Two massless rods are attached to frictionless pivots, with their ends touching. The distances between the pivot points and the endpoints of the rods are shown below.

Neglecting friction between the rods, if a force F is applied at the left end of the left rod, what force $F^{/}$ must be applied at the right end of the right rod to keep the system in equilibrium?

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$\frac{F}{8}$

$\frac{4 F}{7}$

$\frac{F}{2}$

$\frac{6 F}{5}$

answer is B.

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Detailed Solution

By balancing torques on the first rod, we see that the normal force on the right end is twice the force exerted on the left end; that is, the first rod is acting as a lever with mechanical advantage 2. By similar logic, the second rod acts as a lever with mechanical advantage $1 / 4$ . Thus, the combination of the two has mechanical advantage $2 (1 / 4) = 1 / 2$ , so $F^{/} = F / 2$ .