Question

A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?

What is the vertical component of the force exerted by the hi.nge on the beam?

Answer 1

The tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

Tension in the cable

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (21 x 9.8)/(2 x sin66)

T = 112.64 N

Vertical component of the force

T + F = W

F = W - T

F = (9.8 x 21) - 112.64

F = 93.16 N

Thus, the tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

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