Equilibrium of a rigid body

For a point mass, equilibrium is a single condition: zero net force. A rigid body, which can rotate as well as translate, needs two simultaneous conditions.

Law — Conditions for equilibrium

A rigid body is in equilibrium when:

1. $\sum\vec{F} = \vec{0}$ (translational equilibrium);
2. $\sum M = 0$ (rotational equilibrium, about any point).

The second condition has a convenient property: if the first is satisfied (zero net force), then the sum of the moments is the same about any pole. You can therefore choose the most convenient pole — usually the one through which most unknown forces pass, so as to eliminate them from the moment balance.

Warning — One condition alone is not enough

The first condition alone is not enough! A body with $\sum\vec{F}=\vec{0}$ but $\sum M\neq 0$ rotates without translating (this is the case of a couple of forces). Conversely, $\sum M = 0$ with $\sum\vec{F}\neq\vec{0}$ gives translation without rotation. Both are needed for complete rest.

The worked exercise The door with two forces shows how to compute the resultant moment about a pole (the hinge) and check whether the body is in rotational equilibrium.

Links

Topics: Statics and equilibrium Concepts: Static equilibrium · Moment of a force Skills: Moment balance

Related exercises: Worked exercise — The door with two forces · Worked exercise — System with two masses and a pulley · Worked exercise — Square merry-go-round