A bar of length L and uniform density is held horizontally against a wall by a p

Question

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 with the vertical wall. The rope is attached to the bar at a distance 1- L from the wall. A mass M is resting at the end of the bar. If the bar has total mass 2.5kg, the mass M=2.8kg, and the rope makes an angle of 36.6° with the wall, what is the magnitude of the force exerted on the bar by the pivot |Fpivot l? Answer in Newtons Answer: 111.02

Explanation / Answer

Balancing torque about pivot, T cos theta l = MgL + mgL/2

T* cos 36.6 degree /3 = 2.8*9.8 + 2.5*9.8/2 = 39.69

T = 3*39.69/ cos 36.6 degree = 148.3 N

Fx = T sin theta = 148.3* sin 36.6 degree = 88.42 N

Fy = T cos theta - mg -Mg = 148.3 cos 36.6 degree - 2.5*9.8 - 2.8*9.8 = 67.12

F = sqrt(88.42^2+67.12^2) = 111 N answer

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