The drawing shows a rectangular piece of wood. The forces applied to corners B a

Question

The drawing shows a rectangular piece of wood. The forces applied to corners B and D have the same magnitude of 13 N and are directed parallel to the long and short sides of the rectangle. The long side of the rectangle is twice as long as the short side. An axis of rotation is shown perpendicular to the plane of the rectangle at its center. A third force (not shown in the drawing) is applied to corner A, directed along the short side of the rectangle (either toward B or away from B), such that the piece of wood is at equilibrium. Find the magnitude and direction of the force applied to corner A.

Magnitude in N

Direction (toward corner B/away from corner B)

Explanation / Answer

Torque generated by the force at the corner B, b = FL/4 anticlockwise.

Where L = length of longer side of the block.

Torque generated by the force at the corner D, d = FL/2 clockwise.

So net torque acting on the block, n =FL/4 clockwise.

Now in order for block to be in equilibrium, net torque must be zero. So we need to apply an equal and opposite torque to the block. So we need to apply FL/4 anticlockwise torque. Hence force acting at the corner must be F/2 = 6.5 N because arm length would be L/2 and it must be acting towards the corner B so that it can produce anticlockwise torque.

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