Question

Please explain how to derive this and provide the answer.

A bar of length L and uniform density is held horizontally against a wall by a pivot at one end and a rope of tension T which pulls up and to the left making an angle θ with the vertical wall. The rope is attached to the bar at a distance l L from the wall. A mass M is resting at the end of the bar. If the bar has total mass 2.9kg, the mass M=1.6kg, and the rope makes an angle of 27.0° with the wall, what is the magnitude of the force exerted on the bar by the pivot | Fpivot I? Answer in Newtons( N)

Answer 1

considering torque avout the hinge

mg (L/2) + Mg L = T cos x * (L/3)

9.8* ( 2.9 / 2 + 1.6 ). = T cos 27 / 3

T = 100.64 N

net vertical force at hinge

Fy = (2.9 + 1.6)* 9.8 - T cos 27

Fy = 45.57 N

net horizontal force

Fx = T sin x = 45.689 N

net force at hinge

F^2 j= Fx^2 + Fy^2

F = 64.53 N

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