A plank with mass m=3.0kg and length of l=2.0m which is supported at its center,

Question

A plank with mass m=3.0kg and length of l=2.0m which is supported at its center, and allowed to rotate freely about this support. There is a force of F1=75N in the downward direction on the left end of the plank, and a force of F2=125N in the downward direction an unknown distance to the right of the support which is at the center of the plank. What distance from the end of the left end of the plank should the force F2 be placed such that the plank will be in static equilibrium? (the answer is 1.6m.)

Explanation / Answer

plank will be in static equilibrium when net torque(r x F) on plank is zero.

so Net torque = ( 2/2 x 75) - (d x 125) = 0

125d = 75

d = 0.6 m from the centre of plank.

distance from left end = 2/2   + 0.6 = 1.6 m

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