What is the force exerted by the pivot on the stick?

[hidemi]

Homework Statement

A massless 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. A second force F2 (not shown) is applied perpendicularly at the 60-cm mark. The forces are in the plane of the table top. 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|

The answer is B.

Relevant Equations

τ ＝rF

If exerted a force at 60 cm, should the force F2 be the opposite direction of F1? Thanks!

 

[Doc Al]

If exerted a force at 60 cm, should the force F2 be the opposite direction of F1?

If both F1 and F2 were in the same direction, could the stick remain at rest? (Consider moments about the pivot.)

 

[hidemi]

Thanks for your hint. I got it.