 N

=

631.33 ⇒ tan b = 0.03441 ⇒ b = 1.97° 18348.4

FP = 18.4 kN

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31.97° 

PROBLEM 12.128 A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) θ= 90°, (b) θ= 75°, (c) θ= 45°. Indicate in each case the direction of the impending motion.

SOLUTION vC = (r sin θ )φAB

First note

= (0.6 m)(6 rad/s)sin θ = (3.6 m/s)sin θ

(a)

With θ= 90°,

vC = 3.6 m/s ΣF= 0: F − W= 0 y C

or

F = mC g

Now

F = µs N

or

N=

1

µs

= ΣFn mC a= mC n: N

1 or

or

µs

mC g = mC

µ= s

mC g vC2 r

vC2 r

gr (9.81 m/s 2 )(0.6 m) = vC2 (3.6 m/s)2

( µ s )min = 0.454 

or

The direction of the impending motion is downward. 

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PROBLEM 12.128 (Continued)

(b) and (c) First observe that for an arbitrary value of θ, it is not known whether the impending motion will be upward or downward. To consider both possibilities for each value of θ, let Fdown correspond to impending motion downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown, we have

= ΣFy 0: N cos θ ± F sin= θ − WC 0

F = µs N

Now

N cos θ ± µ s N sin θ − mC g = 0

Then or

N=

mC g cos θ ± µ s sin θ

and

F=

µ s mC g cos θ ± µ s sin θ

vC2 = ΣFn mC an : N sin θ =  F cos θ mC = ρ r sin θ

ρ

Substituting for N and F

mC g µ s mC g v2 sin θ  cos θ = mC C cos θ ± µ s sin θ cos θ ± µ s sin θ r sin θ vC2 µs tan θ  = 1 ± µ s tan θ 1 ± µ s tan θ gr sin θ

or

µs = ±

or

1+

vC2 gr sin θ

vC2 gr sin θ

tan θ

vC2 [(3.6 m/s)sin θ ]2 = = 2.2018sin θ gr sin θ (9.81 m/s 2 )(0.6 m)sin θ

Now

µs = ±

Then (b)

tan θ −

tan θ − 2.2018 sin θ 1 + 2.2018sin θ tan θ

θ= 75° tan 75° − 2.2018sin 75° 1 + 2.2018sin 75° tan 75°

± = ± 0.1796 µs =

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PROBLEM 12.128 (Continued)

Then downward:

µ s = + 0.1796

upward:

µs < 0

not possible

( µ s )min = 0.1796  The direction of the impending motion is downward.  (c)

θ= 45°

µ s =±

tan 45° − 2.2018sin 45° =± (− 0.218) 1 + 2.2018sin 45° tan 45°

Then downward:

µs < 0

upward:

µ s = 0.218

not possible

( µ s )min = 0.218  The direction of the impending motion is upward. 

Note: When or

tan θ − 2.2018sin θ = 0

= θ 62.988°,

µ s = 0. Thus, for this value of θ , friction is not necessary to prevent the collar from sliding on the rod.

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PROBLEM 12.129 Telemetry technology is used to quantify kinematic values of a 200-kg roller coaster cart as it passes overhead. According to r = −2 m/s 2 , θ = 90°, the system, r = 25 m, r = −10 m/s,  θ = −0.4 rad/s, θ = −0.32 rad/s 2. At this instant, determine (a) the normal force between the cart and the track, (b) the radius of curvature of the track.

SOLUTION Find the acceleration and velocity using polar coordinates.

vr = r = −10 m/s −10 m/s vθ = rθ = (25 m)( − 0.4 rad/s) = So the tangential direction is

45° and v = 10 2 m/s.

 ar = r − rθ 2 = −2 m/s − (25 m)( − 0.4 rad/s) 2 = −6 m/s 2 a= rθ + 2rθ θ = (25 m)(−0.32) rad/s 2 | +2(−10 m/s)( −0.4 rad/s) =0

So the acceleration is vertical and downward. (a)

To find the normal force use Newton’s second law. y-direction N − mg sin 45° = −ma cos 45° N m( g sin 45° − a cos 45°) =

= (200 kg)(9.81) m/s 2 − 6 m/s 2 )(0.70711) = 538.815 N

N = 539 N 

(b)

Radius l curvature of the track.

an =

ρ =

v2

ρ v2 (10 2)2 = an 6cos 45°

ρ = 47.1 m 

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PROBLEM 12.130 The radius of the orbit of a moon of a given planet is equal to twice the radius of that planet. Denoting by ρ the mean density of the planet, show that the time required by the moon to complete one full revolution about the planet is (24 π /G ρ )1/2 , where G is the constant of gravitation.

SOLUTION For gravitational force and a circular orbit,

= Fr

GMm mv 2 = r r2

v or =

GM r

Let τ be the periodic time to complete one orbit.

GM = vτ 2= πr or τ 2π r r

τ=

Solving for τ,

But

= M

τ=

Then

2π r 3/2 GM

4 π = GM 2 G ρ R3/2 π R3 ρ , hence, 3 3 3π  r  G ρ  R 

3/2

Using r = 2R as a given leads to

3π 3/2 = τ 2= Gρ

24π Gρ

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τ = (24π /G ρ )1/2 

PROBLEM 12.131 At engine burnout on a mission, a shuttle had reached Point A at an altitude of 40 mi above the surface of the earth and had a horizontal velocity v0. Knowing that its first orbit was elliptic and that the shuttle was transferred to a circular orbit as it passed through Point B at an altitude of 170 mi, determine (a) the time needed for the shuttle to travel from A to B on its original elliptic orbit, (b) the periodic time of the shuttle on its final circular orbit.

SOLUTION R= 3960 mi = 20.909 × 106 ft, g = 32.2 ft/s 2

For Earth,

GM =gR 2 =(32.2)(20.909 × 106 ) 2 =14.077 × 1015 ft 3 /s 2 (a)

For the elliptic orbit,

rA= 3960 + 40= 4000 mi= 21.12 × 106 ft rB = 3960 + 170 = 4130 mi = 21.8064 × 106 ft 1 (rA + rB )= 21.5032 × 106 ft 2 = b = rA rB 21.4605 × 106 ft a=

Using Eq. 12.39,

1 GM = + C cos θ A rA h2

and

1 GM = + C cos θ B rB rB2

But θ B =θ A + 180°, so that cos θ A = − cos θ B 1 1 r +r 2a 2GM + = A B = 2 = rA rB rA rB b h2

Adding,

h=

or Periodic time.

= τ

= τ

GMb 2 a 2π ab 2π ab a 2π a 3/ 2 = = h GM GMb 2 2π (21.5032 × 106 )3/ 2 = 5280.6 = s 1.4668 h 14.077 × 1015

The time to travel from A to B is one half the periodic time

τ AB = 0.7334 h

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τ AB = 44.0 min 

PROBLEM 12.131 (Continued)

(b)

For the circular orbit,

a= b= rB= 21.8064 × 106 ft

τ circ =

2π a3/ 2 2π (21.8064 × 106 )3/ 2 = = 5393 s GM 14.077 × 1015

τ circ = 1.498 h

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τ circ = 89.9 min 

PROBLEM 12.132 A space probe in a low earth orbit is inserted into an elliptic transfer orbit to the planet Venus. Knowing that the mass of the 3 sun is 332.8 × 10 times the mass of the earth and assuming that the probe is subjected only to the gravitational attraction of the sun, determine the value of φ , which defines the relative position of Venus with respect to the earth at the time the probe is inserted into the transfer orbit.

SOLUTION First determine the time tprobe for the probe to travel from the earth to Venus: 1 tprobe = τ tr 2

where τ tr is the periodic time of the elliptic transfer orbit. Applying Kepler’s Third Law to the orbits about the sun of the earth and the probe, we obtain

τ tr2

=

2 τ earth

atr3 3 aearth

1 (rE + rv ) 2 1 = (93 × 106 + 67.2 × 106 ) mi 2 = 80.1 × 106 mi

where

= atr

and

aearth ≈ rE

Then

tprobe = =

( Note: ε earth = 0.0167)

1  atr    2  rE 

3/ 2

τ earth

1  80.1 × 106 mi    2  93.0 × 106 mi 

3/ 2

(365.25 days)

= 145.977 days = 12.6124 × 106 s

In time tprobe , Venus travels through the angle θv given by vv = θv θ= tprobe v tprobe rv

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PROBLEM 12.132 (Continued)

Assuming that the orbit of Venus is circular (note: ε venus = 0.0068), then, for a circular orbit vv =

Now

GM sun rv

[Eq. (12.44)]

= GM sun G (332.8 × 103 M earth )

(

2 = 332.8 × 103 gRearth

Then

(

using Eq. (12.30)

3 2 tprobe  332.8 × 10 gRearth  θv = rv  rv 

= tprobe Rearth

where

)

) 

1/ 2

 

(332.8 g × 103 )1/ 2 rv3/ 2

= Rearth 3960 = mi 20.9088 × 106 ft

and

rv = 67.2 × 106 mi = 354.816 × 109 ft

Then

(12.6124 × 106 s)(20.9088 × 106 ft) θv =

(332.8 × 103 × 32.2 ft/s 2 )1/ 2 (354.816 × 109 ft)1/2

= 4.0845 rad = 234.02°

Finally,

φ =θv − 180° =234.02° − 180° = ϕ 54.0° 

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PROBLEM 12.133* Disk A rotates in a horizontal plane about a vertical axis at the constant rate θ0 = 10 rad/s. Slider B has mass 1 kg and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k, which is undeformed when r = 0. Knowing that the slider is released with no radial velocity in the position r = 500 mm, determine the position of the slider and the horizontal force exerted on it by the disk at t = 0.1 s for (a) k = 100 N/m, (b) k = 200 N/m.

SOLUTION First we note

r= 0, xsp =⇒ 0 Fsp = kr

when

= r0 500 = mm 0.5 m

and

 θ= 12 rad/s θ= 0

then

θ = 0 mB ar : − Fsp= mB ( r − rθ02 ) ΣF= r

 k   0 − θ02  r = r +  mB 

ΣFθ= mB aθ : FA= mB (0 + 2rθ0 ) (a)

k = 100 N/m

Substituting the given values into Eq. (1)

100 N/m   r+ − (10 rad/s) 2  r = 0  1 kg   r =0

Then

dr t 0,= r 0 : =  r= 0 and at= dt

∫

r 0

dr =

∫

0.1 0

(0) dt

r = 0

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(1) (2)

PROBLEM 12.133* (Continued)

dr = t 0,= r0 0.5 m = r= 0 and at dt

and

∫

r

dr =

r0

∫

0.1 0

(0) dt

r = r0 r = 0.5 m 

Note: r = 0 implies that the slider remains at its initial radial position. With r = 0, Eq. (2) implies

FH = 0  (b)

k = 200 N/m

Substituting the given values into Eq. (1)

 200   r+ − (10 rad/s) 2  r = 0 1 kg   r + 100 r = 0  = r

Now

d (r) = r vr dt dvr dr

 r = vr

Then so that

dvr 0 + 100r = dr

vr

∫

t 0,= vr 0,= r r0 : At=

d dr d d = = vr dt dt dr dr

vr 0

vr d vr = −100

∫

r

r dr r0

vr2 = −100(r 2 − r02 ) = vr 10 r02 − r 2 v= r

Now

t 0,= r r0 : At=

∫

r r0

dr = r02 − r 2

dr = 10 r02 − r 2 dt t

10 dt ∫= 0

10t

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PROBLEM 12.133* (Continued)

Let

= r r0= sin φ , dr r0 cos φ dφ sin −1 ( r/r0 )

∫π

Then

/2

r0 cos φ dφ r02 − r02 sin 2 φ sin −1 ( r/r0 )

∫π

/2

r sin −1   r0

= 10t

dφ = 10t

 π 10 t − =  2

π  = = r r0 sin 10 t += r0 cos10 t (0.5 ft) cos10 t 2   r = −(5 m/s)sin10 t

Then Finally,

at t = 0.1 s: = r (0.5 ft) cos (10 × 0.1) r = 0.270 m 

Eq. (2)

F= 1 kg × 2 × [−(5 ft/s)sin (10 × 0.1)] (10 rad/s) H FH = −84.1 N 

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PROBLEM 12.CQ1 A 1000 lb boulder B is resting on a 200 lb platform A when truck C accelerates to the left with a constant acceleration. Which of the following statements are true (more than one may be true)? (a) The tension in the cord connected to the truck is 200 lb (b) The tension in the cord connected to the truck is 1200 lb (c) The tension in the cord connected to the truck is greater than 1200 lb (d ) The normal force between A and B is 1000 lb (e) The normal force between A and B is 1200 lb ( f ) None of the above

SOLUTION The boulder and platform are accelerating upwards; therefore it will require a force larger than their weight to achieve this acceleration: Answers: (c) The tension will be greater than 1200 lb and (d) the normal force between A and B is 1000 lb.

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PROBLEM 12.CQ2 Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed from the top, which of the following choices best describes the path of the marble after leaving the tube? (a) 1 (b) 2 (c) 3 (d ) 4 (e) 5

SOLUTION Answer: (d ) The particle will have velocity components along the tube and perpendicular to the tube when it leaves. After it leaves, it will not accelerate in the horizontal plane since there are no forces in that plane; therefore, it will travel in a straight line.

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PROBLEM 12.CQ3 The two systems shown start from rest. On the left, two 40 lb weights are connected by an inextensible cord, and on the right, a constant 40 lb force pulls on the cord. Neglecting all frictional forces, which of the following statements is true? (a) Blocks A and C will have the same acceleration (b) Block C will have a acceleration than block A

larger

(c) Block A will have a acceleration than block C

larger

(d ) Block A will not move (e) None of the above

SOLUTION Answer: (b) If you draw a FBD of B, you will see that since it is accelerating downward, the tension in the cable will be less than 40 lb, so the acceleration of A will be less than the acceleration of C. Also, the system on the left has more inertia, so it will require more force to have the same acceleration as the system on the right.

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PROBLEM 12.CQ4 The system shown is released from rest in the position shown. Neglecting friction, the normal force between block A and the ground is (a) less than the weight of A plus the weight of B (b) equal to the weight of A plus the weight of B (c) greater than the weight of A plus the weight of B

SOLUTION Answer: (a) Since B has an acceleration component downward the normal force between A and the ground will be less than the sum of the weights. To see this, you can draw an FBD of both block A and B. From block A you will find that the normal force on the ground is the sum of the weight of block A plus the vertical component of the normal force of block B on A since block A does not accelerate in the vertical direction. From the FBD of block B and using F=ma, you will see that the vertical component of the normal force of B on A is less than the weight of block B since it is accelerating downwards.

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PROBLEM 12.CQ5 People sit on a Ferris wheel at Points A, B, C and D. The Ferris wheel travels at a constant angular velocity. At the instant shown, which person experiences the largest force from his or her chair (back and seat)? Assume you can neglect the size of the chairs, that is, the people are located the same distance from the axis of rotation. (a) A (b) B (c) C (d ) D (e) The force is the same for all the passengers.

SOLUTION Answer: (c) Draw a FBD and KD at each location and it will be clear that the maximum force will be experienced by the person at Point C. At this point the normal force isdirectly opposite the force of gravity and the acceleration is in the same direction as the normal force making the normal force the sum of mg and ma.

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PROBLEM 12.CQ6 A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about Point D, that is, the upper left corner of the crate? (a) 0 (b) mv1a (c) mv1b (d ) mv1 a 2 + b 2

SOLUTION Answers: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum, mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point D.

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PROBLEM 12.CQ7 A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about Point A, that is, the point of contact between the front tire of the forklift and the ground? (a) 0 (b) mv1d (c) 3mv1 (d ) mv1 32 + d 2

SOLUTION Answer: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum, mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point A.

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PROBLEM 12.F1 Crate A is gently placed with zero initial velocity onto a moving conveyor belt. The coefficient of kinetic friction between the crate and the belt is µk. Draw the FBD and KD for A immediately after it contacts the belt.

SOLUTION Answer:

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PROBLEM 12.F2 Two blocks weighing WA and WB are at rest on a conveyor that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Assuming the coefficient of friction between the boxes and the belt is µk, draw the FBDs and KDs for blocks A and B. How would you determine if A and B remain in contact?

SOLUTION Answer: Block A

Block B

To see if they remain in contact assume aA = aB and then check to see if NAB is greater than zero.

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PROBLEM 12.F3 Objects A, B, and C have masses mA, mB, and mC respectively. The coefficient of kinetic friction between A and B is µk, and the friction between A and the ground is negligible and the pulleys are massless and frictionless. Assuming B slides on A draw the FBD and KD for each of the three masses A, B and C.

SOLUTION Answer: Block A

Block B

Block C

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PROBLEM 12.F4 Blocks A and B have masses mA and mB respectively. Neglecting friction between all surfaces, draw the FBD and KD for each mass.

SOLUTION Block A

Block B

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PROBLEM 12.F5 Blocks A and B have masses mA and mB respectively. Neglecting friction between all surfaces, draw the FBD and KD for the two systems shown.

SOLUTION System 1

System 2

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PROBLEM 12.F6 A pilot of mass m flies a jet in a half vertical loop of radius R so that the speed of the jet, v, remains constant. Draw a FBD and KD of the pilot at Points A, B and C.

SOLUTION

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PROBLEM 12.F7 Wires AC and BC are attached to a sphere which revolves at a constant speed v in the horizontal circle of radius r as shown. Draw a FBD and KD of C.

SOLUTION

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PROBLEM 12.F8 A collar of mass m is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 5 in. and a constant k. Knowing that the collar has a speed v at Point C, draw the FBD and KD of the collar at this point.

SOLUTION

where x = 2/12 ft and r = 5/12 ft.

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PROBLEM 12.F9 Four pins slide in four separate slots cut in a horizontal circular plate as shown. When the plate is at rest, each pin has a velocity directed as shown and of the same constant magnitude u. Each pin has a mass m and maintains the same velocity relative to the plate when the plate rotates about O with a constant counterclockwise angular velocity ω. Draw the FBDs and KDs to determine the forces on pins P1 and P2.

SOLUTION

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PROBLEM 12.F10 At the instant shown, the length of the boom AB is being decreased at the constant rate of 0.2 m/s, and the boom is being lowered at the constant rate of 0.08 rad/s. If the mass of the men and lift connected to the boom at Point B is m, draw the FBD and KD that could be used to determine the horizontal and vertical forces at B.

SOLUTION

Where r = 6 m, r = −0.2 m/s,  −0.08 rad/s, θ = r= 0, θ = 0

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PROBLEM 12.F11 Disk A rotates in a horizontal plane about a vertical axis at the constant rate θ0 . Slider B has a mass m and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k, which is undeformed when r = 0. Knowing that the slider is released with no radial velocity in the position r = r0, draw a FBD and KD at an arbitrary distance r from O.

SOLUTION

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PROBLEM 12.F12 Pin B has a mass m and slides along the slot in the rotating arm OC and along the slot DE which is cut in a fixed horizontal plate. Neglecting friction and knowing that rod OC rotates at the constant rate θ0 , draw a FBD and KD that can be used to determine the forces P and Q exerted on pin B by rod OC and the wall of slot DE, respectively.

SOLUTION

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