Worked Exercise — Square Roundabout

Problem

A square roundabout with side $3\;\text{m}$ is in equilibrium. On one side a force $F_1 = 20\;\text{N}$ acts, perpendicular to the side. On another side a force $F_2 = ?$ acts, also perpendicular. Find the value of $F_2$ for rotational equilibrium about the centre.

Solution

Step 1 — Torque of a force. The torque about the centre is $M = F\cdot b$, where $b$ is the arm, that is, the distance of the centre from the line of application of the force.

Step 2 — The arms. For a square roundabout with side $\ell = 3\;\text{m}$, each side is at a distance from the centre of

$b_1 = b_2 = \frac{\ell}{2} = \frac{3}{2} = 1{,}5\;\text{m}$

Step 3 — Rotational equilibrium. The two forces produce opposite torques about the centre; for equilibrium they must balance:

$\abs{F_1}\,b_1 = \abs{F_2}\,b_2 \;\Rightarrow\; F_2 = F_1\,\frac{b_1}{b_2} = F_1$

Since the arms are equal,

$\ev{F_2 = F_1 = 20\;\text{N}}$

Links

Topics: Statica ed equilibrio Concepts: Equilibrio statico · Momento di una forza Skills: Bilancio dei momenti Objects: Corpo rigido