7. A massless meter stick on a horizontal frictionless table top is pivoted at t

Question

7. 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 is applied perpendicularly to the stick at the 100-cm end of the stick. 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

I think the answer is E, but I'm not sure why. Please explain, thank you!!

Explanation / Answer

Since the stick does not move therefore the stick is in equilibrium therefore
Net Force and Net Torque acting on the stick would be zero

therefore  the force exerted by the pivot on the stick must be directed opposite to F1 and F2 and have magnitude of F2+F1 as then only Net Force would be zero

as Force in downward direction is F1 + F2 therefore the force exerted by the pivot on the stick will be in opposite direction

Hence answer is option E) must be directed opposite to F1 and have magnitude F2+F1

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