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

Question

4. 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 (not shown) is applied perpendicularly at the 100-cm end of the suck. The forces are horizontal. If the stick does not move, the force exerted by the pivot on the stick: A) must be in the same direction as F1 and have magnitude |F2| + |F1| B) must be directed opposite to F1 and have magnitude |F2| + |F1| C) must be zero 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

a) The stick is in equilibrium: the net force and net torque on the stick will be zero. i.e, Force exerted by the pivot on the stick=F1+F2...........(1) and F1*.80=F2*.20=>F2=4F1. hence from (1) Force exerted by the pivot on the stick=F1+F2=F1+4F1=5F1

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