Equilibrium: balanced forces and torques

Key ideas

A torque (turning effect) is produced when a force acts at a distance from a pivot:

$\tau = Fd$

where $d$ is the perpendicular distance from the pivot to the line of the force. Torque is measured in newton metres (N m).

An object is in equilibrium when both:

the forces balance (no acceleration), and
the torques balance — total clockwise torque = total anticlockwise torque about any pivot (no rotation).

A balanced see-saw with a 30 kg child sitting 2.0 metres left of the pivot and a 40 kg child sitting 1.5 metres right of the pivot.

Worked ExampleBalancing a see-saw

A $30$ kg child sits $2.0$ m to the left of a see-saw's pivot. Where must a $40$ kg child sit for it to balance?

Step 1 — Anticlockwise torque from the lighter child

Her weight is $30 \times 9.8 = 294$ N, acting $2.0$ m from the pivot:

$\tau = Fd = (294)(2.0) = 588 \text{ N m}$

Step 2 — Set the clockwise torque equal

The heavier child's weight is $40 \times 9.8 = 392$ N at unknown distance $d$ :

$392\,d = 588 \;\Rightarrow\; d = 1.5 \text{ m}$

The heavier child sits closer to the pivot — the same torque from more force needs less distance.

Step 3 — Force check (for completeness)

The pivot pushes up with $294 + 392 = 686$ N, balancing the total downward weight.

Practice question

A $6.0$ m uniform plank is pivoted at its centre. A $250$ N toolbox sits $1.2$ m from the pivot. What force applied at the far end ( $3.0$ m from the pivot, other side) balances it?

Worked solution: torque to balance $= 250 \times 1.2 = 300$ N m; force needed $= 300 / 3.0 = 100$ N downward on the other side.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $30$ N force is applied at the end of a $0.60$ m long spanner, perpendicular to the handle.

Calculate the torque (turning moment) about the bolt.

Merit

A uniform plank is balanced on a central pivot. A $50$ N weight is placed $1.2$ m to the left of the pivot.

Calculate where a $40$ N weight must be placed on the right-hand side for the plank to balance.

Excellence

A $4.0$ m uniform beam is pivoted at its centre. A $200$ N box sits $1.5$ m to the left of the pivot, and a person pushes straight down at the right-hand end, $2.0$ m from the pivot.

Calculate the force the person must apply to balance the beam, and explain why the beam's own weight does not appear in your torque calculation.

Equilibrium: balanced forces and torques

Covers AS91171

Key ideas

A torque (turning effect) is produced when a force acts at a distance from a pivot:

$\tau = Fd$

where $d$ is the perpendicular distance from the pivot to the line of the force. Torque is measured in newton metres (N m).

An object is in equilibrium when both:

the forces balance (no acceleration), and
the torques balance — total clockwise torque = total anticlockwise torque about any pivot (no rotation).

A balanced see-saw with a 30 kg child sitting 2.0 metres left of the pivot and a 40 kg child sitting 1.5 metres right of the pivot.

Worked ExampleBalancing a see-saw

A $30$ kg child sits $2.0$ m to the left of a see-saw's pivot. Where must a $40$ kg child sit for it to balance?

Step 1 — Anticlockwise torque from the lighter child

Her weight is $30 \times 9.8 = 294$ N, acting $2.0$ m from the pivot:

$\tau = Fd = (294)(2.0) = 588 \text{ N m}$

Step 2 — Set the clockwise torque equal

The heavier child's weight is $40 \times 9.8 = 392$ N at unknown distance $d$ :

$392\,d = 588 \;\Rightarrow\; d = 1.5 \text{ m}$

The heavier child sits closer to the pivot — the same torque from more force needs less distance.

Step 3 — Force check (for completeness)

The pivot pushes up with $294 + 392 = 686$ N, balancing the total downward weight.

Practice question

A $6.0$ m uniform plank is pivoted at its centre. A $250$ N toolbox sits $1.2$ m from the pivot. What force applied at the far end ( $3.0$ m from the pivot, other side) balances it?

Worked solution: torque to balance $= 250 \times 1.2 = 300$ N m; force needed $= 300 / 3.0 = 100$ N downward on the other side.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A $30$ N force is applied at the end of a $0.60$ m long spanner, perpendicular to the handle.

Calculate the torque (turning moment) about the bolt.

Merit

A uniform plank is balanced on a central pivot. A $50$ N weight is placed $1.2$ m to the left of the pivot.

Calculate where a $40$ N weight must be placed on the right-hand side for the plank to balance.

Excellence

A $4.0$ m uniform beam is pivoted at its centre. A $200$ N box sits $1.5$ m to the left of the pivot, and a person pushes straight down at the right-hand end, $2.0$ m from the pivot.

Calculate the force the person must apply to balance the beam, and explain why the beam's own weight does not appear in your torque calculation.