A nonuniform beam 4.60 m long and weighing 1.60 kN makes an angle of 25.0? below

Question

A nonuniform beam 4.60 m long and weighing 1.60 kN makes an angle of 25.0? below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 2.90 m farther down the beam and perpendicular to it in (Figure 1) . The center of gravity of the beam is a distance of 1.90 m down the beam from the pivot. Lighting equipment exerts a downward force of 5.05 kN on the end of the beam.

Part B

Find the tension T in the cable.

Part C

Find the x component of the force exerted on the beam by the pivot.

Part D

Find the y component of the force exerted on the beam by the pivot.

Explanation / Answer

The solution is as follows:

Part B: Let the tension in the cable be T
taking moment abt pivot,
1.60 k * 1.90 cos 25 + 5.05 k * 4.60 cos 25 = T * 2.90
=> 1.60k*1.722 + 5.05k* 4.17 = T*2.90

=> T = 8.21 kN

Part D: let the vertical component of the force exerted on the beam by the pivot be Fy
Fy + T cos 25 = 5.05 k + 1.60 k
Fy = 6.65 k - 8.21 k cos 25 = 6.65 k - 7.44 k = - 0.79 kN

Part C: let the horizontal component of the force exerted on the beam by the pivot be Fx

Fx = T sin 28
Fx = 8.21 k sin 25 N = 3.47 kN

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