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)

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:

TestHub · Accepted Answer

The stick does not move, so net force and net torque are zero. Take torques about the pivot (80 cm). $F_{1}$ acts at 80 cm from the pivot and $F_{2}$ at 20 cm on the same side, so $F_{1} (80) = F_{2} (20)$ with $F_{2}$ opposite to $F_{1}$ : $∣ F_{2} ∣ = 4∣ F_{1} ∣$ .

Force balance: $F_{1} - F_{2} + P = 0$ (taking $F_{1}$ direction as positive), so the pivot force is $P = ∣ F_{2} ∣ - ∣ F_{1} ∣$ , in the same direction as $F_{1}$ .

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