Equilibrium of a Rigid Body and the Principle of Moments

A single particle is in equilibrium when the net force on it is zero. An extended rigid body needs a

SECOND, independent condition as well, because a body can have zero net force acting on it and still begin

to spin. A rigid body is therefore said to be in complete mechanical equilibrium only when BOTH of the

following hold simultaneously:

1. Translational equilibrium: the vector sum of all external forces acting on the body is zero, $\sum \vec F_i = 0$ -- ensuring the body has no tendency to start translating (accelerating linearly).
2. Rotational equilibrium: the vector sum of the torques of all external forces, taken about any one common point, is also zero, $\sum \vec\tau_i = 0$ -- ensuring the body has no tendency to start rotating (angularly accelerating) about that point.

Neither condition alone is enough. A body can satisfy translational equilibrium while still failing

rotational equilibrium -- exactly the situation of a couple: two forces of exactly equal magnitude $F$,

exactly opposite in direction, but acting along two DIFFERENT parallel lines of action, separated by some

perpendicular distance $d$. Because the two forces are equal and opposite, their vector sum is zero, and

the body has no tendency to translate at all; but because their lines of action do not coincide, their

torques about any point do NOT cancel, and instead add up to a net, nonzero turning effect (a moment) of

magnitude $Fd$, directed the same way for both forces. A couple is exactly what is applied to a steering

wheel, a screwdriver, or a tap when turned using two hands pushing in opposite directions on either side.

The principle of moments applies the rotational-equilibrium condition to the everyday case of a rigid

lever -- a light rod free to turn about a fixed pivot, or fulcrum -- with two (or more) forces applied at

different points along it, as in a simple beam balance or a see-saw (see the accompanying figure). Taking

torques about the pivot itself (so that the unknown reaction force at the pivot, whose line of action

passes through the pivot, contributes zero torque and drops out of the equation entirely), rotational

equilibrium for two forces $F_1$ and $F_2$ acting at perpendicular distances $d_1$ and $d_2$ from the …