A meter stick on a horizontal frictionless table top is pivoted at the 80-cm mar

Question

A meter stick on a horizontal frictionless table top is pivoted at the 80-cm mark. A force F_1 is applied perpendicularly to the end of the stick at 0 cm, as shown. A second force F_2 (not shown) is applied perpendicularly at the 60-cm mark. The forces are horizontal. If the stick does not move, the force exerted by the pivot on the stick: must be zero must be in the same direction as F_1 and have magnitude |F_2| - |F_1| must be directed opposite to F_1 and have magnitude |F_2| - |F_1| must be in the same direction as F_1 and have magnitude |F_2| + |F_1| must be directed opposite to F_1 and have magnitude |F_2| + |F_1|

Explanation / Answer

Since the stick is in equilibrium:
so the net force and net torque on the stick will be zero.
Also, the net force applied= F1+F2
So,
Force exerted by the pivot on the stick to keep it in equilibrium=F1+F2 (This is the case when F1 and F2 is in same direction)
This will keep the forces applied in equilibrium.
Now, to keep the torque in equilibrium, piovet's direction should be in opposite
direction to that of F1(since F1 and F2 is creating torque in clockwise direction,
so pivot's force should create torque in anticlockwise direction).
Hense, option E is correct.

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