3. A 4.5m, 35kg steel beam (drawn in grey) is attached to a wall on its left end at an angle of 47° while a cable is attached to its right end at an angle of 22°, and a 22kg mass is also suspended from this end. 47* 22 22kg (a) Draw a freebody diagram showing the forces on the beam and their locations. (b) Determine the tension in the cable and the support force from the wall. List both forces in component and polar form.

3. A 4.5m, 35kg steel beam (drawn in grey) is attached to a wall on its left end at an angle of 47° while a cable is attached to its right end at an angle of 22°, and a 22kg mass is also suspended from this end. 47* 22 22kg (a) Draw a freebody diagram showing the forces on the beam and their locations. (b) Determine the tension in the cable and the support force from the wall. List both forces in component and polar form.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter14: Static Equilibrium, Elasticity, And Fracture

Section: Chapter Questions

Problem 38PQ: At a museum, a 1300-kg model aircraft is hung from a lightweight beam of length 12.0 m that is free...

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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
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