Question

An object that weighs 1160 N rests on a lever at a point 0.410 m from a support. On the same side of the support, at a distance of 3.00 m from it, an upward force with magnitude F is applied. Ignore the weight of the board itself. If the system is in equilibrium, what is F? Express your answer in Newtons.

Answer 1

The system in order to be in equilibrium, the applied force F should be equivalent to the weight force.

$F\cdot3.00m=1160N\cdot0.410m$

Observe that we had to multiply each force by its leverage, which is the distance from the pivot. Solve for F.

$\begin{gathered} F=\frac{475.6}{3}N \\ F\approx158.5N \end{gathered}$

Therefore, the force F is 158.5 N.