• Author / Uploaded
• Eric Castillo Martínez

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• Funciones trigonométricas
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• Física Aplicada e Interdisciplinaria
• Geometría
• Física y matemáticas

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PROBLEM 2.132 Two cables tied together at C are loaded as shown. Knowing that Q = 60 lb, determine the tension (a) in cable AC, (b) in cable BC.

SOLUTION ΣFy = 0: TCA − Q cos 30° = 0

Q = 60 lb

With

TCA = ( 60 lb )( 0.866 )

(a)

TCA = 52.0 lb ΣFx = 0: P − TCB − Q sin 30° = 0

(b)

P = 75 lb

With

TCB = 75 lb − ( 60 lb )( 0.50 ) or TCB = 45.0 lb

152

PROBLEM 2.133 Two cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 60 lb in either cable.

SOLUTION ΣFx = 0: TCA − Q cos 30° = 0

Have

TCA = 0.8660 Q

or

TCA ≤ 60 lb

Then for

0.8660Q < 60 lb Q ≤ 69.3 lb

or

ΣFy = 0: TCB = P − Q sin 30°

From or

TCB = 75 lb − 0.50Q

For

TCB ≤ 60 lb 75 lb − 0.50Q ≤ 60 lb 0.50Q ≥ 15 lb

or

Q ≥ 30 lb

Thus,

30.0 ≤ Q ≤ 69.3 lb

Therefore,

153

PROBLEM 2.134 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 8 kN and FB = 16 kN, determine the magnitudes of the other two forces.

SOLUTION Free-Body Diagram of Connection

ΣFx = 0:

3 3 FB − FC − FA = 0 5 5

FA = 8 kN, FB = 16 kN

With

FC =

4 4 (16 kN ) − (8 kN ) 5 5 FC = 6.40 kN

ΣFy = 0: − FD +

3 3 FB − FA = 0 5 5

With FA and FB as above: FD =

3 3 (16 kN ) − (8 kN ) 5 5 FD = 4.80 kN

154

PROBLEM 2.135 A welded connection is in equilibrium under the action of the four forces shown. Knowing that FA = 5 kN and FD = 6 kN, determine the magnitudes of the other two forces.

SOLUTION Free-Body Diagram of Connection

ΣFy = 0: − FD −

3 3 FA + FB = 0 5 5

FB = FD +

or

3 FA 5

FA = 5 kN, FD = 8 kN

With

FB =

5 3  6 kN + ( 5 kN )  3  5  FB = 15.00 kN

ΣFx = 0: − FC + FC = =

4 4 FB − FA = 0 5 5

4 ( FB − FA ) 5 4 (15 kN − 5 kN ) 5 FC = 8.00 kN

155

PROBLEM 2.136 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the magnitude of the force P required to maintain the equilibrium of the collar when (a) x = 4.5 in., (b) x = 15 in.

SOLUTION Free-Body Diagram of Collar

(a)

Triangle Proportions

ΣFx = 0: − P +

4.5 ( 50 lb ) = 0 20.5 or P = 10.98 lb

(b)

Triangle Proportions

ΣFx = 0: − P +

15 ( 50 lb ) = 0 25 or P = 30.0 lb

156

PROBLEM 2.137 Collar A is connected as shown to a 50-lb load and can slide on a frictionless horizontal rod. Determine the distance x for which the collar is in equilibrium when P = 48 lb.

SOLUTION Free-Body Diagram of Collar

Triangle Proportions

Hence:

ΣFx = 0: − 48 +

xˆ =

or

50 xˆ 400 + xˆ 2

=0

48 400 + xˆ 2 50

(

xˆ 2 = 0.92 lb 400 + xˆ 2

)

xˆ 2 = 4737.7 in 2 xˆ = 68.6 in.

157

PROBLEM 2.138 A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support at D.

SOLUTION The force in cable DB can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with JJJG DB = ( 480 mm ) i − ( 510 mm ) j + ( 320 mm ) k DB = F = F λ DB

( 480 )2 + ( 510 )2 + ( 320 )2

= 770 mm

JJJG DB 385 N ( 480 mm ) i − ( 510 mm ) j + ( 320 mm ) k  = F = DB 770 mm  F = ( 240 N ) i − ( 255 N ) j + (160 N ) k

Fx = +240 N, Fy = −255 N, Fz = +160.0 N

158

PROBLEM 2.139 A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. Determine the magnitude and direction of the resultant of the forces exerted by the cable at B knowing that the tension in the cable is 385 N.

SOLUTION The force in each cable can be written as the product of the magnitude of the force and the unit vector along the cable. That is, with JJJG BD = − ( 0.48 m ) i + ( 0.51 m ) j − ( 0.32 m ) k BD =

( −0.48 m )2 + ( 0.51 m )2 + ( −0.32 m )2

TBD = T λ BD = TBD

= 0.77 m

JJJG BD TBD  − ( 0.48 m ) i + ( 0.51 m ) j − ( 0.32 m ) k  = BD 0.77 m 

TBD = TBD ( −0.6234i + 0.6623j − 0.4156k )

and JJJG BE = − ( 0.27 m ) i + ( 0.40 m ) j − ( 0.6 m ) k

BE =

( −0.27 m )2 + ( 0.40 m )2 + ( −0.6 m )2

TBE = T λ BE = TBE

= 0.770 m

JJJG BD TBE  − ( 0.26 m ) i + ( 0.40 m ) j − ( 0.6 m ) k  = BD 0.770 m 

TBE = TBE ( −0.3506i + 0.5195 j − 0.7792k )

Now, because of the frictionless ring at B, TBE = TBD = 385 N and the force on the support due to the two cables is F = 385 N ( −0.6234i + 0.6623j − 0.4156k − 0.3506i + 0.5195j − 0.7792k )

= − ( 375 N ) i + ( 455 N ) j − ( 460 N ) k

159

PROBLEM 2.139 CONTINUED The magnitude of the resultant is F =

Fx2 + Fy2 + Fz2 =

( −375 N )2 + ( 455 N )2 + ( −460 N )2

= 747.83 N

or F = 748 N The direction of this force is:

θ x = cos −1

−375 747.83

or θ x = 120.1°

θ y = cos −1

455 747.83

or θ y = 52.5°

θ z = cos −1

−460 747.83

or θ z = 128.0°

160

PROBLEM 2.140 A steel tank is to be positioned in an excavation. Using trigonometry, determine (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R.

SOLUTION Force Triangle

(a) For minimum P it must be perpendicular to the vertical resultant R ∴ P = ( 425 lb ) cos 30° or P = 368 lb R = ( 425 lb ) sin 30°

(b)

or R = 213 lb

161