Problem: A 80 kg construction worker is sitting 2.0 m from the right end of a 1450 kg, 6.0 m long beam. What is the tension in the cable attached at an angle of 30 deg. from the beam?

A 5.45-N beam of uniform density is 1.60 m long. The beam is supported at an angle of 35.0 by a cable attached to one end. There is a pin through the other end of the beam (Fig. P14.30). Use the values given in the figure to find the tension in the cable. FIGURE P14.30

A massless, horizontal beam of length L and a massless rope support a sign of mass m (Fig. P14.78). a. What is the tension in the rope? b. In terms of m, g, d, L, and , what are the components of the force exerted by the beam on the wall? FIGURE P14.78

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