Question

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal..
1. What is the horizontal component of the force exerted by the hi.nge on the beam? (Use the `to the right' as + for the horizontal direction.)
2. What is the magnitude of the force that the beam exerts on the hi.nge?

Answer 1

The tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 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 = (40 x 9.8)/(2 x sin31)

T = 380.55 N

Vertical component of the force

T + F = W

F = W - T

F = (9.8 x 40) - 380.55

F = 11.45 N

Thus, the tension in the cable is 380.55 N and the vertical component of the force exerted by the hi.nge on the beam is 11.45 N.

Learn more about tension here: brainly.com/question/24994188

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