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 F1 is applied perpendicularly to the end of the stick at 0 cm, as shown. A second force F2 is applied perpendicularly at the 100-cm end of the stick. The forces are horizontal. If the stick does not move, the force exerted by the pivot on the stick:


A. must be zero
B. must be in the same direction as F1 and have magnitude |F2|-|F1|
C. must be directed opposite to F1 and have magnitude |F2|-|F1|
D. must be in the same direction as F1 and have magnitude |F2|+|F1|
E. must be directed opposite to F1 and have magnitude |F2|+|F1|

Explanation / Answer

For equilbrium net force should be zero. If both the forces are acting in the same direction then The pivot force has magnitude |F2|+|F1| and must be directed opposite to F1 option D If the two forces are acting in opposite direciton The pivot force must be in the same direction as F1 and have magnitude |F2|-|F1|. option B

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