Topics

Equilibrium: balanced forces and torques

The two conditions for equilibrium

An object is in equilibrium when it has no acceleration — neither moving off in a line nor starting to rotate. That requires both conditions:

Condition 1 — the forces balance. The net force in every direction is zero:
- total up = total down,
- total left = total right.
Condition 2 — the torques balance. About any point:
- total clockwise torque = total anticlockwise torque.
Both must hold. Balanced forces alone do not stop an object rotating: two equal and opposite forces applied at different points will spin it while its centre stays put.
Equilibrium does not mean stationary. An object moving at constant velocity is also in equilibrium.

Centre of mass and uniform beams

The weight of a beam acts at its centre of mass, drawn as a single downward arrow at that point.
For a uniform beam (same thickness and material throughout), the centre of mass is at its geometric centre.
If a beam is described as "light" or of "negligible mass", ignore its weight entirely.
A beam pivoted at its centre has its own weight acting through the pivot, so the beam's weight produces no torque — which is why seesaw problems usually ignore it.

Solving an equilibrium problem

The routine works for every seesaw, plank, bridge and shelf question:

Draw the situation with all forces marked, including the beam's weight at its centre if it is not negligible.
Choose a pivot. If a force is unknown, pivot at the point where it acts — its torque is then zero and it disappears from the torque equation.
Write the torque equation: total clockwise = total anticlockwise. Each torque is (force) × (distance from your chosen pivot).
Solve for the first unknown.
Write the force equation, total up = total down, to find any remaining unknown.
Distances in the torque equation are measured from the pivot you chose, not from the end of the beam. Redraw and re-measure if you change pivots.

Stability and toppling

An object topples when its line of weight falls outside its base.
- While the weight acts inside the base, the normal force produces a torque that rights it.
- Once the weight acts outside, the torque about the tipping edge carries it over.
An object is more stable when it has:
- a lower centre of mass,
- a wider base.
This is why a racing car is low and wide, and why a tall bookcase is bolted to the wall.

Worked ExampleA seesaw

A uniform plank is pivoted at its centre. A $30$ kg child sits $1.8$ m to the left of the pivot. How far to the right of the pivot must a $45$ kg child sit to balance the plank?

Step 1 — Find the two forces

$F_1 = m_1g = (30)(9.8) = 294 \text{ N}$ $F_2 = m_2g = (45)(9.8) = 441 \text{ N}$

The plank is uniform and pivoted at its centre, so its own weight acts through the pivot and produces no torque.

Step 2 — Write the torque condition

For balance, anticlockwise torque = clockwise torque:

$F_1 d_1 = F_2 d_2$ $(294)(1.8) = (441)d_2$ $529.2 = 441 d_2$

Step 3 — Solve

$d_2 = \frac{529.2}{441} = 1.2 \text{ m}$

$d_2 = 1.2 \text{ m from the pivot}$

Worked ExampleA plank on two supports

A uniform plank of mass $20$ kg and length $4.0$ m rests on two supports, one at each end. A $60$ kg person stands $1.0$ m from the left support. Find the upward force from each support.

Step 1 — List all the forces

Weight of the person: $(60)(9.8) = 588$ N down, $1.0$ m from the left end.
Weight of the plank: $(20)(9.8) = 196$ N down, at the centre, $2.0$ m from the left end.
Upward force from the left support, $F_L$ , at $0$ m.
Upward force from the right support, $F_R$ , at $4.0$ m.

Step 2 — Choose a pivot that removes an unknown

Take torques about the left support. Then $F_L$ acts through the pivot and contributes no torque, leaving only $F_R$ unknown.

Clockwise (the two weights, both pushing the right side down about this pivot):

$\tau_{\text{clock}} = (588)(1.0) + (196)(2.0) = 588 + 392 = 980 \text{ N m}$

Anticlockwise (the right support lifting):

$\tau_{\text{anti}} = F_R(4.0)$

Step 3 — Balance the torques

$F_R(4.0) = 980 \;\Rightarrow\; F_R = \frac{980}{4.0} = 245 \text{ N}$

Step 4 — Use the force condition for the other support

Total up = total down:

$F_L + F_R = 588 + 196 = 784 \text{ N}$ $F_L = 784 - 245 = 539 \text{ N}$

$F_L = 539 \text{ N}, \qquad F_R = 245 \text{ N}$