Equilibrium of a Rigid Body

Two conditions, not one

The chapter on Objects in Equilibrium treated every body as a point. A point is in equilibrium when the forces on it add to zero, \(\sum\vec F = 0\). A rigid body needs more. The forces on it act at different points, and even when they add to zero they can still turn it. A couple is the plainest example: zero net force, yet the wheel spins.

So a rigid body is in equilibrium when two conditions hold together. The net force is zero, so the centre of mass does not accelerate. The net torque is zero, so the angular velocity does not change. \[ \sum\vec F = 0, \qquad \sum\vec\tau = 0 \]

The two conditions are independent. A couple satisfies the first and breaks the second. Two equal forces pushing the same way, on either side of the centre, can do the opposite. Their torques about the centre cancel, but together they accelerate the body. Neither condition can stand in for the other.

Equilibrium means that nothing changes, not that nothing moves. A grinding wheel turning at a steady rate, with no friction at its bearings, has zero net torque and is in rotational equilibrium. A body at rest that stays at rest is in static equilibrium, the case of every problem in this chapter. For forces in one plane, the conditions become three equations: \[ \sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum \tau = 0 \]

Torques about which point?

Torques depend on the point they are taken about. So the condition \(\sum\vec\tau = 0\) seems to need a choice: torques about where? When the net force is zero, the choice does not matter. Move the reference point from O to a new point O′, at position \(\vec c\) from O. Each position vector becomes \(\vec r_i - \vec c\). The new total torque is \[ \vec\tau\,' = \sum(\vec r_i - \vec c)\times\vec F_i = \sum \vec r_i\times\vec F_i - \vec c\times\sum\vec F_i = \vec\tau - \vec c\times\sum\vec F_i \] If \(\sum\vec F_i = 0\), the last term vanishes and \(\vec\tau\,' = \vec\tau\).

This freedom is the most useful tool of statics. Take torques about the point where the unknown forces act. Their moment arms are zero there, so they vanish from the torque equation. Often only one unknown is left, and one line of algebra finds it. The force equations then give the rest.

The see-saw

A see-saw is a plank on a central pivot. Take torques about the pivot, where the unknown push of the pivot acts. The weight of a child on the left turns the plank anticlockwise, and the weight of a child on the right turns it clockwise. The plank's own weight acts at the pivot, so it has no torque. The plank balances when the two torques are equal.

In the first state of Fig 3.1, the weights are \(250\,\text{N}\) and \(300\,\text{N}\). About the pivot, \(250 \times 1.8 = 450\,\text{N m}\) anticlockwise and \(300 \times 1.5 = 450\,\text{N m}\) clockwise. The net torque is zero, so the plank balances. The heavier child sits nearer the pivot. The distances, \(1.8 : 1.5\), are in the inverse ratio of the weights, \(300 : 250\).

Now the lighter child moves out to \(2.4\,\text{m}\). The anticlockwise torque grows to \(250 \times 2.4 = 600\,\text{N m}\), against \(450\,\text{N m}\). The net torque of \(150\,\text{N m}\), anticlockwise, turns the left end down. To balance again, the heavier child needs \(300\,d = 600\), so \(d = 2.0\,\text{m}\). The pivot, meanwhile, must push up with \(250 + 300 = 550\,\text{N}\) plus the plank's weight, so that the net force is zero too.

A beam on two supports

A plank across a drain, a bench on two legs and a bridge on two piers are all beams on two supports. The supports push up with forces \(N_L\) and \(N_R\), which are unknown. Take torques about one support, and its reaction has no torque. That gives the other reaction at once. The force balance, or a second torque equation, then gives the first.

Walk the person from left to right with the slider. The left reaction falls steadily and the right one rises, by \(100\,\text{N}\) for every metre. With the person at the centre, \(x = 2.5\,\text{m}\), both supports carry \(400\,\text{N}\). At every position the two reactions add to \(800\,\text{N}\), the total weight. The torque condition decides how the load is shared; the force condition fixes the total.

Now return to the see-saw in the park. The girl's weight, \(200\,\text{N}\) at \(2.4\,\text{m}\), gives \(480\,\text{N m}\) about the pivot. The father's \(600\,\text{N}\) must give the same, so he sits \(0.8\,\text{m}\) from the pivot. Then the pivot pushes up with \(800\,\text{N}\) plus the plank's weight, and both conditions of equilibrium hold.