nonuniform beam 4.54 m long and weighing 1.05 kN makes an angle of 25 below the

Question

nonuniform beam 4.54 m long and weighing 1.05 kN makes an angle of 25 below the horizontal. It is held in position by a frictionless pivot at its upper-right end and by a cable a distance of 3.00 m farther down the beam and perpendicular to it (see Figure 11.31 in the textbook). The center of gravity of the beam is a distance of 2.10 m down the beam from the pivot. Lighting equipment exerts a downward force of 5.07 kN on the lower-left end of the beam.

Part A: Find the tension T in the cable.

part B : Find the vertical component of the force exerted on the beam by the pivot. Assume that the positive x and y axes are directed to the right and upward respectively.

Part c : Find the horizontal component of the force exerted on the beam by the pivot. Assume that the positive x and y axes are directed to the right and upward respectively.

Explanation / Answer

a)

here

taking moment abt pivot,

1.05 k * 2.1 cos 25 + 5.07 k * 4.54 cos 25 = T * 3

T= 7.62 k N---answer

b)

let the vertical component of the force exerted on the beam by the pivotbe Fy

Fy + T cos 25 = 5.07 k + 1.05 k

Fy = 6.12 k - 7.62 k cos 25

Fy = -786.06 N

c)

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

Fx = T sin 25

Fx = 7.62 k sin 25

Fx = 3.22 kN

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