H60°Figure 6GM8. A uniform rod HG of length 2L m and mass m = 2 kg is hinged at end H, as shown in Figure 6A mass Mg=100 N is hung at the other end G. A horizontal cable at the midpoint of the rodholds it at an angle of 60° to the horizontal. The cable is under a constant tension T.(a) Draw a diagram showing all the forces acting on the rod and determine the tension T inthe cable.(b) Calculate both the x and y components (Fr and F₁) of the force F at the hinge H.(c) Will the net force F at H act along the rod HG? Justify your answer.

Answered: Image /qna-images/answer/ec14169d-7a06-4509-9292-2554765b4cbd.jpg

Answered: H 60° Figure 6 G M 8. A uniform rod HG…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter10: Rotational Motion

Section: Chapter Questions

Problem 32P

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A uniform horizontal strut weighs 400.0 N. One end of the strut is attached to a hinged support the wall and the other end of the strut is attached to a sign that weighs 200.0 N The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the s find the tension in the cable and the force at the hinge of the strut.
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.
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
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OT SHARE 1. A mass m, is suspended from two cables fixed at point B. The cables are fixed to a rigid support at A and C. If the tension in cable AB is not to exceed 120 N, find the maximum mass, m, that can be suspended. A 30 45' 40°
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