6. A cable supports a uniform boom, which is pivoted at point P. ogble boom P 60.0° If the cable is attached two-thirds of the way along the boom from P, and the tension force in the cable is 840 N, what is the force of gravity on the boom? (ans: 2.24 x 10 N)

6. A cable supports a uniform boom, which is pivoted at point P. ogble boom P 60.0° If the cable is attached two-thirds of the way along the boom from P, and the tension force in the cable is 840 N, what is the force of gravity on the boom? (ans: 2.24 x 10 N)

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter8: Rotational Motion

Section: Chapter Questions

Problem 74A

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In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the other end to a tee 15 m away as shown below. He then pulls on the center of the rope with a force of 400 N, which causes its center to be displaced 0.30 m, as shown. What is the force of the rope on the car?
Find the magnitude of the tension in each supporting cable shown below. In each case, the weight of the suspended body is 100.0 N and the masses of the cables are negligible.
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At a museum, a 1300-kg model aircraft is hung from a lightweight beam of length 12.0 m that is free to pivot about its base and is supported by a massless cable (Fig. P14.38). Ignore the mass of the beam. a. What is the tension in the section of the cable between the beam and the wall? b. What are the horizontal and vertical forces that the pivot exerts on the beam? FIGURE P14.38 (a) From the free-body diagram, the angle that the string tension makes with the beam is = 55.0 + 18.0 = 73.0, and the perpendicular component of the string tension is FT sin73.0. Summing torques around the base of the rod gives (Eq. 14.2): =0:(12.0m)(1300kg)(9.81m/s2)cos55.0+FT(12.0m)sin73.0=0FT=(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0FT=7.65103N Figure P14.38ANS (b) Using force balance (Eq. 14.1): Fx=0:FHFTcos18.0=0FH=FTcos18.0=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]cos18.0=7.27103NFy=0:FVFTsin18.0(1300kg)(9.81m/s2)=0 FV=FTsin18.0+(1300kg)gFV=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]sin18.0+(1300kg)(9.81m/s2)FV=1.51104N
Under what conditions can a rotating body be in equilibrium? Give an example.
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