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CHAPTER 3 PROBLEM 3.1 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the we
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CHAPTER 3
PROBLEM 3.1 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E.
SOLUTION
(a)
By definition, We have
W mg 80 kg(9.81 m/s2 ) 784.8 N M E : M E (784.8 N)(0.25 m) M E 196.2 N m
(b)
For the force at B to be the smallest, resulting in a moment (ME) about E, the line of action of force FB must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have
d (0.85 m) 2 (0.5 m) 2 0.98615 m
M E : 196.2 N m FB (0.98615 m) FB 198.954 N
and or
0.85 m 59.534 0.5 m
tan 1
FB 199.0 N
59.5°
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PROBLEM 3.2 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E.
SOLUTION First note. . .
W mg (80 kg)(9.81 m/s2 ) 784.8 N
(a)
We have
M E rH /EW (0.25 m)(784.8 N) 196.2 N m
(b)
(c)
or M E 196.2 N m
For FA to be minimum, it must be perpendicular to the line joining Points A and E. Then with FA directed as shown, we have ( M E ) rA /E ( FA ) min .
Where
rA /E (0.35 m) 2 (0.5 m) 2 0.61033 m
then
196.2 N m (0.61033 m)( FA ) min
or
( FA ) min 321 N
Also
tan
0.35 m 0.5 m
or
35.0
(FA ) min 321 N
35.0°
For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be maximum. Thus, apply (Fvertical)min at Point D, and then ( M E ) rD / E ( Fvertical ) min 196.2 N m (0.85 m)( Fvertical )min
or (Fvertical ) min 231 N
at Point D
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PROBLEM 3.3 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α 10°, (c) the smallest force P that creates the same moment about B.
SOLUTION (a)
M B rC/B FN
We have
(4 in.)(200 lb) 800 lb in. or MB 800 lb in. (b)
By definition,
M B rA/B P sin
10 (180 70) 120
Then
800 lb in. (18 in.) P sin120
or P 51.3 lb (c)
For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus
M B dPmin d rA/B
or or
800 lb in. (18 in.) Pmin
Pmin 44.4 lb
Pmin 44.4 lb
20°
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PROBLEM 3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D.
SOLUTION (a)
Fx (300 N) cos 25 271.89 N Fy (300 N)sin 25 126.785 N F (271.89 N)i (126.785 N) j r DA (0.1 m)i (0.2 m) j
MD r F M D [(0.1 m)i (0.2 m) j] [(271.89 N)i (126.785 N) j] (12.6785 N m)k (54.378 N m)k (41.700 N m)k M D 41.7 N m
(b)
The smallest force Q at B must be perpendicular to DB at 45° M D Q( DB)
41.700 N m Q(0.28284 m) Q 147.4 N
45.0°
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PROBLEM 3.5 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.
SOLUTION (a)
See Problem 3.3 for the figure and analysis leading to the determination of MD M D 41.7 N m
(b)
Since C is horizontal C C i r DC (0.2 m)i (0.125 m) j M D r C i C (0.125 m)k 41.7 N m (0.125 m)(C ) C 333.60 N
(c)
C 334 N
The smallest force C must be perpendicular to DC; thus, it forms with the vertical
0.125 m 0.2 m 32.0
tan
M D C ( DC ); DC (0.2 m) 2 (0.125 m) 2 0.23585 m 41.70 N m C (0.23585 m)
C 176.8 N
58.0°
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PROBLEM 3.6 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α 25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components.
SOLUTION Free-Body Diagram of Rod AB: x (9 in.) cos 65 3.8036 in. y (9 in.)sin 65 8.1568 in.
F Fx i Fy j (20 lb cos 25)i (20 lb sin 25) j
rA/B
(18.1262 lb)i (8.4524 lb) j BA (3.8036 in.)i (8.1568 in.)j
M B rA /B F (3.8036i 8.1568 j) (18.1262i 8.4524 j) 32.150k 147.852k 115.702 lb-in.
M B 115.7 lb-in.
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PROBLEM 3.7 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α 25°, determine the moment of the force about Point B by resolving the force into components along AB and in a direction perpendicular to AB.
SOLUTION Free-Body Diagram of Rod AB:
90 (65 25) 50
Q (20 lb) cos 50 12.8558 lb M B Q(9 in.) (12.8558 lb)(9 in.) 115.702 lb-in.
M B 115.7 lb-in.
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PROBLEM 3.8 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb · in. clockwise, determine the value of α.
SOLUTION Free-Body Diagram of Rod AB:
25
Q (20 lb) cos
and Therefore,
M B (Q )(9 in.)
120 lb-in. (20 lb)(cos )(9 in.) cos
120 lb-in. 180 lb-in.
or
48.190
Therefore,
48.190 25
23.2
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PROBLEM 3.9 Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c = 360 mm, determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied (a) at point A, (b) at point C.
SOLUTION Free-Body Diagram of Rod AB:
(a) F 1350 N
cos
AC (450) 2 (600) 2 750 mm
450 0.6 750
sin
600 0.8 750
F F cos i + F sin j (1350 N)0.6i (1350 N)0.8 j
M B r A/ B F (0.45i 0.24 j) (810i 1080 j)
(810 N)i (1080N)j r A/ B (0.45 m)i (0.24 m)j
486k 194.4k (291.6 N m)k
(b)
M B 292 N m
M B 292 N m
F (810 N)i (1080 N)j
rC / B (0.36 m) j
M B rC / B F 0.36 j (810i 1080 j) (291.6 N m)k
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PROBLEM 3.10 Rod AB is held in place by the cord AC. Knowing that c = 840 mm and that the moment about B of the force exerted by the cord at point A is 756 N·m, determine the tension in the cord.
SOLUTION Free-Body Diagram of Rod AB: M B (756 N m )k
cos
AC (450) 2 (1080) 2 1170 mm
450 1170
sin
1080 1170
F F cos i + F sin j
450 1080 Fi Fj 1170 1170
r A/ B (0.45 m)i (0.24 m)j F M B r A/ B F (0.45i 0.24 j) (450i 1080 j) 1170
(486k 108k ) 378 F k 1170
F 1170 Substituting for M B :
756
378 F 1170
F 2340 N
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PROBLEM 3.11 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.
SOLUTION First note
dCB (12.0 in.) 2 (2.33 in.) 2 12.2241 in.
Then
cos
12.0 in. 12.2241 in.
sin
2.33 in. 12.2241 in.
FCB FCB cos i FCB sin j
and
125 lb [(12.0 in.) i (2.33 in.) j] 12.2241 in.
Now
M A rB/A FCB
where
rB/A (15.3 in.) i (12.0 in. 2.33 in.) j (15.3 in.) i (14.33 in.) j
Then
M A [(15.3 in.)i (14.33 in.) j]
125 lb (12.0i 2.33j) 12.2241 in.
(1393.87 lb in.)k (116.156 lb ft)k
or M A 116.2 lb ft
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PROBLEM 3.12 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.
SOLUTION First note
dCB (17.2 in.) 2 (7.62 in.) 2 18.8123 in.
17.2 in. 18.8123 in. 7.62 in. sin 18.8123 in.
cos
Then
FCB ( FCB cos )i ( FCB sin ) j
and
125 lb [(17.2 in.)i (7.62 in.) j] 18.8123 in.
Now
M A rB/A FCB
where
rB/A (20.5 in.)i (4.38 in.) j
Then
MA [(20.5 in.)i (4.38 in.) j] (1538.53 lb in.)k (128.2 lb ft)k
125 lb (17.2i 7.62 j) 18.8123 in. or M A 128.2 lb ft
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PROBLEM 3.13 It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.
SOLUTION Using (a): M C y1 ( FAB ) x x1 ( FAB ) y 7 24 (0.056 m) 2500 N (0.042 m) 2500 N 25 25 140.0 N m
(a) MC 140.0 N m
Using (b): M C y2 ( FAB ) x 7 (0.2 m) 2500 N 25 140.0 N m
(b) M C 140.0 N m
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PROBLEM 3.14 It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.
SOLUTION Using (a): M C y1 ( FAB ) x x1 ( FAB ) y 7 24 (0.056 m) 2500 N (0.042 m) 2500 N 25 25 61.6 N m
(a) MC 61.6 N m
Using (b): M C y2 ( FAB ) x 7 (0.088 m) 2500 N 25 61.6 N m
(b) M C 61.6 N m
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PROBLEM 3.15 Form the vector products B × C and B × C, where B B, and use the results obtained to prove the identity
sin cos
1 1 sin ( ) sin ( ). 2 2
SOLUTION Note:
B B (cos i sin j) B B (cos i sin j) C C (cos i sin j)
By definition,
Now
| B C | BC sin ( )
(1)
| B C | BC sin ( )
(2)
B C B (cos i sin j) C (cos i sin j) BC (cos sin sin cos )k
and
(3)
B C B (cos i sin j) C (cos i sin j) BC (cos sin sin cos ) k
(4)
Equating the magnitudes of B C from Equations (1) and (3) yields: BC sin( ) BC (cos sin sin cos )
(5)
Similarly, equating the magnitudes of B C from Equations (2) and (4) yields: BC sin( ) BC (cos sin sin cos )
(6)
Adding Equations (5) and (6) gives: sin( ) sin( ) 2 cos sin
or sin cos
1 1 sin( ) sin( ) 2 2
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PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = –8i + 4j – 4k and Q = 3i + 3j + 6k, (b) P = 7i – 6j – 3k and Q = –3i + 6j – 2k
SOLUTION (a)
We have
A |P Q|
where
P 8i 4 j 4k Q 3i 3j 6k
Then
i j k P Q 8 4 4 3
3
6
[(24 12)i (12 48) j (24 12)k ] (36)i (36) j (36)k A (36) 2 (36) 2 ( 36) 2
(b)
We have
A |P Q|
where
P 7i 6 j 3k
or A 62.4
Q 3i 6 j 2k
Then
i PQ 7 3
j k 6 3 6
2
[(12 18)i (9 14) j (42 18)k ] (30)i (23) j (24)k A (30) 2 (23) 2 (24) 2
or A 44.8
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PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 2i + 3j – 6k and 5i – 8j – 6k, (b) 4i – 4j + 3k and –3i + 7j – 5k.
SOLUTION (a)
AB |A B|
We have
λ
where
A 2i 3j 6k B 5i 8 j 6k
Then
i AB 2
j 3
k 6
5 8 6 (18 15)i (30 12) j (16 15)k (33i 18 j 31k ) and
|A B | ( 33) 2 ( 18) 2 ( 31) 2 2374
λ
(b)
(33i 18 j 31k ) 2374
or λ 0.677i 0.369 j 0.636k
AB |A B|
We have
λ
where
A 4i 4 j 3k B 3i 7 j 5k
Then
i AB 4
j 4
k 3
3
7
5
(20 21)i (9 20) j (28 12)k (i 11j 16k ) and
|A B| ( 1) (11) 2 (16) 2 378
λ
(i 11j 16k ) 378
or λ 0.0514i 0.566 j 0.823k
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PROBLEM 3.18 A line passes through the points (12 m, 8 m) and (–3 m, –5 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.
SOLUTION d AB [12 m (3 m)]2 [8 m ( 5 m)]2 394 m
Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B. F F AB
Then where
AB
By definition, where
rB rA d AB 1 394
(15i 13j)
M O | rA F | dF rA ( 3 m)i + ( 5 m) j
M O [(3 m)i + (5 m) j]
Then
F 394
F 394m
[(15 m)i (13 m) j]
[39k 75k ] N m
36 F k N m 394
Finally,
36 F d (F ) 394 36 d m 394
d 1.184 m
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PROBLEM 3.19 Determine the moment about the origin O of the force F 4i 3j 5k that acts at a Point A. Assume that the position vector of A is (a) r 2i 3j 4k, (b) r 8i 6j 10k, (c) r 8i 6j 5k.
SOLUTION MO r F
i (a)
j
k
M O 2 3 4 4 3 5 (15 12)i (16 10) j (6 12)k
(b)
i j k MO 8 6 10 4 3 5 (30 30)i (40 40) j (24 24)k
(c)
M O 3i 26 j 18k
MO 0
i j k MO 8 6 5 4 3 5 (30 15)i (20 40) j (24 24)k
M O 15i 20 j
Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the determinant are proportional.
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PROBLEM 3.20 Determine the moment about the origin O of the force F 2i 3j 4k that acts at a Point A. Assume that the position vector of A is (a) r 3i 6j 5k, (b) r i 4j 2k, (c) r 4i 6j 8k.
SOLUTION MO r F
(a)
i j k M O 3 6 5 2 3 4 (24 15)i (10 12) j (9 12)k
(b)
i j k M O 1 4 2 2 3 4 (16 6)i (4 4) j (3 8)k
i (c)
M O 9i 22 j 21k
j
M O 22i 11k
k
M O 4 6 8 2 3 4 (24 24)i (16 16) j (12 12)k
MO 0
Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the determinant are proportional.
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PROBLEM 3.21 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.
SOLUTION
d BA (0.75 m) 2 (7 m)2 (6 m)2 9.25 m d BC (4.25 m)2 (7 m) 2 (1 m) 2 8.25 m
Have
TB A TB A
BA 555 N (0.75 m i 7 m j 6 m k ) dB A 9.25 m
TBA (45 N)i (420 N)j (360 N)k
TBC TBC
BC 660 N (4.25 m i 7 m j k ) d BC 8.25 m
TBC (340 N)i (540 N) j (80 N)k R TBA TBC
R 295 N i 980 N j 440 N k M O rB / O R where rB / O (7 m)j
i MO
j
k
0 7 0 Nm 295 980 440
3080 N m i 2065 N m k MO 3080 N m i 2070 N m k
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PROBLEM 3.22 The 12-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 380 lb, determine the moment about A of the force exerted by the cable at B.
SOLUTION First note
d BC (12)2 (4.8)2 (8)2 15.2 ft 380 lb (12i 4.8 j 8k ) 15.2
Then
TBC
We have
M A rB/A TBC
where
rB /A (12 ft)i
Then
M A (12 ft)i
380 lb (12i 4.8 j 8k ) 15.2 or M A (2400 lb ft) j (1440 lb ft)k
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PROBLEM 3.23 A 200-N force is applied as shown to the bracket ABC. Determine the moment of the force about A.
SOLUTION We have
M A rC/A FC
where
rC/A (0.06 m)i (0.075 m) j FC (200 N) cos 30 j (200 N)sin 30k
Then
i j k 0.075 0 M A 200 0.06 0 cos30 sin 30 200[(0.075sin 30)i (0.06sin 30) j (0.06cos 30)k ] or M A (7.50 N m)i (6.00 N m) j (10.39 N m)k
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PROBLEM 3.24 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.
SOLUTION AE (0.21 m)i (0.16 m) j (0.12 m)k
(a)
AE (0.21 m) 2 ( 0.16 m) 2 (0.12 m) 2 0.29 m AE FA FA AE F AE 0.21i 0.16 j 0.12k (435 N) 0.29 (315 N)i (240 N) j (180 N)k
rA /O (0.09 m)i (0.16 m) j
j k i 0 MO 0.09 0.16 315 240 180 28.8i 16.20 j (21.6 50.4)k
(b)
M O (28.8 N m)i (16.20 N m) j (28.8 N m)k
FE FA (315 N)i (240 N) j (180 N)k rE / O (0.12 m)i (0.12 m)k
i j k 0 0.12 M O 0.12 315 240 180 28.8i (37.8 21.6) j 28.8k
M O (28.8 N m)i (16.20 N m) j (28.8 N m)k
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PROBLEM 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION We have
Txz (6 lb) cos 8 5.9416 lb
Then
Tx Txz sin 30 2.9708 lb Ty TBC sin 8 0.83504 lb Tz Txz cos 30 5.1456 lb
Now
M A rB/A TBC
where
rB/A (6sin 45) j (6 cos 45)k
Then
or
6 ft 2
(j k)
i j k 0 1 MA 1 2 2.9708 0.83504 5.1456 6 6 6 (5.1456 0.83504)i (2.9708) j (2.9708)k 2 2 2 6
M A (25.4 lb ft)i (12.60 lb ft) j (12.60 lb ft)k
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PROBLEM 3.26 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.
SOLUTION FF
BD BD
where F 900 N
BD (1 m)i (2 m) j (2 m)k BD (1 m)2 (2 m)2 (2 m)2 3m i 2 j 2k 3 (300 N)i (600 N) j (600 N)k (2.5 m)i (2 m) j
F (900 N) rB /O M O rB /O F
i j k 2.5 2 0 300 600 600 1200i 1500 j (1500 600)k MO (1200 N m)i (1500 N m) j (900 N m)k
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PROBLEM 3.27 In Prob. 3.22, determine the perpendicular distance from point A to cable BC. PROBLEM 3.22 The 12-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 380 lb, determine the moment about A of the force exerted by the cable at B.
SOLUTION M A (2400)2 (1440)2
From the solution to problem 3.22
2798.9 lb ft But
M A FB d
d
MA FB
d
2798.9 lb ft 380 lb or d 7.37 ft
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PROBLEM 3.28 In Prob. 3.24, determine the perpendicular distance from point O to wire AE. PROBLEM 3.24 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.
SOLUTION From the solution to Prob. 3.24 M O (28.8 N m)i (16.20 N m) j (28.8 N m)k M O (28.8) 2 (16.20) 2 (28.8) 2 43.8329 N m
But
M O FA d
or
MO FA 43.8329 N m d 435 N 0.100765 m d
d 100.8 mm
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PROBLEM 3.29 In Prob. 3.24, determine the perpendicular distance from point B to wire AE. PROBLEM 3.24 The wire AE is stretched between the corners A and E of a bent plate. Knowing that the tension in the wire is 435 N, determine the moment about O of the force exerted by the wire (a) on corner A, (b) on corner E.
SOLUTION From the solution to Prob. 3.24 FA (315 N)i (240 N) j (180 N)k rA /B (0.210 m)i M B rA /B FA 0.21i (315i 240 j 180k ) 50.4k 37.8 j M B (50.4) 2 (37.8) 2 63.0 N m
M B FA d
or
MB FA 63.0 N m d 435 N 0.144829 m
d
d 144.8 mm
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PROBLEM 3.30 In Prob. 3.25, determine the perpendicular distance from point A to a line drawn through points B and C. PROBLEM 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION From the solution to Prob. 3.25:
M A (25.4 lb ft)i (12.60 lb ft) j (12.60 lb ft)k M A (25.4)2 (12.60)2 (12.60)2 31.027 lb ft M A TBC d
or
d
MA TBC
31.027 lb ft 6 lb 5.1712 ft
d 5.17 ft
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PROBLEM 3.31 In Prob. 3.25, determine the perpendicular distance from point D to a line drawn through points B and C. PROBLEM 3.25 A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the line is 6 lb. Determine the moment about A of the force exerted by the line at B.
SOLUTION
AB 6 ft TBC 6 lb
We have
| M D | TBC d
where d perpendicular distance from D to line BC. M D rB /D TBC
rB /D (6sin 45 ft) j (4.2426 ft)
TBC : (TBC ) x (6 lb) cos8 sin 30 2.9708 lb
(TBC ) y (6 lb)sin 8 0.83504 lb (TBC ) z (6 lb) cos8 cos 30 5.1456 lb
TBC (2.9708 lb)i (0.83504 lb) j (5.1456 lb)k i j k 0 4.2426 0 MD 2.9708 0.83504 5.1456 (21.831 lb ft)i (12.6039 lb ft) | M D | (21.831) 2 ( 12.6039) 2 25.208 lb ft
25.208 lb ft (6 lb)d
d 4.20 ft
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PROBLEM 3.32 In Prob. 3.26, determine the perpendicular distance from point O to cable BD. PROBLEM 3.26 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.
SOLUTION From the solution to Prob. 3.26 we have
M O (1200 N m)i (1500 N m) j (900 N m)k M O (1200)2 (1500)2 (900)2 2121.3 N m M O Fd
MO F 2121.3 N m 900 N
d
d 2.36 m
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PROBLEM 3.33 In Prob. 3.26, determine the perpendicular distance from point C to cable BD. PROBLEM 3.26 A precast concrete wall section is temporarily held by two cables as shown. Knowing that the tension in cable BD is 900 N, determine the moment about Point O of the force exerted by the cable at B.
SOLUTION From the solution to Prob. 3.26 we have F (300 N)i (600 N) j (600 N)k rB /C (2 m) j M C rB /C F (2 m) j (300 Ni 600 Nj 600 Nk ) (600 N m)k (1200 N m)i M C (600) 2 (1200) 2 1341.64 N m
M C Fd
MC F 1341.64 N m 900 N
d
d 1.491 m
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PROBLEM 3.34 Determine the value of a that minimizes the perpendicular distance from Point C to a section of pipeline that passes through Points A and B.
SOLUTION Assuming a force F acts along AB, |M C | |rA / C F| F (d )
d perpendicular distance from C to line AB
where
F λ AB F
(24 ft) i (24 ft) j (28 ft) k (24)2 (24)2 (28) 2 ft
F
F (6) i (6) j (7) k 11 (3 ft)i (10 ft) j (a 10 ft)k
rA /C
i j k F M C 3 10 10a 11 6 6 7 [(10 6a )i (81 6a ) j 78 k ] |M C | |rA/C F 2 |
Since
or
F 11 |rA/C F 2 | ( dF ) 2
1 (10 6a ) 2 (81 6a ) 2 (78) 2 d 2 121 Setting
d da
(d 2 ) 0 to find a to minimize d: 1 [2(6)(10 6a ) 2(6)(81 6a)] 0 121
Solving
a 5.92 ft
or a 5.92 ft
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PROBLEM 3.35 Given the vectors P = 2i + 3j – k, Q = 5i – 4j + 3k, and S = –3i + 2j – 5k, compute the scalar products P · Q, P · S, and Q · S.
SOLUTION P Q (2i 3j k ) (5i 4 j 3k ) (2)(5) (3)(4) (1)(3) 10 12 3
P Q 5 P S (2i 3j k ) (3i 2 j 5k ) (2)(3) (3)(2) (1)(5) 6 6 5
P S 5 Q S (5i 4 j 3k ) (3i 2 j 5k ) (5)(3) (4)(2) (3)(5) 15 8 15
Q S 38
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PROBLEM 3.36 Form the scalar product B · C and use the result obtained to prove the identity cos (α β) cos α cos β sin α sin β .
SOLUTION B B cos i B sin j
(1)
C C cos i C sin j
(2)
By definition: B C BC cos( )
(3)
From (1) and (2): B C ( B cos i B sin j) (C cos i C sin j) BC (cos cos sin sin )
(4)
Equating the right-hand members of (3) and (4), cos( ) cos cos sin sin
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PROBLEM 3.37 Three cables are used to support a container as shown. Determine the angle formed by cables AB and AD.
SOLUTION First note:
AB (450 mm)2 (600 mm)2 750 mm AD (500 mm)2 (600 mm)2 (360 mm)2
and
By definition,
860 mm AB (450 mm)i (600 mm) j AD (500 mm)i (600 mm) j (360 mm)k AB AD ( AB )( AD) cos (450i 600 j) (500i 600 j 360k ) (750)(860) cos (450)(500) (600)(600) (0)(360) (750)(860) cos
or
cos 0.20930
77.9
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PROBLEM 3.38 Three cables are used to support a container as shown. Determine the angle formed by cables AC and AD.
SOLUTION First note:
AC (600 mm) 2 (320 mm)2 680 mm AD (500 mm)2 (600 mm)2 (360 mm)2
and
By definition,
860 mm AC (600 mm)j (320 mm)k AD (500 mm)i (600 mm) j (360 mm)k AC AD ( AC )( AD ) cos (600 j 320k ) (500i 600 j 360k ) (680)(860) cos 0(500) (600)(600) (320)(360) (680)(860) cos
cos 0.41860
65.3
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PROBLEM 3.39 Knowing that the tension in cable AC is 280 lb, determine (a) the angle between cable AC and the boom AB, (b) the projection on AB of the force exerted by cable AC at point A.
SOLUTION (a)
First note
AC (6)2 (2) 2 (3) 2 7.00 ft AB (6)2 (4.5)2 (0)2
and
By definition or or
7.50 ft AC (6 ft)i (2 ft) j (3 ft)k AB (6 ft)i (4.5 ft) j AC AB ( AC )( AB)cos (6i 2 j 3k ) (6i 4.5 j) (7.00)(7.50) cos (6)(6) (2)(4.5) (3)(0) 52.50 cos cos 0.514 29
or (b)
We have
or 59.0
(TAC ) AB TAC AB TAC cos (280 lb)(0.51429)
or (TAC ) AB 144.0 lb
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PROBLEM 3.40 Knowing that the tension in cable AD is 180 lb, determine (a) the angle between cable AD and the boom AB, (b) the projection on AB of the force exerted by cable AD at point A.
SOLUTION (a)
First note
AD (6) 2 (3)2 (6) 2 9.00 ft AB (6) 2 (4.5)2 (0) 2
and
By definition,
7.50 ft AD (6 ft)i (3 ft) j (6 ft)k AB (6 ft)i (4.5 ft) j AD AB ( AD)( AB)cos (6i 3j 6k ) (6i 4.5 j) (9.00)(7.50) cos (6)(6) (3)(4.5) (6)(0) 67.50 cos
cos (b)
1 3
70.5
(TAD ) AB TAD AB TAD cos
1 (180 lb) 3
(TAD ) AB 60.0 lb
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PROBLEM 3.41 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope AB is 540 N, determine (a) the angle between rope AB and the stake, (b) the projection on the stake of the force exerted by rope AB at Point B.
SOLUTION First note:
BA (3) 2 (3) 2 (1.5) 2 4.5 m BD (0.08) 2 (0.38) 2 (0.16)2 0.42 m
BD
(a)
TBA (3i 3j 1.5k ) 4.5 T BA (2i 2 j k ) 3 1 BD (0.08i 0.38 j 0.16k ) BD 0.42 1 (4i 19 j 8k ) 21
TBA
Then
We have or
or
TBA BD TBA cos
TBA 1 (2i 2 j k ) (4i 19 j 8k ) TBA cos 3 21 1 [(2)(4) (2)(19) ( 1)(8)] 63 0.60317
cos
or 52.9 (b)
We have
(TBA ) BD TBA BD TBA cos (540 N)(0.60317)
or (TBA ) BD 326 N
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PROBLEM 3.42 Ropes AB and BC are two of the ropes used to support a tent. The two ropes are attached to a stake at B. If the tension in rope BC is 490 N, determine (a) the angle between rope BC and the stake, (b) the projection on the stake of the force exerted by rope BC at Point B.
SOLUTION First note:
BC (1)2 (3) 2 (1.5) 2 3.5 m BD (0.08) 2 (0.38) 2 (0.16)2 0.42 m TBC (i 3j 1.5k ) 3.5 T BC (2i 6 j 3k ) 7 1 BD (0.08i 0.38 j 0.16k ) BD 0.42 1 (4i 19 j 8k ) 21
TBC
BD
(a)
TBC BD TBC cos
TBC 1 (2i 6 j 3k ) (4i 19 j 8k ) TBC cos 7 21 1 [(2)(4) (6)(19) (3)(8)] 147 0.55782
cos
56.1 (b)
(TBC ) BD TBC BD TBC cos (490 N)(0.55782) (TBC ) BD 273 N
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PROBLEM 3.43 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC.
SOLUTION (a)
Using Eq. (3.32): CA 11i 12 j 24k CA (11)2 (12) 2 (24) 2 29 in. CB 31i 12 j 24k CB (31) 2 (12)2 (24)2 41 in.
CA CB cos (CA)(CB) (11i 12 j 24k ) (31i 12 j 24k ) (29)(41) (11)(31) (12)(12) (24)(24) (29)(41) 0.89235 (b)
26.8
Law of cosines: ( AB ) 2 (CA) 2 (CB ) 2 2(CA)(CB ) cos (20) 2 (29) 2 (41) 2 2(29)(41) cos cos 0.89235
26.8
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PROBLEM 3.44 Solve Prob. 3.43 for the position corresponding to x 4 in. PROBLEM 3.43 The 20-in. tube AB can slide along a horizontal rod. The ends A and B of the tube are connected by elastic cords to the fixed point C. For the position corresponding to x 11 in., determine the angle formed by the two cords (a) using Eq. (3.32), (b) applying the law of cosines to triangle ABC.
SOLUTION (a)
Using Eq. (3.32): CA 4i 12 j 24k CA (4) 2 (12) 2 (24)2 27.129 in. CB 24i 12 j 24k CB (24)2 (12)2 (24) 2 36 in. CA CB cos (CA)(CB)
(4i 12 j 24k ) (24i 12 j 24k ) (27.129)(36) 0.83551
(b)
33.3
Law of cosines: ( AB ) 2 (CA) 2 (CB) 2 2(CA)(CB ) cos (20)2 (27.129) 2 (36) 2 2(27.129)(36) cos cos 0.83551
33.3
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PROBLEM 3.45 Determine the volume of the parallelepiped of Fig. 3.25 when (a) P 4i 3j 2k, Q 2i 5j k, and S 7i j k, (b) P 5i j 6k, Q 2i 3j k, and S 3i 2j 4k.
SOLUTION Volume of a parallelepiped is found using the mixed triple product. (a)
Vol. P (Q S) 4 3 2 2 5 1 in.3 7 1 1 (20 21 4 70 6 4) 67 or Volume 67.0
(b)
Vol. P (Q S) 5 1 6 3 1 in.3 2 3 2 4 (60 3 24 54 8 10) 111 or Volume 111.0
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PROBLEM 3.46 Given the vectors P = 3i – j + k, Q = 4i + Qyj – 2k, and S = 2i – 2j + 2k, determine the value of Qy for which the three vectors are coplanar.
SOLUTION If P, Q, and S are coplanar, then P must be perpendicular to (Q S). P (Q S) 0
(or, the volume of a parallelepiped defined by P, Q, and S is zero). Then
3 1 4 Qy 2
or
2
1 2 0 2
6Qy 4 8 2Qy 8 12 0
Qy 2
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PROBLEM 3.47 A crane is oriented so that the end of the 25-m boom AO lies in the yz plane. At the instant shown, the tension in cable AB is 4 kN. Determine the moment about each of the coordinate axes of the force exerted on A by cable AB.
SOLUTION OC (OA) 2 ( AC )2 (25 m) 2 (15 m) 2 20 m
rA (15 m)j (20 m)k
AB (2.5 m)2 (15 m) 2 15.2069 m AB P P AB 2.5i 15 j (4 kN) 15.2069 (0.65760 kN)i (3.9456 kN) j
Using Eq. (3.11): i
j k M O rA P 0 15 20 0.65760 3.9456 0 M O (78.912 kN m)i + (13.1520 kN m) j - (9.8640 kN m)k
But Therefore,
MO M x i M y j M z k M x 78.9 kN m, M y 13.15 kN m, M z 9.86 kN m
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PROBLEM 3.48 The 25-m crane boom AO lies in the yz plane. Determine the maximum permissible tension in cable AB if the absolute value of moments about the coordinate axes of the force exerted on A by cable AB must be
|Mx| ≤ 60 kN·m, |My| ≤ 12 kN·m, |Mz| ≤ 8 kN·m
OC (OA)2 ( AC )2 (25 m)2 (15 m)2 20 m rA (15 m)j (20 m)k AB (2.5 m) 2 (15 m) 2 15.2069 m AB P P AB 2.5i 15 j ( P) 15.2069 (0.164399 P kN)i (0.98639 P kN) j
Using Eq. (3.11): i
j
k
M O rA P
0 15 20 0.163499 P 0.98639 P 0 M O (19.7278 P )i + (3.2700 P) j - (2.4525 P )k
But
M x < 60 kN m:
19.7278P 60
M y 12 kN m:
3.270P 12
M z 8 kN m:
2.4525P 8
Therefore, the maximum permissible tension is
P 3.04 kN P 3.67 kN P 3.15 kN P 3.04 kN
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PROBLEM 3.49 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the handle of the valve. Knowing that 25, Mx 61 lb ft, and M z 43 lb ft, determine and d.
SOLUTION We have
M O : rA/O F M O
where
rA/O (4 in.)i (11 in.) j (d )k F F (cos cos i sin j cos sin k )
F 70 lb, 25
For
F (70 lb)[(0.90631cos )i 0.42262 j (0.90631sin )k ] i 4 M O (70 lb) 0.90631cos
j k 11 d in. 0.42262 0.90631sin
(70 lb)[(9.9694sin 0.42262d ) i (0.90631d cos 3.6252sin ) j (1.69048 9.9694cos )k ] in. and
M x (70 lb)(9.9694sin 0.42262d ) in. (61 lb ft)(12 in./ft)
(1)
M y (70 lb)(0.90631d cos 3.6252sin ) in.
(2)
M z (70 lb)(1.69048 9.9694 cos ) in. 43 lb ft(12 in./ft)
(3)
634.33 24.636 697.86
From Equation (3):
cos 1
or
From Equation (1):
1022.90 d 34.577 in. 29.583
or d 34.6 in.
24.6
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PROBLEM 3.50 When a force F is applied to the handle of the valve shown, its moments about the x and z axes are, respectively, M x 77 lb ft and M z 81 lb ft. For d 27 in., determine the moment My of F about the y axis.
SOLUTION We have
M O : rA/O F M O
Where
rA/O (4 in.)i (11 in.) j (27 in.)k F F (cos cos i sin j cos sin k )
i MO F 4 cos cos
j 11 sin
k 27 lb in. cos sin
F [(11cos sin 27sin )i (27 cos cos 4 cos sin ) j (4sin 11cos cos )k ](lb in.) and
M x F (11cos sin 27 sin )(lb in.)
(1)
M y F (27cos cos 4cos sin ) (lb in.)
(2)
M z F (4sin 11cos cos ) (lb in.)
(3)
Now, Equation (1)
cos sin
1 Mx 27 sin 11 F
(4)
and Equation (3)
cos cos
M 1 4sin z F 11
(5)
Substituting Equations (4) and (5) into Equation (2),
1 M 1 M M y F 27 4sin z 4 x 27sin F 11 11 F or
My
1 (27 M z 4 M x ) 11
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PROBLEM 3.50 (Continued)
27 and 114 are the ratios of lengths, have Noting that the ratios 11
27 4 (81 lb ft) (77 lb ft) 11 11 226.82 lb ft
My
or M y 227 lb ft
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PROBLEM 3.51 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the moments about the y and the z axes of the force exerted at B by portion AB of the rope are, respectively, 120 N m and 460 N m, determine the distance a.
SOLUTION First note
BA (2.2 m)i (3.2 m) j (a m)k
Now
M D rA/D TBA
where
rA/D (2.2 m)i (1.6 m) j TBA
TBA (2.2i 3.2 j ak ) (N) d BA
i j k TBA MD 2.2 1.6 0 d BA 2.2 3.2 a
Then
TBA {1.6a i 2.2a j [(2.2)(3.2) (1.6)(2.2)]k} d BA
M y 2.2
Thus
TBA a d BA
M z 10.56
Then forming the ratio
TBA d BA
(N m) (N m)
My Mz T
2.2 dBA (N m) 120 N m BA 460 N m 10.56 TdBA (N m)
or a 1.252 m
BA
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PROBLEM 3.52 To lift a heavy crate, a man uses a block and tackle attached to the bottom of an I-beam at hook B. Knowing that the man applies a 195-N force to end A of the rope and that the moment of that force about the y axis is 132 N m, determine the distance a.
SOLUTION d BA (2.2)2 (3.2)2 (a) 2
First note
15.08 a 2 m 195 N (2.2i 3.2 j a k ) d BA
and
TBA
Now
M y j (rA/D TBA )
where
rA/0 (2.2 m)i (1.6 m) j
Then
0 1 0 195 2.2 1.6 0 My d BA 2.2 3.2 a
195 (2.2a) (N m) d BA
Substituting for My and dBA 132 N m
or
195 15.08 a 2
(2.2a )
0.30769 15.08 a 2 a
Squaring both sides of the equation
0.094675(15.08 a2 ) a2 or a 1.256 m
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PROBLEM 3.53 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ft, determine the magnitude of TDE when TAB 255 lb.
SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at their intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have
M x : (TAB ) z ( y A ) (TDE ) z ( yD ) M x
where
(TAB ) z k TAB k (TAB AB ) i 12 j 12k k 255 lb 17 180 lb (TDE ) z k TDE k (TDE DE ) 1.5i 14 j 12k k TDE 18.5 0.64865TDE y A 12 ft yD 14 ft M x 4728 lb ft (180 lb)(12 ft) (0.64865TDE )(14 ft) 4728 lb ft
and
TDE 282.79 lb
or TDE 283 lb
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PROBLEM 3.54 Solve Problem 3.53 when the tension in cable AB is 306 lb. PROBLEM 3.53 A farmer uses cables and winch pullers B and E to plumb one side of a small barn. If it is known that the sum of the moments about the x-axis of the forces exerted by the cables on the barn at Points A and D is equal to 4728 lb ft, determine the magnitude of TDE when TAB 255 lb.
SOLUTION The moment about the x-axis due to the two cable forces can be found using the z components of each force acting at the intersection with the xy plane (A and D). The x components of the forces are parallel to the x-axis, and the y components of the forces intersect the x-axis. Therefore, neither the x or y components produce a moment about the x-axis. We have
M x : (TAB ) z ( y A ) (TDE ) z ( yD ) M x
Where
(TAB ) z k TAB k (TAB λAB ) i 12 j 12k k 306 lb 17 216 lb (TDE ) z k TDE k (TDE λDE ) 1.5i 14 j 12k k TDE 18.5 0.64865TDE y A 12 ft yD 14 ft M x 4728 lb ft (216 lb)(12 ft) (0.64865TDE )(14 ft) 4728 lb ft
and
TDE 235.21 lb
or TDE 235 lb
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PROBLEM 3.55 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P.
SOLUTION AB (32 in.)i (30 in.) j (24 in.)k AB (32) 2 (30) 2 (24) 2 50 in. AB AB 0.64i 0.60 0.48k AB
We shall apply the force P at Point G: rG /B (5 in.)i (30 in.)k DG (21 in.)i (38 in.) j (18 in.)k DG (21) 2 ( 38) 2 (18) 2 47 in. DG 21i 38 j 18k PP (235 lb) DG 47
P (105 lb)i (190 lb) j (90 lb)k
The moment of P about AB is given by Eq. (3.46):
MAB
0.64 0.60 0.48 AB (rG /B P) 5 in. 0 30 in. 105 lb 190 lb 90 lb
M AB 0.64[0 (30 in.)(190 lb)] 0.60[(30 in.)(105 lb) (5 in.)(90 lb)] 0.48[(5 in.)(190 lb) 0] 2484 lb in. M AB 207 lb ft
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PROBLEM 3.56 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q.
SOLUTION AB (32 in.)i (30 in.) j (24 in.)k AB (32) 2 (30) 2 (24) 2 50 in. AB AB 0.64i 0.60 j 0.48k AB
We shall apply the force Q at Point H: rH /B (32 in.)i (17 in.) j DH (16 in.)i (21 in.) j (12 in.)k DH (16) 2 (21) 2 ( 12) 2 29 in. DH 16i 21j 12k Q (174 lb) DH 29
Q (96 lb)i (126 lb) j (72 lb)k
The moment of Q about AB is given by Eq. (3.46):
MAB
0.64 0.60 0.48 AB (rH /B Q) 32 in. 17 in. 0 96 lb 126 lb 72 lb
M AB 0.64[(17 in.)(72 lb) 0]
0.60[(0 (32 in.)( 72 lb)] 0.48[(32 in.)(126 lb) (17 in.)( 96 lb)] 2119.7 lb in. M AB 176.6 lb ft
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PROBLEM 3.57 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.
SOLUTION M AD AD (rB/A TBH )
Where
1 AD (4i 3k ) 5 rB/A (0.5 m)i
d BH (0.375)2 (0.75)2 (0.75)2
and
1.125 m 450 N (0.375i 0.75 j 0.75k ) 1.125 (150 N)i (300 N) j (300 N)k
TBH
Then
Finally,
MAD
4 0 3 1 0 0.5 0 5 150 300 300 1 [(3)(0.5)(300)] 5
or M AD 90.0 N m
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PROBLEM 3.58 In Problem 3.57, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. PROBLEM 3.57 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.
SOLUTION M AD AD (rB/A TBG )
Where
1 AD (4i 3k ) 5 rB/A (0.5 m) j
BG (0.5)2 (0.925)2 (0.4)2
and
1.125 m 450 N (0.5i 0.925 j 0.4k ) 1.125 (200 N)i (370 N) j (160 N)k
TBG
Then
Finally,
MAD
4 0 3 1 0.5 0 0 5 200 370 160 1 [(3)(0.5)(370)] 5
M AD 111.0 N m
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PROBLEM 3.59 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.
SOLUTION First note:
TAE TAE
AE AE
AE (0.9) 2 (0.6) 2 (0.2) 2 1.1 m
55 N (0.9i 0.6 j 0.2k ) 1.1 5[(9 N)i (6 N) j (2 N)k ]
TAE
Then
DB (1.2)2 (0.35)2 (0)2
Also,
DB
Then
M DB DB (rA/D TAE )
Now where Then
1.25 m DB DB 1 (1.2i 0.35 j) 1.25 1 (24i 7 j) 25
rA /D (0.1 m) j (0.2 m)k
M DB
24 7 0 1 (5) 0 0.1 0.2 25 9 2 6 1 (4.8 12.6 28.8) 5
or M DB 2.28 N m
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PROBLEM 3.60 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.
SOLUTION First note:
TCF TCF
CF CF
CF (0.6) 2 (0.9) 2 (0.2) 2 1.1 m
33 N (0.6i 0.9 j 0.2k ) 1.1 3[(6 N)i (9 N) j (2 N)k ]
Then
TCF
Also,
DB (1.2)2 (0.35)2 (0)2 1.25 m DB DB 1 (1.2i 0.35 j) 1.25 1 (24i 7 j) 25
Then
DB
Now
M DB DB (rC/D TCF )
where
rC/D (0.2 m) j (0.4 m)k
Then
M DB
24 7 0 1 (3) 0 0.2 0.4 25 6 9 2
3 (9.6 16.8 86.4) 25
or M DB 9.50 N m
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PROBLEM 3.61 A regular tetrahedron has six edges of length a. A force P is directed as shown along edge BC. Determine the moment of P about edge OA.
SOLUTION We have From triangle OBC:
M OA OA (rC /O P )
(OA) x
a 2
(OA) z (OA) x tan 30 Since
a 1 a 2 3 2 3
(OA) 2 (OA) 2x (OA) 2y (OAz ) 2 2
or
a a a (OA) 2y 2 2 3
2
2
(OA) y a 2
a2 a2 2 a 4 12 3
a 2 a ia j k 2 3 2 3
Then
rA/O
and
OA i
1 2
2 1 j k 3 2 3
P BC P
(a sin 30)i (a cos 30)k P ( P ) (i 3k ) a 2
rC /O ai
M OA
1 2 1
2 3 0
2 3 P (a ) 0 2
1
0
3
1
aP 2 aP (1)( 3) 2 3 2
M OA
aP
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2
PROBLEM 3.62 A regular tetrahedron has six edges of length a. (a) Show that two opposite edges, such as OA and BC, are perpendicular to each other. (b) Use this property and the result obtained in Problem 3.61 to determine the perpendicular distance between edges OA and BC.
SOLUTION (a)
For edge OA to be perpendicular to edge BC, OA BC 0 From triangle OBC:
(OA) x
a 2
(OA) z (OA) x tan 30
and
a a OA i (OA) y j k 2 2 3 BC (a sin 30) i (a cos 30) k
Then
or
so that (b)
We have M OA
a 1 a 2 3 2 3
a a 3 a i k (i 3 k ) 2 2 2
a a a i (OA) y j k (i 3k ) 0 2 2 3 2 a2 a2 (OA) y (0) 0 4 4 OA BC 0 OA is perpendicular to BC. Pd , with P acting along BC and d the perpendicular distance from OA to BC.
From the results of Problem 3.57, M OA Pa 2
Pa 2
Pd
or d
a
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2
PROBLEM 3.63 Two forces F1 and F2 in space have the same magnitude F. Prove that the moment of F1 about the line of action of F2 is equal to the moment of F2 about the line of action of F1 .
SOLUTION
First note that
F1 F11 and F2 F2 2
Let M 1 moment of F2 about the line of action of F1 and M 2 moment of F1 about the line of action of F2 . Now, by definition,
M 1 1 (rB /A F2 ) 1 (rB /A 2 ) F2 M 2 2 (rA /B F1 ) 2 (rA /B 1 ) F1
Since
F1 F2 F
and rA /B rB /A
M 1 1 (rB /A 2 ) F M 2 2 ( rB /A 1 ) F
Using Equation (3.39): so that
1 (rB /A 2 ) 2 (rB /A 1 ) M 2 1 (rB /A 2 ) F
M 12 M 21
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PROBLEM 3.64 In Prob. 3.55, determine the perpendicular distance between rod AB and the line of action of P. PROBLEM 3.55 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 235-lb force P.
SOLUTION AB (32 in.)i (30 in.) j (24 in.)k AB (32) 2 ( 30) 2 ( 24) 2 50 in. AB AB 0.64i 0.60 j 0.48k AB
P
P 105i 190 j 90k 235 P
Angle between AB and P:
cos AB P (0.64i 0.60 j 0.48k )
105i 190 j 90k 235
0.58723 54.039
The moment of P about AB may be obtained by multiplying the projection of P on a plane perpendicular to AB by the perpendicular distance d from AB to P: M AB ( P sin )d
From the solution to Prob. 3.55: M AB 207 lb ft 2484 lb in. We have
2484 lb in. (235 lb)(sin 54.039) d
d 13.06 in.
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PROBLEM 3.65 In Prob. 3.56, determine the perpendicular distance between rod AB and the line of action of Q. PROBLEM 3.56 The 23-in. vertical rod CD is welded to the midpoint C of the 50-in. rod AB. Determine the moment about AB of the 174-lb force Q.
SOLUTION AB (32 in.)i (30 in.) j (24 in.)k AB (32) 2 ( 30) 2 ( 24) 2 50 in. AB AB 0.64i 0.60 j 0.48k AB
Q
Q 96i 126 j 72k Q 174
Angle between AB and Q: cos AB Q (0.64i 0.60 j 0.48k )
(96i 126 j 72k ) 174
0.28000 73.740
The moment of Q about AB may be obtained by multiplying the projection of Q on a plane perpendicular to AB by the perpendicular distance d from AB to Q: M AB (Q sin )d
From the solution to Prob. 3.56: M AB 176.6 lb ft 2119.2 lb in. 2119.2 lb in. (174 lb)(sin 73.740) d
d 12.69 in.
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PROBLEM 3.66 In Problem 3.57, determine the perpendicular distance between portion BH of the cable and the diagonal AD. PROBLEM 3.57 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.
SOLUTION From the solution to Problem 3.57:
TBH 450 N TBH (150 N)i (300 N) j (300 N)k
| M AD | 90.0 N m
1 AD (4i 3k ) 5 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBH will contribute to the moment of TBH about line AD. (TBH )parallel TBH AD
Now
1 (150i 300 j 300k ) (4i 3k ) 5 1 [(150)(4) (300)(3)] 5 300 N
TBH (TBH )parallel (TBH )perpendicular
Also, so that
(TBH ) perpendicular (450) 2 (300) 2 335.41 N
Since AD and (TBH )perpendicular are perpendicular, it follows that
M AD d (TBH )perpendicular or
90.0 N m d (335.41 N)
d 0.26833 m
d 0.268 m
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PROBLEM 3.67 In Problem 3.58, determine the perpendicular distance between portion BG of the cable and the diagonal AD. PROBLEM 3.58 In Problem 3.57, determine the moment about the diagonal AD of the force exerted on the frame by portion BG of the cable. PROBLEM 3.57 The frame ACD is hinged at A and D and is supported by a cable that passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable.
SOLUTION From the solution to Problem 3.58:
BG 450 N TBG (200 N)i (370 N) j (160 N)k
| M AD | 111 N m
1 AD (4i 3k ) 5 Based on the discussion of Section 3.11, it follows that only the perpendicular component of TBG will contribute to the moment of TBG about line AD. (TBG ) parallel TBG AD
Now
1 (200i 370 j 160k ) (4i 3k ) 5 1 [(200)(4) (160)(3)] 5 64 N
TBG (TBG )parallel (TBG )perpendicular
Also, so that
(TBG ) perpendicular (450) 2 ( 64) 2 445.43 N
Since AD and (TBG )perpendicular are perpendicular, it follows that
M AD d (TBG )perpendicular or
111 N m d (445.43 N)
d 0.24920 m
d 0.249 m
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PROBLEM 3.68 In Problem 3.59, determine the perpendicular distance between cable AE and the line joining Points D and B. PROBLEM 3.59 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable AE at A is 55 N, determine the moment of that force about the line joining Points D and B.
SOLUTION From the solution to Problem 3.59:
AE 55 N TAE 5[(9 N)i (6 N) j (2 N)k ]
| M DB | 2.28 N m
DB
1 (24i 7 j) 25
Based on the discussion of Section 3.11, itfollows that only the perpendicular component of TAE will contribute to the moment of TAE about line DB. (TAE ) parallel TAE DB
Now
5(9i 6 j 2k )
1 (24i 7 j) 25
1 [(9)(24) (6)(7)] 5 51.6 N
TAE (TAE )parallel (TAE )perpendicular
Also, so that
(TAE ) perpendicular (55) 2 (51.6) 2 19.0379 N
Since DB and (TAE )perpendicular are perpendicular, it follows that
M DB d (TAE )perpendicular or
2.28 N m d (19.0379 N)
d 0.119761
d 0.1198 m
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PROBLEM 3.69 In Problem 3.60, determine the perpendicular distance between cable CF and the line joining Points D and B. PROBLEM 3.60 The triangular plate ABC is supported by ball-and-socket joints at B and D and is held in the position shown by cables AE and CF. If the force exerted by cable CF at C is 33 N, determine the moment of that force about the line joining Points D and B.
SOLUTION CF 33 N
From the solution to Problem 3.60:
TCF 3[(6 N)i (9 N) j (2 N)k ]
| M DB | 9.50 N m
DB
1 (24i 7 j) 25
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCF will contribute to the moment of TCF about line DB. (TCF ) parallel TCF DB
Now
3(6i 9 j 2k )
1 (24i 7 j) 25
3 [(6)(24) (9)(7)] 25 24.84 N
TCF (TCF )parallel (TCF )perpendicular
Also, so that
(TCF )perpendicular (33)2 (24.84)2 21.725 N
Since DB and (TCF )perpendicular are perpendicular, it follows that
| M DB | d (TCF )perpendicular or
9.50 N m d 21.725 N or d 0.437 m
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PROBLEM 3.70 Two 80-N forces are applied as shown to the corners B and D of a rectangular plate. (a) Determine the moment of the couple formed by the two forces by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples. (b) Use the result obtained to determine the perpendicular distance between lines BE and DF.
SOLUTION (a)
Resolving forces into components:
P (80 N)sin 50 61.284 N Q (80 N)cos50 51.423 N M (51.423 N)(0.5 m) (61.284 N)(0.3 m) 7.3263 N m
M 7.33 N m
(b)
Distance between lines BE and DF
M Fd or
7.3263 N m=(80 N)d d 0.091579 m
d 91.6 mm
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PROBLEM 3.71 Two parallel 40-N forces are applied to a lever as shown. Determine the moment of the couple formed by the two forces (a) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples, (b) by using the perpendicular distance between the two forces, (c) by summing the moments of the two forces about Point A.
SOLUTION (a)
We have
M B : d1Cx d2C y M
where
d1 (0.270 m)sin 55° 0.22117 m d 2 (0.270 m) cos 55 0.154866 m C x (40 N) cos 20 37.588 N C y (40 N)sin 20 13.6808 N M (0.22117 m)(37.588 N)k (0.154866 m)(13.6808 N)k or M 6.19 N m (6.1946 N m)k
(b)
We have
M Fd (k ) 40 N[(0.270 m)sin(55 20)]( k )
(6.1946 N m)k
(c)
We have
or M 6.19 N m
or M 6.19 N m
M A : (rA F ) rB/A FB rC/A FC M
i j k M (0.390 m)(40 N) cos 55 sin 55 0 cos 20 sin 20 0 i j k (0.660 m)(40 N) cos 55 sin 55 0 cos 20 sin 20 0 (8.9478 N m 15.1424 N m)k (6.1946 N m)k
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PROBLEM 3.72 Four 1 12 -in.-diameter pegs are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that minimum tension?
SOLUTION M (60 lb)(10.5 in.) (40 lb)(13.5 in.)
(a)
630 lb in. 540 lb in.
M 1170 lb in. (b)
With only one string, pegs A and D, or B and C should be used. We have
tan
9 12
36.9
90 53.1
Direction of forces:
(c)
With pegs A and D:
53.1
With pegs B and C:
53.1
The distance between the centers of the two pegs is 122 92 15 in.
Therefore, the perpendicular distance d between the forces is 3 d 15 in. 2 in. 4 16.5 in. We must have
M Fd 1170 lb in. F (16.5 in.)
F 70.9 lb
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PROBLEM 3.73 Four pegs of the same diameter are attached to a board as shown. Two strings are passed around the pegs and pulled with the forces indicated. Determine the diameter of the pegs knowing that the resultant couple applied to the board is 1132.5 lb·in. counterclockwise.
SOLUTION M d AD FAD d BC FBC 1132.5 lb in. [(9 d ) in.](60 lb) [(12 d ) in.](40 lb)
d 1.125 in.
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PROBLEM 3.74 A piece of plywood in which several holes are being drilled successively has been secured to a workbench by means of two nails. Knowing that the drill exerts a 12-N·m couple on the piece of plywood, determine the magnitude of the resulting forces applied to the nails if they are located (a) at A and B, (b) at B and C, (c) at A and C.
SOLUTION (a)
M Fd 12 N m F (0.45 m)
F 26.7 N
(b)
M Fd 12 N m F (0.24 m)
F 50.0 N
(c)
M Fd
d (0.45 m)2 (0.24 m)2 0.510 m
12 N m F (0.510 m)
F 23.5 N
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PROBLEM 3.75 The two shafts of a speed-reducer unit are subjected to couples of magnitude M1 15 lb·ft and M2 3 lb·ft, respectively. Replace the two couples with a single equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION M 1 (15 lb ft)k M 2 (3 lb ft)i
M M12 M 22 (15) 2 (3) 2 15.30 lb ft 15 5 tan x 3
x 78.7 y 90
z 90 78.7 11.30 M 15.30 lb ft; x 78.7, y 90.0, z 11.30
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PROBLEM 3.76 If P 0, replace the two remaining couples with a single equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION M M1 M 2 ; F1 16 lb, F2 40 lb M1 rC F1 (30 in.)i [(16 lb) j] (480 lb in.)k M 2 rE/B F2 ; rE/B (15 in.)i (5 in.) j
d DE (0)2 (5)2 (10)2 5 5 in. F2
40 lb 5 5
(5 j 10k )
8 5[(1 lb) j (2 lb)k ] i j k M 2 8 5 15 5 0 0 1 2 8 5[(10 lb in.)i (30 lb in.) j (15 lb in.)k ] M (480 lb in.)k 8 5[(10 lb in.)i (30 lb in.) j (15 lb in.)k ]
(178.885 lb in.)i (536.66 lb in.) j (211.67 lb in.)k M (178.885)2 (536.66) 2 (211.67) 2
M 604 lb in.
603.99 lb in
M 0.29617i 0.88852 j 0.35045k M cos x 0.29617 axis
cos y 0.88852 cos z 0.35045
x 72.8 y 27.3 z 110.5
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PROBLEM 3.77 If P 20 lb, replace the three couples with a single equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION From the solution to Problem. 3.78: 16-lb force:
M 1 (480 lb in.)k
40-lb force:
M 2 8 5[(10 lb in.)i (30 lb in.) j (15 lb in.)k ]
P 20 lb
M 3 rC P (30 in.)i (20 lb)k (600 lb in.) j
M M1 M 2 M 3 (480)k 8 5 (10i 30 j 15k ) 600 j (178.885 lb in.)i (1136.66 lb in.) j (211.67 lb in.)k M (178.885) 2 (113.66)2 (211.67) 2
M 1170 lb in.
1169.96 lb in. M 0.152898i 0.97154 j 0.180921k M cos x 0.152898 axis
cos y 0.97154 cos z 0.180921
x 81.2 y 13.70 z 100.4
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PROBLEM 3.78 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown: P (50 N)
CD 160 mm (50 N) 40 N CG 200 mm
Q (50 N)
CF 120 mm (50 N) 30 N CG 200 mm
Couple vector M1 perpendicular to plane ABCD: M 1 (40 N)(0.24 m) (30 N)(0.16 m) 4.80 N m
Couple vector M2 in the xy plane: M 2 (12.5 N)(0.192 m) 2.40 N m
144 mm 36.870 192 mm M1 (4.80 cos 36.870) j (4.80 sin 36.870)k 3.84 j 2.88k
tan
M 2 2.40 j M M1 M 2 1.44 j 2.88k M 3.22 N m; x 90.0, y 53.1, z 36.9
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PROBLEM 3.79 Solve Prob. 3.78, assuming that two 10-N vertical forces have been added, one acting upward at C and the other downward at B. PROBLEM 3.78 Replace the two couples shown with a single equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown. P (50 N)
CD 160 mm (50 N) 40 N CG 200 mm
Q (50 N)
CF 120 mm (50 N) 30 N CG 200 mm
Couple vector M1 perpendicular to plane ABCD. M 1 (40 N)(0.24 m) (30 N)(0.16 m) 4.80 N m
144 mm 36.870 192 mm M1 (4.80 cos 36.870) j (4.80sin 36.870)k 3.84 j 2.88k
tan
M 2 (12.5 N)(0.192 m) 2.40 N m 2.40 j
M 3 rB /C M 3 ; rB /C (0.16 m)i (0.144 m) j (0.192 m)k (0.16 m)i (0.144 m) j (0.192 m)k (10 N) j 1.92i 1.6k
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PROBLEM 3.79 (Continued)
M M 1 M 2 M 3 (3.84 j 2.88k ) 2.40 j (1.92i 1.6k ) (1.92 N m)i (1.44 N m) j (1.28 N m)k M ( 1.92) 2 (1.44) 2 (1.28) 2 2.72 N m
M 2.72 N m
cos x 1.92/2.72 cos y 1.44/2.72 cos z 1.28/2.72
x 134.9 y 58.0 z 61.9
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PROBLEM 3.80 Shafts A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and B lie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts by a single equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION M A 1200sin 20 j 1200cos 20 k 410.42 j 1127.63k M B 900sin 20 j 900cos 20 k 307.82 j 845.72k M C 840i
The single equivalent couple is the sum of the three moments M (840 lb ft)i (102.60 lb ft) j (1973.35 lb ft )k M ( 840) 2 (102.60) 2 (1973.35) 2
M 2150 lb ft
2147.3 lb ft
M 0.39119i 0.047921j 0.91899k M cos x 0.39119 axis
cos y 0.047921 cos z 0.91899
x 113.0 y 92.7 z 23.2
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PROBLEM 3.81 A 500-N force is applied to a bent plate as shown. Determine (a) an equivalent force-couple system at B, (b) an equivalent system formed by a vertical force at A and a force at B.
SOLUTION
(a)
Force-couple system at B F (500 N) sin 30 i (500 N) cos 30 k F (250.0 N)i (433.01 N)k
M B (0.3i 0.175 j) (250i 433.01j) 86.153k N m The equivalent force-couple system at B is FB 500 N
(b)
60.0
M B 86.2 N m
Require
Equivalence requires M B M B 86.153 N m A (0.125 m); A (689.22 N)j F A + B or B F - A; B (250 N)i (433.01 N)j (689.22 N)j B (250 N)i (1122.23 N)j 1149.74 N
77.4
B 1150 N
77.4
A 689 N
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PROBLEM 3.82 The tension in the cable attached to the end C of an adjustable boom ABC is 560 lb. Replace the force exerted by the cable at C with an equivalent force-couple system (a) at A, (b) at B.
SOLUTION (a)
Based on
F : FA T 560 lb FA 560 lb
or
20.0°
M A : M A (T sin 50)(d A )
(560 lb)sin 50(18 ft) 7721.7 lb ft M A 7720 lb ft
or (b)
Based on
F : FB T 560 lb
or
FB 560 lb
20.0°
M B : M B (T sin 50)(d B )
(560 lb)sin 50°(10 ft) 4289.8 lb ft or
M B 4290 lb ft
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PROBLEM 3.83 A dirigible is tethered by a cable attached to its cabin at B. If the tension in the cable is 1040 N, replace the force exerted by the cable at B with an equivalent system formed by two parallel forces applied at A and C.
SOLUTION Require the equivalent forces acting at A and C be parallel and at an angle of with the vertical. Then for equivalence, Fx : (1040 N)sin 30 FA sin FB sin
(1)
Fy : (1040 N) cos 30 FA cos FB cos
(2)
Dividing Equation (1) by Equation (2), ( FA FB )sin (1040 N)sin 30 (1040 N) cos 30 ( FA FB ) cos Simplifying yields 30. Based on M C : [(1040 N) cos 30](4 m) ( FA cos 30)(10.7 m) FA 388.79 N
or
FA 389 N
60.0°
Based on M A : [(1040 N) cos 30](6.7 m) ( FC cos 30)(10.7 m) FC 651.21 N
or
FC 651 N
60.0°
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PROBLEM 3.84 A 30-lb vertical force P is applied at A to the bracket shown, which is held by screws at B and C. (a) Replace P with an equivalent force-couple system at B. (b) Find the two horizontal forces at B and C that are equivalent to the couple obtained in part a.
SOLUTION (a)
M B (30 lb)(5 in.)
150.0 lb in.
F 30.0 lb , M B 150.0 lb in. (b)
BC
150 lb in. 50.0 lb 3.0 in.
B 50.0 lb
; C 50.0 lb
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PROBLEM 3.85 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. (a) Replace that force with an equivalent force-couple system at D. (b) Two workers attempt to move the same rock by applying a vertical force at A and another force at D. Determine these two forces if they are to be equivalent to the single force of Part a.
SOLUTION (a)
(a)
We have
(b)
F : 360 N( sin 40°i cos 40 j) (231.40 N)i (275.78 N) j F or F 360 N
We have where
50°
M D : rB/D R M
rB/D [(0.65 m) cos 30]i [(0.65 m)sin 30]j (0.56292 m)i (0.32500 m) j i j k M 0.56292 0.32500 0 N m 231.40 275.78 0 [155.240 75.206)N m]k (230.45 N m)k
(b)
We have where
or M 230 N m
M D : M rA/D FA
rB/D [(1.05 m) cos 30]i [(1.05 m) sin 30]j (0.90933 m)i (0.52500 m) j i j k FA 0.90933 0.52500 0 N m 1 0 0 [230.45 N m]k
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PROBLEM 3.85 (Continued)
(0.90933FA )k 230.45k
or
FA 253.42 N
We have
or FA 253 N
F : F FA FD
(231.40 N)i (275.78 N) j (253.42 N) j FD ( cos i sin j)
From
i : 231.40 N FD cos
(1)
22.36 N FD sin
(2)
j:
Equation (2) divided by Equation (1) tan 0.096629 5.5193 or 5.52
Substitution into Equation (1) FD
231.40 232.48 N cos 5.5193
or FD 232 N
5.52°
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PROBLEM 3.86 A worker tries to move a rock by applying a 360-N force to a steel bar as shown. If two workers attempt to move the same rock by applying a force at A and a parallel force at C, determine these two forces so that they will be equivalent to the single 360-N force shown in the figure.
SOLUTION FX : -360sin40 = FA sin FC sin (1) FY : -360 cos 40 FA cos FC cos (2)
Dividing (1) by (2) yields 40 and then FA FC 360 N (3)
M B : 0 (0.4 m)FA cos10 (0.35 m)FC cos10 7 FC (4) 8 Substituting into (3) ....
FA
7 FC FC 360 8 Finally
FA 168.0 N
50.0°; FC 192.0 N
50.0°;
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PROBLEM 3.87 The shearing forces exerted on the cross section of a steel channel can be represented by a 900-N vertical force and two 250-N horizontal forces as shown. Replace this force and couple with a single force F applied at Point C, and determine the distance x from C to line BD. (Point C is defined as the shear center of the section.)
SOLUTION Replace the 250-N forces with a couple and move the 900-N force to Point C such that its moment about H is equal to the moment of the couple
M H (0.18)(250 N) 45 N m Then or
M H x(900 N)
45 N m x(900 N) x 0.05 m
F 900 N
x 50.0 mm
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PROBLEM 3.88 A force and a couple are applied as shown to the end of a cantilever beam. (a) Replace this system with a single force F applied at Point C, and determine the distance d from C to a line drawn through Points D and E. (b) Solve part a if the directions of the two 360-N forces are reversed.
SOLUTION (a)
We have
F : F (360 N) j (360 N) j (600 N)k
or F (600 N)k and
M D : (360 N)(0.15 m) (600 N)(d )
d 0.09 m
or d 90.0 mm below ED (b)
We have from part a: and
F (600 N)k
M D : (360 N)(0.15 m) (600 N)( d )
d 0.09 m
or d 90.0 mm above ED
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PROBLEM 3.89 Three control rods attached to a lever ABC exert on it the forces shown. (a) Replace the three forces with an equivalent force-couple system at B. (b) Determine the single force that is equivalent to the force-couple system obtained in Part a, and specify its point of application on the lever.
SOLUTION (a)
First note that the two 20-lb forces form A couple. Then
F 48 lb
where
180 (60 55) 65
and
M M B (30 in.)(48 lb) cos 55 (70 in.)(20 lb) cos 20 489.62 lb in
The equivalent force-couple system at B is
F 48.0 lb (b)
65.0°;
M 490 lb in.
The single equivalent force F is equal to F. Further, since the sense of M is clockwise, F must be applied between A and B. For equivalence. M B : M aF cos 55
where a is the distance from B to the point of application of F. Then 489.62 lb in. a(48.0 lb) cos 55
or
a 17.78 in.
F 48.0 lb
65.0°
and is applied to the lever 17.78 in. to the left of pin B
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PROBLEM 3.90 A rectangular plate is acted upon by the force and couple shown. This system is to be replaced with a single equivalent force. (a) For 40, specify the magnitude and the line of action of the equivalent force. (b) Specify the value of if the line of action of the equivalent force is to intersect line CD 300 mm to the right of D.
SOLUTION (a)
The given force-couple system (F, M ) at B is
F 48 N and
M M B (0.4 m)(15 N) cos 40 (0.24 m)(15 N) sin 40
or
M 6.9103 N m
The single equivalent force F is equal to F. Further for equivalence M B : M dF
6.9103 N m d 48 N
or
d 0.14396 m
or
F 48.0 N
and the line of action of F intersects line AB 144.0 mm to the right of A. (b)
Following the solution to Part a but with d 0.1 m and unknown, have M B : (0.4 m)(15 N) cos (0.24 m)(15 N)sin
(0.1 m)(48 N)
or
5cos 3sin 4
Rearranging and squaring
25 cos 2 (4 3 sin ) 2
Using cos 2 1 sin 2 and expanding 25(1 sin 2 ) 16 24 sin 9 sin 2
or Then
34 sin 2 24 sin 9 0
24 (24) 2 4(34)(9) 2(34) sin 0.97686 or sin 0.27098 sin
77.7 or 15.72
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PROBLEM 3.91 While tapping a hole, a machinist applies the horizontal forces shown to the handle of the tap wrench. Show that these forces are equivalent to a single force, and specify, if possible, the point of application of the single force on the handle.
SOLUTION Since the forces at A and B are parallel, the force at B can be replaced with the sum of two forces with one of the forces equal in magnitude to the force at A except with an opposite sense, resulting in a forcecouple. We have FB 2.9 lb 2.65 lb 0.25 lb, where the 2.65-lb force is part of the couple. Combining the two parallel forces, M couple (2.65 lb)[(3.2 in. 2.8 in.) cos 25] 14.4103 lb in. M couple 14.4103 lb in.
and
A single equivalent force will be located in the negative z direction. Based on
M B : 14.4103 lb in. [(0.25 lb) cos 25](a)
a 63.600 in.
F (0.25 lb)(cos 25i sin 25k ) F (0.227 lb)i (0.1057 lb)k and is applied on an extension of handle BD at a distance of 63.6 in. to the right of B.
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PROBLEM 3.92 A hexagonal plate is acted upon by the force P and the couple shown. Determine the magnitude and the direction of the smallest force P for which this system can be replaced with a single force at E.
SOLUTION From the statement of the problem, it follows that M E 0 for the given force-couple system. Further, for Pmin, we must require that P be perpendicular to rB/E . Then M E : (0.2 sin 30 0.2)m 300 N (0.2 m)sin 30 300 N (0.4 m) Pmin 0
or
Pmin 300 N Pmin 300 N
30.0°
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PROBLEM 3.93 An eccentric, compressive 250-kN force P is applied to the end of a column. Replace P with an equivalent force-couple system at G.
SOLUTION
F :
Have
250 kN j F
or F 250 kN j Also have
M G :
rP P M
i j k 0.030 0 0.060 kN m = M 0 250 0 M 15 kN m i 7.5 kN m k
or M 15.00 kN m i 7.50 kN m k
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PROBLEM 3.94 A 2.6-kip force is applied at Point D of the cast iron post shown. Replace that force with an equivalent force-couple system at the center A of the base section.
SOLUTION DE (12 in.) j (5 in.)k ; DE 13.00 in. DE F (2.6 kips) DE
F (2.6 kips)
12 j 5k 13 F (2.40 kips) j (1.000 kip)k
M A rD /A F
where
rD /A (6 in.)i (12 in.) j i j k 12 in. 0 M A 6 in. 0 2.4 kips 1.0 kips M A (12.00 kip in.)i (6.00 kip in.) j (14.40 kip in.)k
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PROBLEM 3.95 Replace the 150-N force with an equivalent force-couple system at A.
SOLUTION Equivalence requires
F : F (150 N)( cos 35 j sin 35k ) (122.873 N) j (86.036 N)k MA : M rD/A F
where
Then
rD/A (0.18 m)i (0.12 m) j (0.1 m)k
i j k 0.1 N m 0.12 M 0.18 0 122.873 86.036 [(0.12)(86.036) (0.1)(122.873)]i [(0.18)(86.036)]j [(0.18)(122.873)]k (22.6 N m)i (15.49 N m) j (22.1 N m)k
The equivalent force-couple system at A is F (122.9 N) j (86.0 N)k M (22.6 N m)i (15.49 N m) j (22.1 N m)k
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PROBLEM 3.96 To keep a door closed, a wooden stick is wedged between the floor and the doorknob. The stick exerts at B a 175-N force directed along line AB. Replace that force with an equivalent force-couple system at C.
SOLUTION We have F : PAB FC
where PAB λAB PAB
(33 mm)i (990 mm) j (594 mm)k (175 N) 1155.00 mm
or FC (5.00 N)i (150.0 N) j (90.0 N)k We have
M C : rB/C PAB M C
i j k M C 5 0.683 0.860 0 N m 1 30 18 (5){( 0.860)(18)i (0.683)(18) j [(0.683)(30) (0.860)(1)]k} or M C (77.4 N m)i (61.5 N m) j (106.8 N m)k
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PROBLEM 3.97 A 46-lb force F and a 2120-lbin. couple M are applied to corner A of the block shown. Replace the given force-couple system with an equivalent force-couple system at corner H.
SOLUTION We have
d AJ (18) 2 (14) 2 (3) 2 23 in.
46 lb (18i 14 j 3k ) 23 (36 lb)i (28 lb) j (6 lb)k
F
Then
d AC ( 45) 2 (0) 2 ( 28) 2 53 in.
Also
2120 lb in. (45i 28k ) 53 (1800 lb in.)i (1120 lb in.)k
Then
M
Now
M M rA/H F
where Then
rA/H (45 in.)i (14 in.) j i j k 0 M (1800i 1120k ) 45 14 36 28 6
(1800i 1120k ) {[(14)( 6)]i [ (45)( 6)]j [(45)(28) (14)(36)]k} (1800 84)i (270) j (1120 1764)k (1884 lb in.)i (270 lb in.)j (2884 lb in.)k (157 lb ft)i (22.5 lb ft) j (240 lb ft)k
The equivalent force-couple system at H is
F (36.0 lb)i (28.0 lb) j (6.00 lb)k
M (157.0 lb ft)i (22.5 lb ft) j (240 lb ft)k
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PROBLEM 3.98 A 110-N force acting in a vertical plane parallel to the yzplane is applied to the 220-mm-long horizontal handle AB of a socket wrench. Replace the force with an equivalent force-couple system at the origin O of the coordinate system.
SOLUTION We have
F : PB F
where
PB 110 N[ (sin15) j (cos15)k ] (28.470 N) j (106.252 N)k
or F (28.5 N) j (106.3 N)k
We have where
M O : rB/O PB M O rB /O [(0.22 cos 35)i (0.15) j (0.22sin 35)k ] m (0.180213 m)i (0.15 m) j (0.126187 m)k
i j k 0.180213 0.15 0.126187 N m M O 0 28.5 106.3 M O [(12.3487)i (19.1566) j (5.1361)k ] N m or M O (12.35 N m)i (19.16 N m)j (5.13 N m)k
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PROBLEM 3.99 An antenna is guyed by three cables as shown. Knowing that the tension in cable AB is 288 lb, replace the force exerted at A by cable AB with an equivalent force-couple system at the center O of the base of the antenna.
SOLUTION We have
d AB (64) 2 ( 128) 2 (16) 2 144 ft
Then
TAB
Now
288 lb (64i 128 j 16k ) 144 (32 lb)(4i 8 j k )
M M O rA / O TAB 128 j 32( 4i 8 j k ) (4096 lb ft)i (16,384 lb ft)k
The equivalent force-couple system at O is F (128.0 lb)i (256 lb) j (32.0 lb)k M (4.10 kip ft)i (16.38 kip ft)k
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PROBLEM 3.100 An antenna is guyed by three cables as shown. Knowing that the tension in cable AD is 270 lb, replace the force exerted at A by cable AD with an equivalent force-couple system at the center O of the base of the antenna.
SOLUTION We have
d AD (64) 2 (128) 2 (128) 2
192 ft
Then
Now
270 lb (64i 128 j 128k ) 192 (90 lb)(i 2 j 2k )
TAD
M M O rA/O TAD 128 j 90(i 2 j 2k ) (23, 040 lb ft)i (11,520 lb ft)k
The equivalent force-couple system at O is F (90.0 lb)i (180.0 lb) j (180.0 lb)k M (23.0 kip ft)i (11.52 kip ft)k
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PROBLEM 3.101 A 3-m-long beam is subjected to a variety of loadings. (a) Replace each loading with an equivalent forcecouple system at end A of the beam. (b) Which of the loadings are equivalent?
SOLUTION
(a)
(a) We have
FY : 300 N 200 N Ra
or R a 500 N and
M A : 400 N m (200 N)(3 m) M a
or M a 1000 N m (b) We have
FY : 200 N 300 N Rb
or R b 500 N and
M A : 400 N m (300 N)(3 m) M b
or M b 500 N m (c) We have
FY : 200 N 300 N Rc
or R c 500 N and
M A : 400 N m (300 N)(3 m) M c
or M c 500 N m
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PROBLEM 3.101 (Continued)
(d) We have
FY : 500 N Rd
or R d 500 N and
M A : 400 N m (500 N)(3 m) M d
or M d 1100 N m (e) We have
FY : 300 N 800 N Re
or R e 500 N and
M A : 400 N m 1000 N m (800 N)(3 m) M e
or M e 1000 N m (f ) We have
FY : 300 N 200 N R f
or R f 500 N and
M A : 400 N m (200 N)(3 m) M f
or M f 200 N m (g) We have
FY : 800 N 300 N Rg
or R g 500 N and
M A : 1000 N m 400 N m (300 N)(3 m) M g
or M g 2300 N m (h) We have
FY : 250 N 250 N Rh
or R h 500 N and
M A : 1000 N m 400 N m (250 N)(3 m) M h
or M h 650 N m (b)
Therefore, loadings (a) and (e) are equivalent.
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PROBLEM 3.102 A 3-m-long beam is loaded as shown. Determine the loading of Prob. 3.101 that is equivalent to this loading.
SOLUTION
FY : 200 N 300 N R
We have
R 500 N
or
M A : 500 N m 200 N m (300 N)(3 m) M
and
M 200 N m
or
Problem 3.101 equivalent force-couples at A:
Case
R
M
(a)
500 N
1000 Nm
(b)
500 N
500 Nm
(c)
500 N
500 Nm
(d)
500 N
1100 Nm
(e)
500 N
1000 Nm
(f )
500 N
200 Nm
(g)
500 N
2300 Nm
(h)
500 N
650 Nm Equivalent to case (f ) of Problem 3.101
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PROBLEM 3.103 Determine the single equivalent force and the distance from Point A to its line of action for the beam and loading of (a) Prob. 3.101a, (b) Prob. 3.101b, (c) Prob. 3.102.
SOLUTION For equivalent single force at distance d from A: (a)
We have
FY : 300 N 200 N R
or R 500 N and
M C : 400 N m (300 N)( d ) (200 N)(3 d ) 0
or d 2.00 m
(b)
We have
FY : 200 N 300 N R
or R 500 N and
M C : 400 N m (200 N)( d ) (300 N)(3 d ) 0
or d 1.000 m
(c)
We have
FY : 200 N 300 N R
or R 500 N and
M C : 500 N m 200 N m (200 N)(d ) (300 N)(3 d ) 0
or d 0.400 m
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PROBLEM 3.104 Five separate force-couple systems act at the corners of a piece of sheet metal, which has been bent into the shape shown. Determine which of these systems is equivalent to a force F (10 lb)i and a couple of moment M (15 lb ft)j (15 lb ft)k located at the origin.
SOLUTION First note that the force-couple system at F cannot be equivalent because of the direction of the force [The force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces remain unchanged. A: M A M O (5 lb ft) j (15 lb ft)k (2 ft)k (10 lb)i
(25 lb ft) j (15 lb ft)k D : M D M O (5 lb ft) j (25 lb ft)k
[(4.5 ft)i (1 ft) j (2 ft)k ] 10 lb)i (15 lb ft)i (15 lb ft)k G : M G M O (15 lb ft)i (15 lb ft) j I : M I M I (15 lb ft) j (5 lb ft)k
[(4.5 ft)i (1 ft) j] (10 lb) j (15 lb ft) j (15 lb ft)k
The equivalent force-couple system is the system at corner D.
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PROBLEM 3.105 The weights of two children sitting at ends A and B of a seesaw are 84 lb and 64 lb, respectively. Where should a third child sit so that the resultant of the weights of the three children will pass through C if she weighs (a) 60 lb, (b) 52 lb.
SOLUTION
(a)
For the resultant weight to act at C, Then
M C 0 WC 60 lb
(84 lb)(6 ft) 60 lb(d ) 64 lb(6 ft) 0 d 2.00 ft to the right of C
(b)
For the resultant weight to act at C, Then
M C 0 WC 52 lb
(84 lb)(6 ft) 52 lb(d ) 64 lb(6 ft) 0 d 2.31 ft to the right of C
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PROBLEM 3.106 Three stage lights are mounted on a pipe as shown. The lights at A and B each weigh 4.1 lb, while the one at C weighs 3.5 lb. (a) If d 25 in., determine the distance from D to the line of action of the resultant of the weights of the three lights. (b) Determine the value of d so that the resultant of the weights passes through the midpoint of the pipe.
SOLUTION
For equivalence Fy : 4.1 4.1 3.5 R or R 11.7 lb
FD : (10 in.)(4.1 lb) (44 in.)(4.1 lb) [(4.4 d )in.](3.5 lb) ( L in.)(11.7 lb) 375.4 3.5d 11.7 L (d , L in in.)
or
d 25 in.
(a) We have
375.4 3.5(25) 11.7 L or
L 39.6 in.
The resultant passes through a Point 39.6 in. to the right of D.
L 42 in.
(b) We have
375.4 3.5d 11.7(42)
or d 33.1 in.
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PROBLEM 3.107 A beam supports three loads of given magnitude and a fourth load whose magnitude is a function of position. If b 1.5 m and the loads are to be replaced with a single equivalent force, determine (a) the value of a so that the distance from support A to the line of action of the equivalent force is maximum, (b) the magnitude of the equivalent force and its point of application on the beam.
SOLUTION
For equivalence,
Fy : 1300 400
a 400 600 R b
a R 2300 400 N b
or
M A :
a a 400 a (400) (a b)(600) LR b 2
L
or
1000a 600b 200 2300 400
Then with
(1)
b 1.5 m
a b
a2 b
4 10a 9 a 2 3 L 8 23 a 3
where a, L are in m. (a)
Find value of a to maximize L. 8 8 4 8 10 a 23 a 10a 9 a 2 dL 3 3 3 3 2 da 8 23 3 a
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(2)
PROBLEM 3.107 (Continued)
230
or
16a 2 276a 1143 0 276 (276) 2 4(16)(1143) 2(16)
Then
a
or
a 10.3435 m and a 6.9065 m
Since (b)
184 80 64 2 80 32 a a a a 24 a 2 0 3 3 9 3 9
or
AB 9 m, a must be less than 9 m 6.9065 1.5
a 6.91 m or R 458 N
Using Eq. (1),
R 2300 400
and using Eq. (2),
4 10(6.9065) 9 (6.9065)2 3 L 3.16 m 8 23 (6.9065) 3 R is applied 3.16 m to the right of A.
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PROBLEM 3.108 A 6 × 12-in. plate is subjected to four loads as shown. Find the resultant of the four loads and the two points at which the line of action of the resultant intersects the edge of the plate.
SOLUTION We have
For equivalence,
R (40 lb)i (20 lb)j
R 44.721 lb
R 44.7 lb
26.6°
M C rB / C B = (6 in.)i (50 lb)(cos45 i sin 45 j) = (212.13 lb in.)k
Intersection with edge AC M C ( xi ) (20) j 212.13k (20 x)k x 10.61 in. 10.61 in. to the left of C. Intersection with edge CD M C ( yj) (40i ) 213.13k (40 y )k y 5.30 in.
5.30 in. below C.
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PROBLEM 3.109 A 32-lb motor is mounted on the floor. Find the resultant of the weight and the forces exerted on the belt, and determine where the line of action of the resultant intersects the floor.
SOLUTION We have F : (60 lb)i (32 lb) j (140 lb)(cos 30i sin 30 j) R R (181.244 lb)i (38.0 lb) j
or R 185.2 lb We have
11.84°
M O : M O xR y
[(140 lb) cos 30][(4 2 cos 30)in.] [(140 lb)sin 30][(2 in.)sin 30] (60 lb)(2 in.) x(38.0 lb)
x and
1 ( 694.97 70.0 120) in. 38.0
x 23.289 in.
Or resultant intersects the base (x-axis) 23.3 in. to the left of the vertical centerline (y-axis) of the motor.
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PROBLEM 3.110 To test the strength of a 625 500-mm suitcase, forces are applied as shown. If P = 88 N, (a) determine the resultant of the applied forces, (b) locate the two points where the line of action of the resultant intersects the edge of the suitcase.
SOLUTION We have
First replace the applied forces and couples with an equivalent force-couple system at B. 45 Fx : 100 (212) 180 Rx ; Rx 100 N (a )Thus, 53 28 Fy : (212) 88 Ry ; R y 200 N 53 R (100 N)i (200 N)j R 224 N
63.4°
(b) 28 M B : (0.1 m)(100 N) (0.53 m) (212 N) 53 (0.08 m)(88 N) (0.28 m)(180 N) M B M B 26.0 N m
(1)
The single equivalent force R must then act as indicated. Then with R at E: M B : 26.0 N m x(200 N) x 130 mm
y 2 or y 260 mm x 1 Therefore, the line of action of R intersects top AB 130.0 mm to the left of B and intersects side BC 260 mm below B.
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PROBLEM 3.111 Solve Prob. 3.110, assuming that P = 138 N.
Problem 3.110 To test the strength of a 625 500-mm suitcase, forces are applied as shown. If P = 88 N, (a) determine the resultant of the applied forces, (b) locate the two points where the line of action of the resultant intersects the edge of the suitcase.
SOLUTION We have
First replace the applied forces and couples with an equivalent force-couple system at B. (a) Thus,
Fx : 100 Fy :
45 (212) 180 Rx ; Rx 100 N 53
28 (212) 138 Ry ; 53
R y 250 N
R 269 N
68.2°
(b) 28 M B : (0.1 m)(100 N) (0.53 m) (212 N) 53 (0.08 m)(138 N) (0.28 m)(180 N) M B M B 30.0 N m
The single equivalent force R must then act as indicated. Then with R at E: M B : 30 N m x(250 N)
x 120 mm y 5 or y 300 mm x 2 Therefore, the line of action of R intersects top AB 120 mm to the left of B and intersects side BC 300 mm below B.
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PROBLEM 3.112 Pulleys A and B are mounted on bracket CDEF. The tension on each side of the two belts is as shown. Replace the four forces with a single equivalent force, and determine where its line of action intersects the bottom edge of the bracket.
SOLUTION Equivalent force-couple at A due to belts on pulley A We have
F : 120 lb 160 lb RA R A 280 lb
We have
M A :
40 lb(2 in.) M A M A 80 lb in.
Equivalent force-couple at B due to belts on pulley B We have
F: (210 lb 150 lb)
25 R B R B 360 lb
We have
25°
M B : 60 lb(1.5 in.) M B M B 90 lb in.
Equivalent force-couple at F We have
F: R F ( 280 lb) j (360 lb)(cos 25i sin 25 j)
(326.27 lb)i (127.857 lb) j R RF 2 2 RFx RFy
(326.27) 2 (127.857) 2 350.43 lb RFy RFx 127.857 tan 1 326.27 21.399
tan 1
or R F R 350 lb
21.4°
PROBLEM 3.112 (Continued)
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We have
M F : M F (280 lb)(6 in.) 80 lb in.
[(360 lb) cos 25](1.0 in.) [(360 lb)sin 25](12 in.) 90 lb in. M F (350.56 lb in.)k
To determine where a single resultant force will intersect line FE,
M F dRy d
MF Ry 350.56 lb in. 127 857 lb
2.7418 in.
or d 2.74 in.
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PROBLEM 3.113 A truss supports the loading shown. Determine the equivalent force acting on the truss and the point of intersection of its line of action with a line drawn through Points A and G.
SOLUTION We have
R F R (240 lb)(cos 70i sin 70 j) (160 lb) j (300 lb)( cos 40i sin 40 j) (180 lb) j
R (147.728 lb)i (758.36 lb) j R Rx2 Ry2 (147.728) 2 (758.36)2 772.62 lb Ry Rx 758.36 tan 1 147.728 78.977
tan 1
We have
M A dR y
where
M A [240 lb cos 70](6 ft) [240 lb sin 70](4 ft)
or
R 773 lb
79.0
(160 lb)(12 ft) [300 lb cos 40](6 ft) [300 lb sin 40](20 ft) (180 lb)(8 ft) 7232.5 lb ft
7232.5 lb ft 758.36 lb 9.5370 ft
d
or
d 9.54 ft to the right of A
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PROBLEM 3.114 A couple of magnitude M = 80 lb·in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC.
SOLUTION (a)
We have
F : R (10 j) (25 cos 60)i (25 sin 60) j (40i ) (27.50 lb)i (11.6506 lb) j
or R 29.9 lb (b)
23.0°
First reduce the given forces and couple to an equivalent force-couple system ( R , M B ) at B. We have
M B : M B (80 lb in) (12 in.)(10 lb) (8 in.)(40 lb) 120.0 lb in.
Then with R at D, or and with R at E, or
M B : 120.0 lb in a (11.6506 lb)
a 10.30 in. M B : 120.0 lb in c (27.5 lb)
c 4.36 in.
The line of action of R intersects line AB 10.30 in. to the left of B and intersects line BC 4.36 in. below B.
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PROBLEM 3.115 A couple M and the three forces shown are applied to an angle bracket. Find the moment of the couple if the line of action of the resultant of the force system is to pass through (a) point A, (b) point B, (c) point C.
SOLUTION In each case, we must have M1R 0 (a)
M AB M A M (12 in.)[(25 lb) sin 60] (8 in.)(40 lb) 0
M 60.192 lb in. (b)
M 200 lb in.
M 20.0 lb in.
M BR M B M (12 in.)(10 lb) (8 in.)(40 lb) 0
M 200 lb in. (c)
M 60.2 lb in.
M CR M C M (12 in.)(10 lb) (8 in.)[(25 lb) cos 60] 0
M 20.0 lb in.
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PROBLEM 3.116 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.
SOLUTION We have
First replace the applied forces and couples with an equivalent force-couple system at G. Fx : 200 cos 15 120 cos 70 P Rx
Thus,
Rx (152.142 P ) N
or
Fy : 200sin 15 120sin 70 80 R y R y 244.53 N
or
M G : (0.47 m)(200 N) cos15 (0.05 m)(200 N) sin15 (0.47 m)(120 N) cos 70 (0.19 m)(120 N) sin 70 (0.13 m)( P N) (0.59 m)(80 N) 42 N m 40 N m M G M G (55.544 0.13P ) N m
or
(1)
Setting P 0 in Eq. (1): Now with R at I,
M G : 55.544 N m a (244.53 N)
a 0.227 m
or and with R at J,
M G : 55.544 N m b(152.142 N)
b 0.365 m
or (a)
The rivet hole is 0.365 m above G.
(b)
The rivet hole is 0.227 m to the right of G.
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PROBLEM 3.117 Solve Problem 3.116, assuming that P 60 N. PROBLEM 3.116 A machine component is subjected to the forces and couples shown. The component is to be held in place by a single rivet that can resist a force but not a couple. For P 0, determine the location of the rivet hole if it is to be located (a) on line FG, (b) on line GH.
SOLUTION See the solution to Problem 3.116 leading to the development of Equation (1): M G (55.544 0.13P ) N m Rx (152.142 P ) N
and
P 60 N
For we have
Rx (152.142 60) 212.14 N M G [55.544 0.13(60)] 63.344 N m
Then with R at I,
M G : 63.344 N m a (244.53 N)
a 0.259 m
or and with R at J,
M G : 63.344 N m b(212.14 N)
b 0.299 m
or (a)
The rivet hole is 0.299 m above G.
(b)
The rivet hole is 0.259 m to the right of G.
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PROBLEM 3.118 As follower AB rolls along the surface of member C, it exerts a constant force F perpendicular to the surface. (a) Replace F with an equivalent force-couple system at Point D obtained by drawing the perpendicular from the point of contact to the x-axis. (b) For a 1 m and b 2 m, determine the value of x for which the moment of the equivalent forcecouple system at D is maximum.
SOLUTION (a)
The slope of any tangent to the surface of member C is
dy d x2 b 1 2 dx dx a
2b 2 x a
Since the force F is perpendicular to the surface, dy tan dx
1
a2 1 2b x
For equivalence, F : F R M D : ( F cos )( y A ) M D
where cos
2bx 2 2
(a ) (2bx)2
x2 y A b 1 2 a
x3 2 Fb 2 x 2 a MD a 4 4b 2 x 2 Therefore, the equivalent force-couple system at D is
RF
a2 tan 1 2bx
x3 2 Fb2 x 2 a M 4 2 2 a 4b x
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PROBLEM 3.118 (Continued)
(b)
To maximize M, the value of x must satisfy
dM 0 dx
a 1 m, b 2 m
where for
M
8 F ( x x3 ) 1 16 x 2
1 1 16 x 2 (1 3x 2 ) ( x x3 ) (32 x)(1 16 x 2 )1/ 2 dM 2 0 8F 2 dx (1 16 x ) (1 16 x 2 )(1 3 x 2 ) 16 x( x x3 ) 0 32 x 4 3 x 2 1 0
or x2
3 9 4(32)( 1) 0.136011 m 2 2(32)
Using the positive value of x2:
x 0.36880 m
and 0.22976 m 2
or x 0.369 m
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PROBLEM 3.119 A machine component is subjected to the forces shown, each of which is parallel to one of the coordinate axes. Replace these forces with an equivalent force-couple system at A.
SOLUTION
For equivalence F : FB FC FD R A R A 240 N j 125 N k 300 N i 150 N k R A 300 N i 240 N j 25.0 N k
Also for equivalence : rB/ A FB rC/ A FC rD/ A FD M A
or M A
i j k i j k i j k 0 0.12 m 0 0.06 m 0.03 m 0.075 m 0.06 m 0.08 m 0.75 m 0 240 N 125 N 300 N 0 0 0 0 150 N 15 N m i 22.5 N m j 9 N m k 12 N m i 9 N m j
or
M A 3.00 N m i 13.50 N m j 9.00 N m k
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PROBLEM 3.120 Two 150-mm-diameter pulleys are mounted on line shaft AD. The belts at B and C lie in vertical planes parallel to the yz-plane. Replace the belt forces shown with an equivalent force-couple system at A.
SOLUTION Equivalent force-couple at each pulley: Pulley B:
R B (145 N)( cos 20 j sin 20k ) 215 Nj (351.26 N) j (49.593 N)k
M B (215 N 145 N)(0.075 m)i (5.25 N m)i
Pulley C:
R C (155 N 240 N)( sin10 j cos10k ) (68.591 N) j (389.00 N)k
M C (240 N 155 N)(0.075 m)i (6.3750 N m)i
Then
R R B R C (419.85 N) j (339.41)k
or R (420 N) j (339 N)k
M A M B M C rB/ A R B rC/ A R C i j k (5.25 N m)i (6.3750 N m)i 0.225 0 0 Nm 351.26 49.593 0 i j k + 0.45 0 0 Nm 68.591 389.00 0 (1.12500 N m)i (163.892 N m) j (109.899 N m)k or M A (1.125 N m)i (163.9 N m) j (109.9 N m)k
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PROBLEM 3.121 As an adjustable brace BC is used to bring a wall into plumb, the force-couple system shown is exerted on the wall. Replace this force-couple system with an equivalent force-couple system at A if R 21.2 lb and M 13.25 lb · ft.
SOLUTION
F : R R A RBC
We have where λBC
RA
(42 in.)i (96 in.) j (16 in.)k 106 in.
21.2 lb (42i 96 j 16k ) 106 or
R A (8.40 lb)i (19.20 lb) j (3.20 lb)k
M A : rC/A R M M A
We have
where rC/A (42 in.)i (48 in.)k
1 (42i 48k )ft 12
(3.5 ft)i (4.0 ft)k R (8.40 lb)i (19.50 lb) j (3.20 lb)k M BC M 42i 96 j 16k (13.25 lb ft) 106 (5.25 lb ft)i (12 lb ft) j (2 lb ft)k
Then
i j k 3.5 0 4.0 lb ft (5.25i 12 j 2k ) lb ft M A 8.40 19.20 3.20 M A (71.55 lb ft)i (56.80 lb ft)j (65.20 lb ft)k
or M A (71.6 lb ft)i (56.8 lb ft)j (65.2 lb ft)k
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PROBLEM 3.122 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied.
SOLUTION We first reduce the given forces to force-couple systems at A and B, noting that | M A | | M B | (40 lb)(10 in.) 400 lb in.
We now determine the equivalent force-couple system at C. R (40 lb)(1 cos )i (40 lb)sin j
(1)
M CR M A M B (15 in.)k [(40 lb) cos i (40 lb)sin j] (7.5 in.)k (40 lb)i 400 400 600 cos j 600sin i 300 j (600 lb in.)sin i (300 lb in.)(1 2cos ) j (a)
(2)
For no rotation about vertical, y component of M CR must be zero. 1 2 cos 0 cos 1/2
60.0 (b)
For 60.0 in Eqs. (1) and (2), R (20.0 lb)i (34.641 lb) j; M CR (519.62 lb in.)i R (20.0 lb)i (34.6 lb) j; M CR (520 lb in.)i
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PROBLEM 3.123 Assuming θ 60° in Prob. 3.122, replace the two 40-lb forces with an equivalent forcecouple system at D and determine whether the plumber’s action tends to tighten or loosen the joint between (a) pipe CD and elbow D, (b) elbow D and pipe DE. Assume all threads to be right-handed. PROBLEM 3.122 In order to unscrew the tapped faucet A, a plumber uses two pipe wrenches as shown. By exerting a 40-lb force on each wrench, at a distance of 10 in. from the axis of the pipe and in a direction perpendicular to the pipe and to the wrench, he prevents the pipe from rotating, and thus avoids loosening or further tightening the joint between the pipe and the tapped elbow C. Determine (a) the angle θ that the wrench at A should form with the vertical if elbow C is not to rotate about the vertical, (b) the force-couple system at C equivalent to the two 40-lb forces when this condition is satisfied.
SOLUTION The equivalent force-couple system at C for 60 was obtained in the solution to Prob. 3.122: R (20.0 lb)i (34.641 lb) j M CR (519.62 lb in.)i
The equivalent force-couple system at D is made of R and M RD where M RD M CR rC /D R (519.62 lb in.)i (25.0 in.) j [(20.0 lb)i (34.641 lb) j] (519.62 lb in.)i (500 lb in.)k
Equivalent force-couple at D: R (20.0 lb)i (34.6 lb) j; M CR (520 lb in.)i (500 lb in.)k
(a) (b)
Since M RD has no component along the y-axis, the plumber’s action will neither loosen nor tighten the joint between pipe CD and elbow.
Since the x component of M RD is , the plumber’s action will tend to tighten the joint between elbow and pipe DE.
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PROBLEM 3.124 Four forces are applied to the machine component ABDE as shown. Replace these forces with an equivalent force-couple system at A.
SOLUTION R (50 N) j (300 N)i (120 N)i (250 N)k R (420 N)i (50 N)j (250 N)k rB (0.2 m)i rD (0.2 m)i (0.16 m)k rE (0.2 m)i (0.1 m) j (0.16 m)k
M RA rB [(300 N)i (50 N) j] rD (250 N)k r ( 120 N)i i j k i j k 0.2 m 0 0 0.2 m 0 0.16 m 300 N 50 N 0 0 0 250 N i j k 0.2 m 0.1 m 0.16 m 120 N 0 0 (10 N m)k (50 N m) j (19.2 N m) j (12 N m)k Force-couple system at A is R (420 N)i (50 N) j (250 N)k
M RA (30.8 N m) j (22.0 N m)k
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PROBLEM 3.125 A blade held in a brace is used to tighten a screw at A. (a) Determine the forces exerted at B and C, knowing that these forces are equivalent to a forcecouple system at A consisting of R = (25 N)i + Ryj + Rzk and M RA = – (13.5 N·m)i. (b) Find the corresponding values of Ry and Rz. (c) What is the orientation of the slot in the head of the screw for which the blade is least likely to slip when the brace is in the position shown?
SOLUTION (a)
Equivalence requires or Equating the i coefficients: Also, or
F : R B C (25 N)i R y j Rz k Bk ( C x i C y j C z k )
i : 25 N C x
or C x 25 N
M A : M RA rB/A B rC/A C
(13.5 N m)i [(0.2 m)i (0.15 m)j] ( B )k (0.4 m)i [(25 N)i C y j C z k ]
Equating coefficients:
i : 13.5 N m (0.15 m) B k : 0 (0.4 m)C y
or or
j: 0 (0.2 m)(75 N) (0.4 m)C z or
B 75 N Cy 0 C z 37.5 N
B (75.0 N)k C (25.0 N)i (37.5 N)k
(b)
Now we have for the equivalence of forces (25 N)i R y j Rz k (75 N)k [( 25 N)i (37.5 N)k ]
Equating coefficients:
j: R y 0
k : Rz 75 37.5
(c)
R y 0
or
Rz 37.5 N
First note that R (25 N)i (37.5 N)k. Thus, the screw is best able to resist the lateral force Rz when the slot in the head of the screw is vertical.
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PROBLEM 3.126 A mechanic uses a crowfoot wrench to loosen a bolt at C. The mechanic holds the socket wrench handle at Points A and B and applies forces at these points. Knowing that these forces are equivalent to a force-couple system at C consisting of the force C (8 lb)i + (4 lb)k and the couple M C (360 lb · in.)i, determine the forces applied at A and at B when Az 2 lb.
SOLUTION F :
We have or
ABC
Fx : Ax Bx 8 lb Bx ( Ax 8 lb)
(1)
Fy : Ay By 0 Ay B y (2)
or
Fz : 2 lb Bz 4 lb or
Bz 2 lb
(3) M C : rB/C B rA/C A M C
We have
or
i 8
j 0
k i 2 8
j 0
k 8
Bx
By
2
Ay
2
Ax
lb in. (360 lb in.)i
(2 B y 8 Ay )i (2 Bx 16 8 Ax 16) j (8 B y 8 Ay )k (360 lb in.)i
From i-coefficient:
2 B y 8 Ay 360 lb in.
(4)
j-coefficient:
2 Bx 8 Ax 32 lb in.
(5)
k-coefficient:
8 B y 8 Ay 0
(6)
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PROBLEM 3.126 (Continued) From Equations (2) and (4): B y 36 lb
2 By 8( B y ) 360
Ay 36 lb
From Equations (1) and (5):
2( Ax 8) 8 Ax 32
Ax 1.6 lb From Equation (1):
Bx (1.6 8) 9.6 lb
A (1.600 lb)i (36.0 lb) j (2.00 lb)k B (9.60 lb)i (36.0 lb) j (2.00 lb)k
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PROBLEM 3.127 Three children are standing on a 5 5-m raft. If the weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively, determine the magnitude and the point of application of the resultant of the three weights.
SOLUTION
We have
F : FA FB FC R
(375 N) j (260 N) j (400 N) j R (1035 N) j R
We have
or
R 1035 N
or
z D 3.05 m
or
xD 2.57 m
M x : FA ( z A ) FB ( z B ) FC ( zC ) R ( z D )
(375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m) (1035 N)(z D )
z D 3.0483 m
We have
M z : FA ( x A ) FB ( xB ) FC ( xC ) R ( xD )
375 N(1 m) (260 N)(1.5 m) (400 N)(4.75 m) (1035 N)( xD )
xD 2.5749 m
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PROBLEM 3.128 Three children are standing on a 5 5-m raft. The weights of the children at Points A, B, and C are 375 N, 260 N, and 400 N, respectively. If a fourth child of weight 425 N climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft.
SOLUTION
We have
F : FA FB FC R
(375 N) j (260 N) j (400 N) j (425 N) j R R (1460 N) j
We have
M x : FA ( z A ) FB ( z B ) FC ( zC ) FD ( z D ) R ( z H )
(375 N)(3 m) (260 N)(0.5 m) (400 N)(4.75 m) (425 N)(z D ) (1460 N)(2.5 m)
z D 1.16471 m
We have
or
z D 1.165 m
M z : FA ( x A ) FB ( xB ) FC ( xC ) FD ( xD ) R ( xH )
(375 N)(1 m) (260 N)(1.5 m) (400 N)(4.75 m) (425 N)(xD ) (1460 N)(2.5 m)
xD 2.3235 m
or
xD 2.32 m
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PROBLEM 3.129 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine the magnitude and the point of application of the resultant of the four wind forces when a 1 ft and b 12 ft.
SOLUTION We have
Assume that the resultant R is applied at Point P whose coordinates are (x, y, 0). Equivalence then requires Fz : 105 90 160 50 R
or R 405 lb M x : (5 ft)(105 lb) (1 ft)(90 lb) (3 ft)(160 lb) (5.5 ft)(50 lb) y (405 lb)
or
y 2.94 ft M y : (5.5 ft)(105 lb) (12 ft)(90 lb) (14.5 ft)(160 lb) (22.5 ft)(50 lb) x (405 lb)
or
x 12.60 ft
R acts 12.60 ft to the right of member AB and 2.94 ft below member BC.
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PROBLEM 3.130 Four signs are mounted on a frame spanning a highway, and the magnitudes of the horizontal wind forces acting on the signs are as shown. Determine a and b so that the point of application of the resultant of the four forces is at G.
SOLUTION Since R acts at G, equivalence then requires that M G of the applied system of forces also be zero. Then at G : M x : (a 3) ft (90 lb) (2 ft)(105 lb) (2.5 ft)(50 lb) 0
or a 0.722 ft
M y : (9 ft)(105 ft) (14.5 b) ft (90 lb) (8 ft)(50 lb) 0
or b 20.6 ft
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PROBLEM 3.131 A concrete foundation mat of 5-m radius supports four equally spaced columns, each of which is located 4 m from the center of the mat. Determine the magnitude and the point of application of the resultant of the four loads.
SOLUTION Have: F :
FA FC FD FE R R 100 kN j 125 kN j 25 kN j 75 kN j 325 kN j
R P (325 kN)j
Have:
M x :
or R 325 kN
FC zC FE z E R zG 125 kN (4 m) (75 kN)(4 m) 325 kN zG
zG 0.61539 m
or zG 0.615 m M z :
FA x A FD xD R xG
100 kN (4 m) (25 kN) 4 m 325 kN xG or xG 0.923 m
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PROBLEM 3.132 Determine the magnitude and the point of application of the smallest additional load that must be applied to the foundation mat of Prob. 3.131 if the resultant of the five loads is to pass through the center of the mat.
SOLUTION The smallest load P will be placed on the edge of the mat at a radius of 5 m. xP 2 z P 2 (5 m) 2 (1)
Have:
M x :
Therefore
FC zC FE z E P z P R zG 125 kN 4 m 75 kN (4 m) P( z P ) R(0)
P( zG ) 200 kN m
Also
M z :
(2)
FA x A FD xD P( xP ) R xG
100 kN (4 m) (25 kN)(4 m) P( xP )
R(0)
P( xP ) 300 kN m (3) 2
Substituting for x and z from equations (2) and (3) into (1), P2
(300) 2 (200) 2 5200 (5) 2
Finally, from (2) and (3)
or P 72.111 kN
2
300 200 2 (5) P P
P 72.1 kN xP 4.16 m ; z P 2.77 m
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PROBLEM 3.133* Three forces of the same magnitude P act on a cube of side a as shown. Replace the three forces by an equivalent wrench and determine (a) the magnitude and direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.
SOLUTION Force-couple system at O: R Pi Pj Pk P (i j k ) M OR aj Pi ak Pj ai Pk Pak Pai Paj M OR Pa (i j k )
Since R and M OR have the same direction, they form a wrench with M1 M OR . Thus, the axis of the wrench is the diagonal OA. We note that cos x cos y cos z
a a 3
1 3
R P 3 x y z 54.7 M1 M OR Pa 3 Pitch p
M 1 Pa 3 a R P 3
(a)
R P 3 x y z 54.7
(b)
– a
(c)
Axis of the wrench is diagonal OA.
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PROBLEM 3.134* A piece of sheet metal is bent into the shape shown and is acted upon by three forces. If the forces have the same magnitude P, replace them with an equivalent wrench and determine (a) the magnitude and the direction of the resultant force R, (b) the pitch of the wrench, (c) the axis of the wrench.
SOLUTION
First reduce the given forces to an equivalent force-couple system R, MOR at the origin. We have F : Pj Pj Pk R
or
R Pk 5 M O : (aP ) j (aP)i aP k M OR 2
or (a)
5 M OR aP i j k 2
Then for the wrench,
RP λ axis
and
R k R
cos x 0 cos y 0 cos z 1
x 90 y 90 z 0
or (b)
Now M 1 axis M OR
5 k aP i j k 2 5 aP 2 Then
P
M1 52 aP R P
or P
5 a 2
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PROBLEM 3.134* (Continued)
(c)
The components of the wrench are ( R , M1 ), where M1 M 1 axis , and the axis of the wrench is assumed to intersect the xy-plane at Point Q, whose coordinates are (x, y, 0). Thus, we require M z rQ R R
M z M O M1
where Then
5 5 aP i j k aPk ( xi yj) Pk 2 2
Equating coefficients: i : aP yP
or
j: aP xP or
y a xa
The axis of the wrench is parallel to the z-axis and intersects the xy-plane at
x a, y a.
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PROBLEM 3.135* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.
SOLUTION First, reduce the given force system to a force-couple system. We have
F : (20 N)i (15 N) j R
We have
M O : (rO F ) M C M OR
R 25 N
M OR 20 N(0.1 m)j (4 N m)i (1 N m)j (4 N m)i (3 N m) j
(a)
R (20.0 N)i (15.00 N)j
(b)
We have
Pitch:
R R (0.8i 0.6 j) [(4 N m)i (3 N m)j] 5 Nm
M1 R M OR
p
M1 5 N m 0.200 m R 25 N
or p 0.200 m (c)
From above, note that M1 M OR
Therefore, the axis of the wrench goes through the origin. The line of action of the wrench lies in 3 y x the xy-plane with a slope of 4
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PROBLEM 3.136* The forces and couples shown are applied to two screws as a piece of sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin. We have
F : (10 lb) j (11 lb) j R
R (21 lb) j
We have
M O : (rO F ) M C M OR
i j k i j k M OR 0 0 20 lb in. 0 0 15 lb in. (12 lb in) j 0 10 0 0 11 0 (35 lb in.)i (12 lb in.) j R (21 lb) j
(a) (b)
We have
and pitch
or R (21.0 lb) j
R R ( j) [(35 lb in.)i (12 lb in.) j] 12 lb in. and M1 (12 lb in.) j
M1 λ R M OR
p
λR
M 1 12 lb in. 0.57143 in. 21 lb R
or
p 0.571 in.
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PROBLEM 3.136* (Continued)
(c)
We have
M OR M1 M 2 M 2 MOR M1 (35 lb in.)i
We require
M 2 rQ/O R (35 lb in.)i ( xi zk ) [(21 lb) j] 35i (21x)k (21z )i
From i: From k:
35 21z z 1.66667 in.
0 21x z0
The axis of the wrench is parallel to the y-axis and intersects the xz-plane at x 0, z 1.667 in.
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PROBLEM 3.137* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin. We have
F : (84 N) j (80 N)k R
and
M O : (rO F ) M C M OR
R 116 N
i j k i j k 0.6 0 0.1 0.4 0.3 0 (30 j 32k ) N m M OR 0 84 0 0 0 80 M OR (15.6 N m)i (2 N m) j (82.4 N m)k
R (84.0 N) j (80.0 N)k
(a) (b)
We have
M 1 λ R M OR
λR
R R
84 j 80k [ (15.6 N m)i (2 N m) j (82.4 N m)k ] 116 55.379 N m
and Then pitch
M1 M 1R (40.102 N m) j (38.192 N m)k
p
M 1 55.379 N m 0.47741 m 116 N R
or p 0.477 m
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PROBLEM 3.137* (Continued)
(c)
We have
M OR M1 M 2 M 2 M OR M1 [(15.6i 2 j 82.4k ) (40.102 j 38.192k )] N m (15.6 N m)i (42.102 N m) j (44.208 N m)k
We require
M 2 rQ/O R ( 15.6i 42.102 j 44.208k ) ( xi zk ) (84 j 80k ) (84 z )i (80 x) j (84 x)k
From i: or From k: or
15.6 84 z z 0.185714 m z 0.1857 m
44.208 84 x x 0.52629 m x 0.526 m
The axis of the wrench intersects the xz-plane at x 0.526 m
y 0 z 0.1857 m
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PROBLEM 3.138* Two bolts at A and B are tightened by applying the forces and couples shown. Replace the two wrenches with a single equivalent wrench and determine (a) the resultant R, (b) the pitch of the single equivalent wrench, (c) the point where the axis of the wrench intersects the xz-plane.
SOLUTION
First, reduce the given force system to a force-couple at the origin at B. 15 8 (a) We have F : (26.4 lb)k (17 lb) i j R 17 17 R (8.00 lb)i (15.00 lb) j (26.4 lb)k
and We have
M RB
R 31.4 lb M B : rA/B FA M A M B M RB
i j k 15 8 0 220k 238 i j 264i 220k 14(8i 15 j) 0 10 17 17 0 0 26.4
M RB (152 lb in.)i (210 lb in.)j (220 lb in.)k (b)
We have
R R 8.00i 15.00 j 26.4k [(152 lb in.)i (210 lb in.) j (220 lb in.)k ] 31.4 246.56 lb in.
M1 λ R M OR
λR
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PROBLEM 3.138* (Continued)
and Then pitch (c)
We have
M1 M 1R (62.818 lb in.)i (117.783 lb in.) j (207.30 lb in.)k
p
M 1 246.56 lb in. 7.8522 in. 31.4 lb R
or p 7.85 in.
M RB M1 M 2 M 2 M RB M1 (152i 210 j 220k ) ( 62.818i 117.783j 207.30k ) (214.82 lb in.)i (92.217 lb in.) j (12.7000 lb in.)k
We require
M 2 rQ/B R
i j k 214.82i 92.217 j 12.7000k x 0 z 8 15 26.4 (15 z )i (8 z ) j (26.4 x) j (15 x)k From i:
214.82 15 z
z 14.3213 in.
From k:
12.7000 15 x
x 0.84667 in.
The axis of the wrench intersects the xz-plane at
x 0.847 in. y 0 z 14.32 in.
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PROBLEM 3.139* Two ropes attached at A and B are used to move the trunk of a fallen tree. Replace the forces exerted by the ropes with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the yz-plane.
SOLUTION (a)
First replace the given forces with an equivalent force-couple system R, MOR at the origin. We have d AC (6)2 (2)2 (9) 2 11 m d BD (14)2 (2)2 (5)2 15 m Then 1650 N (6i 2 j 9k ) 11 (900 N)i (300 N) j (1350 N)k
TAC
and 1500 N (14i 2 j 5k ) 15 (1400 N)i (200 N) j (500 N)k
TBD
Equivalence then requires F : R TAC TBD (900i 300 j 1350k ) (1400i 200 j 500k ) (2300 N)i (500 N) j (1850 N)k M O : M OR rA TAC rB TBD
(12 m)k [(900 N)i (300 N)j (1350 N)k ] (9 m)i [(1400 N)i (200 N)j (500 N)k ] (3600)i (10,800 4500) j (1800)k (3600 N m)i (6300 N m)j (1800 N m)k
The components of the wrench are ( R , M1 ), where R (2300 N)i (500 N) j (1850 N)k
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PROBLEM 3.139* (Continued) (b)
We have R 100 (23) 2 (5) 2 (18.5) 2 2993.7 N
Let
axis Then
R 1 (23i 5 j 18.5k ) R 29.937
M1 axis M OR 1 (23i 5 j 18.5k ) (3600i 6300 j 1800k ) 29.937 1 [(23)(36) (5)(63) (18.5)(18)] 0.29937 601.26 N m
Finally,
P
M 1 601.26 N m R 2993.7 N or P 0.201 m
(c)
We have
M 1 M1 λ axis (601.26 N m)
1 (23i 5 j 18.5k ) 29.937
or
M1 (461.93 N m)i (100.421 N m) j (371.56 N m)k
Now
M 2 M OR M1 (3600i 6300 j 1800k ) (461.93i 100.421j 371.56k ) (3138.1 N m)i (6400.4 N m)j (2171.6 N m)k
For equivalence:
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PROBLEM 3.139* (Continued)
Thus, we require
M 2 rP R
r ( yj zk )
Substituting: 3138.1i 6400.4 j 2171.6k
i 0
j y
k z
2300 500 1850
Equating coefficients: j : 6400.4 2300 z
or
k : 2171.6 2300 y or
The axis of the wrench intersects the yz-plane at
z 2.78 m y 0.944 m
y 0.944 m
z 2.78 m
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PROBLEM 3.140* A flagpole is guyed by three cables. If the tensions in the cables have the same magnitude P, replace the forces exerted on the pole with an equivalent wrench and determine (a) the resultant force R, (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz-plane.
SOLUTION
(a)
First reduce the given force system to a force-couple at the origin. We have
F : P λ BA P λ DC P λDE R 4 3 3 4 9 4 12 R P j k i j i j k 5 5 5 25 5 25 5
R R
We have
3P (2i 20 j k ) 25
3P 27 5 (2)2 (20)2 (1) 2 P 25 25 M : (rO P ) M OR
3P 4P 4P 12 P 4 P 3P 9 P j k (20a ) j i j (20a ) j i j k M OR (24a ) j 5 5 5 5 25 5 25
M OR
24 Pa ( i k ) 5
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PROBLEM 3.140* (Continued)
(b)
We have
M 1 R M OR
where
R
Then
M1
R 3P 25 1 (2i 20 j k ) (2i 20 j k ) R 25 27 5 P 9 5
p
and pitch
1 9 5
(2i 20 j k )
M 1 8 Pa 25 8a R 15 5 27 5 P 81
M1 M 1 R
(c) Then
M 2 M OR M1
24 Pa 8Pa ( i k ) 5 15 5
or p 0.0988a
8 Pa 1 8 Pa ( 2i 20 j k ) (2i 20 j k ) 675 15 5 9 5
24 Pa 8Pa 8Pa ( i k ) (2i 20 j k ) (430i 20 j 406k ) 5 675 675 M 2 rQ/O R
We require
8 Pa 3P (403i 20 j 406k ) ( xi zk ) (2i 20 j k ) 675 25 3P [20 zi ( x 2 z ) j 20 xk ] 25
From i:
8( 403)
Pa 3P 20 z 675 25
From k:
8(406)
Pa 3P 20 x x 2.0049a 675 25
z 1.99012a
The axis of the wrench intersects the xz-plane at x 2.00a, z 1.990a
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PROBLEM 3.141* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane.
SOLUTION First, reduce the given force system to a force-couple at the origin. We have
F : FA FG R (40 mm)i (60 mm) j (120 mm)k R (50 N)k 70 N 140 mm (20 N)i (30 N) j (10 N)k
R 37.417 N
and We have
M O : (rO F ) M C M OR
M OR [(0.12 m) j (50 N)k ] {(0.16 m)i [(20 N)i (30 N) j (60 N)k ]} (160 mm)i (120 mm) j (10 N m) 200 mm (40 mm)i (120 mm) j (60 mm)k (14 N m) 140 mm M 0R (18 N m)i (8.4 N m) j (10.8 N m)k
To be able to reduce the original forces and couples to a single equivalent force, R and M must be perpendicular. Thus, R M 0. Substituting ?
(20i 30 j 10k ) (18i 8.4 j 10.8k ) 0 ?
or
(20)(18) (30)(8.4) (10)(10.8) 0
or
0 0
R and M are perpendicular so that the given system can be reduced to the single equivalent force. R (20.0 N)i (30.0 N) j (10.00 N)k
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PROBLEM 3.141* (Continued) Then for equivalence,
Thus, we require
M OR rp R rp yj zk
Substituting: i j k 18i 8.4 j 10.8k 0 y z 20 30 10
Equating coefficients: j: 8.4 20 z k:
or
z 0.42 m
10.8 20 y or
y 0.54 m
The line of action of R intersects the yz-plane at
x0
y 0.540 m
z 0.420 m
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PROBLEM 3.142* Determine whether the force-and-couple system shown can be reduced to a single equivalent force R. If it can, determine R and the point where the line of action of R intersects the yz-plane. If it cannot be so reduced, replace the given system with an equivalent wrench and determine its resultant, its pitch, and the point where its axis intersects the yz-plane.
SOLUTION First determine the resultant of the forces at D. We have d DA (12) 2 (9)2 (8) 2 17 in. d ED (6) 2 (0)2 (8) 2 10 in. Then 34 lb (12i 9 j 8k ) 17 (24 lb)i (18 lb) j (16 lb)k
FDA
and 30 lb (6i 8k ) 10 (18 lb)i (24 lb)k
FED
Then F : R FDA FED
(24i 18 j 16k (18i 24k ) (42 lb)i (18 lb)j (8 lb)k
For the applied couple d AK ( 6) 2 (6) 2 (18) 2 6 11 in.
Then M
160 lb in.
(6i 6 j 18k ) 6 11 160 [(1 lb in.)i (1 lb in.)j (3 lb in.)k ] 11
To be able to reduce the original forces and couple to a single equivalent force, R and M must be perpendicular. Thus ?
R M 0
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PROBLEM 3.142* (Continued)
Substituting (42i 18 j 8k ) 160
or
11
160 11
?
( i j 3k ) 0 ?
[(42)(1) (18)(1) ( 8)(3)] 0
0 0
or
R and M are perpendicular so that the given system can be reduced to the single equivalent force. R (42.0 lb)i (18.00 lb) j (8.00 lb)k
Then for equivalence,
M rP/D R
Thus, we require where
rP/D (12 in.)i [( y 3)in.]j ( z in.)k
Substituting:
160 11
i j (i j 3k ) 12 ( y 3) 42
18
k z 8
[( y 3)(8) ( z )(18)]i [( z )(42) (12)(8)]j [(12)(18) ( y 3)(42)]k Equating coefficients: j: k:
160
42 z 96 or 11 480 216 42( y 3) or 11
The line of action of R intersects the yz-plane at
x0
z 1.137 in. y 11.59 in.
y 11.59 in. z 1.137 in.
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PROBLEM 3.143* Replace the wrench shown with an equivalent system consisting of two forces perpendicular to the y-axis and applied respectively at A and B.
SOLUTION Express the forces at A and B as A Ax i Az k B Bx i Bz k
Then, for equivalence to the given force system, Fx : Ax Bx 0
(1)
Fz : Az Bz R
(2)
M x : Az (a ) Bz (a b) 0
(3)
M z : Ax ( a ) Bx (a b) M
(4)
Bx Ax
From Equation (1), Substitute into Equation (4):
Ax (a ) Ax (a b) M M M Ax and Bx b b
From Equation (2),
Bz R Az
and Equation (3),
Az a ( R Az )(a b) 0
a Az R 1 b
and
Then
a Bz R R 1 b a Bz R b a M A i R 1 k b b
M B b
a i b R k
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PROBLEM 3.144* Show that, in general, a wrench can be replaced with two forces chosen in such a way that one force passes through a given point while the other force lies in a given plane.
SOLUTION
First, choose a coordinate system so that the xy-plane coincides with the given plane. Also, position the coordinate system so that the line of action of the wrench passes through the origin as shown in Figure a. Since the orientation of the plane and the components (R, M) of the wrench are known, it follows that the scalar components of R and M are known relative to the shown coordinate system. A force system to be shown as equivalent is illustrated in Figure b. Let A be the force passing through the given Point P and B be the force that lies in the given plane. Let b be the x-axis intercept of B. The known components of the wrench can be expressed as R Rx i R y j Rz k
and M M x i M y j M z k
while the unknown forces A and B can be expressed as A Ax i Ay j Az k
and B Bx i Bz k
Since the position vector of Point P is given, it follows that the scalar components (x, y, z) of the position vector rP are also known. Then, for equivalence of the two systems, Fx : Rx Ax Bx
(1)
Fy : R y Ay
(2)
Fz : Rz Az Bz
(3)
M x : M x yAz zAy
(4)
M y : M y zAx xAz bBz
(5)
M z : M z xAy yAx
(6)
Based on the above six independent equations for the six unknowns ( Ax , Ay , Az , Bx , Bz , b), there exists a unique solution for A and B. From Equation (2),
Ay R y
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PROBLEM 3.144* (Continued)
1 Ax ( xRy M z ) y
Equation (6):
Equation (1):
1 Bx Rx ( xR y M z ) y
Equation (4):
1 Az ( M x zRy ) y
Equation (3):
1 Bz Rz ( M x zRy ) y
Equation (5):
b
( xM x yM y zM z ) ( M x yRz zR y )
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PROBLEM 3.145* Show that a wrench can be replaced with two perpendicular forces, one of which is applied at a given point.
SOLUTION
First, observe that it is always possible to construct a line perpendicular to a given line so that the constructed line also passes through a given point. Thus, it is possible to align one of the coordinate axes of a rectangular coordinate system with the axis of the wrench while one of the other axes passes through the given point. See Figures a and b. We have
R Rj and M Mj
and are known.
The unknown forces A and B can be expressed as A Ax i Ay j Az k
and B Bx i B y j Bz k
The distance a is known. It is assumed that force B intersects the xz-plane at (x, 0, z). Then for equivalence, Fx :
0 Ax Bx
(1)
Fy :
R Ay B y
(2)
Fz :
0 Az Bz
(3)
0 zB y
(4)
M x :
M y : M aAz xBz zBx
(5)
M z :
(6)
0 aAy xB y
Since A and B are made perpendicular, A B 0 or
There are eight unknowns:
Ax Bx Ay B y Az Bz 0
Ax , Ay , Az , Bx , B y , Bz , x, z
But only seven independent equations. Therefore, there exists an infinite number of solutions. Next, consider Equation (4): If By 0, Equation (7) becomes Using Equations (1) and (3), this equation becomes
0 zB y
Ax Bx Az Bz 0 Ax2 Az2 0
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(7)
PROBLEM 3.145* (Continued) Since the components of A must be real, a nontrivial solution is not possible. Thus, it is required that By 0, so that from Equation (4), z 0. To obtain one possible solution, arbitrarily let Ax 0. (Note: Setting Ay , Az , or Bz equal to zero results in unacceptable solutions.) The defining equations then become 0 Bx
(1)
R Ay B y
(2)
0 Az Bz
(3)
M aAz xBz
(5)
0 aAy xB y
(6)
Ay B y Az Bz 0
(7)
Then Equation (2) can be written
Ay R B y
Equation (3) can be written
Bz Az
Equation (6) can be written
x
aAy By
Substituting into Equation (5),
R By M aAz a By M Az By aR
or
( Az )
Substituting into Equation (7), M M ( R By ) By By By 0 aR aR By
or
a 2 R3 a2 R2 M 2
Then from Equations (2), (8), and (3),
a2 R2 RM 2 a2 R2 M 2 a2 R2 M 2 M a 2 R3 aR 2 M Az 2 2 aR a R M 2 a2 R2 M 2 Ay R
Bz
aR 2 M a2 R2 M 2
PROBLEM 3.145* (Continued)
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(8)
In summary, A
RM ( Mj aRk ) a R2 M 2
B
aR 2 (aRj Mk ) a2 R2 M 2
2
Which shows that it is possible to replace a wrench with two perpendicular forces, one of which is applied at a given point. Lastly, if R 0 and M 0, it follows from the equations found for A and B that Ay 0 and B y 0. From Equation (6), x 0 (assuming a 0). Then, as a consequence of letting Ax 0, force A lies in a plane parallel to the yz-plane and to the right of the origin, while force B lies in a plane parallel to the yzplane but to the left to the origin, as shown in the figure below.
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PROBLEM 3.146* Show that a wrench can be replaced with two forces, one of which has a prescribed line of action.
SOLUTION
First, choose a rectangular coordinate system where one axis coincides with the axis of the wrench and another axis intersects the prescribed line of action (AA). Note that it has been assumed that the line of action of force B intersects the xz-plane at Point P(x, 0, z). Denoting the known direction of line AA by
A x i y j z k it follows that force A can be expressed as A A A A(x i y j z k )
Force B can be expressed as B Bx i B y j Bz k
Next, observe that since the axis of the wrench and the prescribed line of action AA are known, it follows that the distance a can be determined. In the following solution, it is assumed that a is known. Then for equivalence, Fx : 0 Ax Bx
(1)
Fy : R A y B y
(2)
Fz : 0 Az Bz
(3)
M x : 0 zB y
(4)
M y : M aAz zBx xBz
(5)
M x : 0 aA y xB y
(6)
Since there are six unknowns (A, Bx, By, Bz, x, z) and six independent equations, it will be possible to obtain a solution.
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PROBLEM 3.146* (Continued) Case 1: Let z 0 to satisfy Equation (4). Now Equation (2): Equation (3): Equation (6):
A y R By Bz Az
x
aA y By
a By
( R By )
Substitution into Equation (5):
a M aAz By
( R By )( Az )
1 M
A
B z aR y
R
By y By z aR
Substitution into Equation (2):
By
1 M
z aR 2 z aR y M
MR R z aR y M aR y z M x MR Bx Ax z aR y M A
Then
Bz Az
z MR z aR y M A
In summary,
B
and
R x a 1 By z aR y M a 1 R 2 z aR
R
z aR y M
P A aR y z M
(x M i z aRj z M k )
or x
y M z R
Note that for this case, the lines of action of both A and B intersect the x-axis.
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PROBLEM 3.146* (Continued)
Case 2: Let B y 0 to satisfy Equation (4). Now Equation (2):
A
R
y
Equation (1):
Bx R x y
Equation (3):
Bz R z y
Equation (6):
aA y 0
which requires a 0
Substitution into Equation (5):
M z R x y
x R z y
M or z x x z R
y
This last expression is the equation for the line of action of force B. In summary,
R A y
A
R B y
( x i x k )
Assuming that x , y , z . 0, the equivalent force system is as shown below.
Note that the component of A in the xz-plane is parallel to B.
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PROBLEM 3.147 A 300-N force P is applied at Point A of the bell crank shown. (a) Compute the moment of the force P about O by resolving it into horizontal and vertical components. (b) Using the result of part (a), determine the perpendicular distance from O to the line of action of P.
SOLUTION
x (0.2 m) cos 40 0.153209 m y (0.2 m)sin 40 0.128558 m rA /O (0.153209 m)i (0.128558 m) j
(a)
Fx (300 N) sin 30 150 N
Fy (300 N) cos 30 259.81 N F (150 N)i (259.81 N) j M O rA / O F (0.153209i 0.128558 j) m (150i 259.81j) N (39.805k 19.2837k ) N m (20.521 N m)k
(b)
M O 20.5 N m
M O Fd 20.521 N m (300 N)(d )
d 0.068403 m
d 68.4 mm
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PROBLEM 3.148 A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and length d is 1.90 m, determine the moment about D of the force exerted by the cable at C by resolving that force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
SOLUTION (a)
Slope of line: Then
and
Then
EC
0.875 m 5 1.90 m 0.2 m 12
12 (TAB ) 13 12 (1040 N) 13 960 N 5 TABy (1040 N) 13 400 N TABx
(a)
M D TABx (0.875 m) TABy (0.2 m) (960 N)(0.875 m) (400 N)(0.2 m)
760 N m
(b)
We have
or M D 760 N m
M D TABx ( y ) TABx ( x) (960 N)(0) (400 N)(1.90 m)
760 N m
(b) or M D 760 N m
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PROBLEM 3.149 A small boat hangs from two davits, one of which is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant force RA exerted on the davit at A.
SOLUTION We have
R A 2FAB FAD
where and
FAB (82 lb) j AD 6i 7.75 j 3k (82 lb) FAD FAD 10.25 AD FAD (48 lb)i (62 lb) j (24 lb)k
Thus
R A 2FAB FAD (48 lb)i (226 lb) j (24 lb)k
Also
rA/C (7.75 ft) j (3 ft)k
Using Eq. (3.21):
i j k M C 0 7.75 3 48 226 24 (492 lb ft)i (144.0 lb ft) j (372 lb ft)k M C (492 lb ft)i (144.0 lb ft) j (372 lb ft)k
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PROBLEM 3.150 Consider the volleyball net shown. Determine the angle formed by guy wires AB and AC.
SOLUTION First note:
and
By definition, or or
AB (6.5) 2 (8) 2 (2)2 10.5 ft AC (0) 2 (8)2 (6)2 10 ft AB (6.5 ft)i (8 ft) j (2 ft)k AC (8 ft) j (6 ft)k AB AC ( AB )( AC ) cos (6.5i 8 j 2k ) (8 j 6k ) (10.5)(10) cos (6.5)(0) (8)(8) (2)(6) 105cos cos 0.72381
or 43.6
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PROBLEM 3.151 A single force P acts at C in a direction perpendicular to the handle BC of the crank shown. Determine the moment Mx of P about the x-axis when θ 65°, knowing that My 15 N · m and Mz 36 N · m.
SOLUTION See the solution to Prob. 3.53 for the derivation of the following equations: M x (0.2) P sin( ) M tan z My
(1) (4)
P 4 M y2 M z2
(5)
Substituting for known data gives: tan
36 N m 15 N m
67.380 P 4 (15)2 (36) 2 P 156.0 N M x 0.2 m(156.0 N)sin(65 67.380) 23.047 N m M x 23.0 N m
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PROBLEM 3.152 A small boat hangs from two davits, one of which is shown in the figure. It is known that the moment about the z-axis of the resultant force RA exerted on the davit at A must not exceed 279 lbft in absolute value. Determine the largest allowable tension in line ABAD when x 6 ft.
SOLUTION First note:
R A 2TAB TAD
Also note that only TAD will contribute to the moment about the z-axis. AD (6) 2 (7.75) 2 (3) 2
Now
10.25 ft AD T AD T (6i 7.75 j 3k ) 10.25
Then
TAD
Now
M z k (rA/C TAD )
where
rA/C (7.75 ft) j (3 ft)k
Then for Tmax ,
0 0 1 Tmax 279 0 7.75 3 10.25 6 7.75 3
Tmax | (1)(7.75)(6)| 10.25
or Tmax 61.5 lb
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PROBLEM 3.153 In a manufacturing operation, three holes are drilled simultaneously in a workpiece. If the holes are perpendicular to the surfaces of the workpiece, replace the couples applied to the drills with a single equivalent couple, specifying its magnitude and the direction of its axis.
SOLUTION M M1 M 2 M 3 (1.5 N m)( cos 20 j sin 20k ) (1.5 N m) j (1.75 N m)( cos 25 j sin 25k ) (4.4956 N m) j (0.22655 N m)k M (0)2 (4.4956) 2 (0.22655)2 4.5013 N m
M 4.50 N m
M (0.99873 j 0.050330k ) M cos x 0
axis
cos y 0.99873 cos z 0.050330
x 90.0, y 177.1, z 87.1
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PROBLEM 3.154 A 260-lb force is applied at A to the rolled-steel section shown. Replace that force with an equivalent force-couple system at the center C of the section.
SOLUTION AB (2.5 in.) 2 (6.0 in.) 2 6.50 in.
2.5 in. 5 6.5 in. 13 6.0 in. 12 cos 6.5 in. 13 sin
22.6
F F sin i F cos j 5 12 i (260 lb) j 13 13 (100.0 lb)i (240 lb) j (260 lb)
M C rA /C F (2.5i 4.0 j) (100.0i 240 j) 400k 600k (200 lb in.)k
F 260 lb
67.4°; M C 200 lb in.
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PROBLEM 3.155 The force and couple shown are to be replaced by an equivalent single force. Knowing that P 2Q, determine the required value of α if the line of action of the single equivalent force is to pass through (a) Point A, (b) Point C.
SOLUTION
(a)
We must have M A 0 ( P sin )a Q(a ) 0 sin
Q Q 1 P 2Q 2
30.0 (b)
MC 0
We must have
( P sin )a ( P cos )a Q (a ) 0 sin cos
Q Q 1 P 2Q 2
sin cos
1 2
(1)
1 4 1 1 cos 2 cos 2 cos 4 2 2 cos cos 0.75 0 sin 2 cos 2 cos
(2)
Solving the quadratic in cos : cos
1 7 4
65.7 or 155.7
Only the first value of satisfies Eq. (1), therefore 65.7
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PROBLEM 3.156 A 77-N force F1 and a 31-N m couple M1 are applied to corner E of the bent plate shown. If F1 and M1 are to be replaced with an equivalent force-couple system (F2, M2) at corner B and if (M2)z 0, determine (a) the distance d, (b) F2 and M2.
SOLUTION (a)
M Bz : M 2 z 0
We have
k (rH /B F1 ) M 1z 0
where
(1)
rH /B (0.31 m)i (0.0233) j
F1 λ EH F1
(0.06 m)i (0.06 m) j (0.07 m)k (77 N) 0.11 m
(42 N)i (42 N) j (49 N)k M1z k M1 M1 λEJ M1
di (0.03 m) j (0.07 m)k d 2 0.0058 m
(31 N m)
Then from Equation (1), 0 0 1 (0.07 m)(31 N m) 0.31 0.0233 0 0 2 d 0.0058 42 42 49
Solving for d, Equation (1) reduces to (13.0200 0.9786)
from which
2.17 N m d 2 0.0058
d 0.1350 m
0
or d 135.0 mm
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PROBLEM 3.156 (Continued)
(b)
F2 F1 (42i 42 j 49k ) N or F2 (42.0 N)i (42.0 N) j (49.0 N)k
M 2 rH /B F1 M1 i j k (0.1350)i 0.03j 0.07k 0.31 0.0233 0 (31 N m) 0.155000 49 42 42 (1.14170i 15.1900 j 13.9986k ) N m (27.000i 6.0000 j 14.0000k ) N m
M 2 (25.858 N m)i (21.190 N m) j
or M 2 (25.9 N m)i (21.2 N m) j
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PROBLEM 3.157 Three horizontal forces are applied as shown to a vertical cast iron arm. Determine the resultant of the forces and the distance from the ground to its line of action when (a) P 200 N, (b) P 2400 N, (c) P 1000 N.
SOLUTION (a)
RD 200 N 600 N 400 N 800 N M D (200 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m) 150.0 N m
y
M D 150 N m 0.1875 m 800 N R
R 800 N
; y 187.5 mm
(b)
RD 2400 N 600 N 400 N 1400 N M D (2400 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m) 840 N m
y
M D 840 N m 0.600 m 1400 N R
R 1400 N
; y 600 mm
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PROBLEM 3.157 (Continued)
(c) RD 1000 600 400 0 M D (1000 N)(0.450 m) (600 N)(0.300 m) (400 N)(0.1500 m) 210 N m
y System reduces to a couple. M D 210 N m
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PROBLEM 3.158 While using a pencil sharpener, a student applies the forces and couple shown. (a) Determine the forces exerted at B and C knowing that these forces and the couple are equivalent to a force-couple system at A consisting of the force R (2.6 lb)i + R y j (0.7 lb)k and the couple M RA M x i (1.0 lb · ft)j (0.72 lb · ft)k. (b) Find the corresponding values of R y and M x.
SOLUTION (a)
From the statement of the problem, equivalence requires F : B C R or
Fx : Bx C x 2.6 lb
Fy : C y R y
(2)
(1)
Fz : C z 0.7 lb or C z 0.7 lb
and
M A : (rB/A B M B ) rC/A C M AR
or
1.75 ft (C y ) M x M x : (1 lb ft) 12
(3)
3.75 1.75 3.5 ft ( Bx ) ft (C x ) ft (0.7 lb) 1 lb ft M y : 12 12 12
or Using Eq. (1):
3.75 Bx 1.75C x 9.55 3.75 Bx 1.75(2.6 Bx ) 9.55
or
Bx 2.5 lb
and
C x 0.1 lb
3.5 M z : ft (C y ) 0.72 lb ft 12
or
C y 2.4686 lb
B (2.50 lb)i C (0.1000 lb)i (2.47 lb) j (0.700 lb)k
(b)
Eq. (2) Using Eq. (3):
Ry 2.47 lb
1.75 1 (2.4686) M x 12
or M x 1.360 lb ft
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