Cambridge Notes, Past Papers, Revision Questions

IGCSE Physics 0625 – Turning Effect of Forces

1.5.2 Turning Effect of Forces

Objective

Describe an experiment that demonstrates there is no resultant moment (net torque) on an object that is in static equilibrium.

Key Concepts

Moment (torque) \$M = F \times d\$, where \$F\$ is the force and \$d\$ is the perpendicular distance from the line of action of the force to the pivot.

Clockwise moments are taken as positive, anticlockwise moments as negative (or vice‑versa, provided consistency is maintained).

For an object in equilibrium the vector sum of forces is zero and the algebraic sum of moments about any point is zero:
\$\sum F = 0 \qquad \text{and} \qquad \sum M = 0\$

Apparatus

Rigid wooden board (≈ 0.5 m long) with a low‑friction pivot at its centre.

Set of calibrated masses (e.g., 50 g, 100 g, 200 g).

Two identical strings with hooks at the ends.

Meter rule or measuring tape.

Clamp stand to hold the board horizontally when required.

Experimental Arrangement

The board is placed on the pivot so that it can rotate freely about the centre. Two strings are attached to the board at known distances from the pivot (one on the left, one on the right). Masses are hung from the strings to create forces at different lever arms.

Suggested diagram: Board on a central pivot with two strings at distances \$d1\$ and \$d2\$ from the pivot, each supporting a hanging mass.

Method

Measure and record the distances \$d1\$ (left side) and \$d2\$ (right side) from the centre of the pivot to the points where the strings are attached.

Hang a mass \$m_1\$ on the left string. Allow the board to come to rest and note the direction of rotation (if any).

Adjust the mass on the right string (\$m2\$) until the board remains perfectly horizontal and does not rotate. Record the value of \$m2\$.

Repeat the procedure for at least three different pairs of distances (e.g., \$d1 = 0.10\,\$m, \$d2 = 0.20\,\$m; \$d1 = 0.15\,\$m, \$d2 = 0.25\,\$m; etc.).

For each trial calculate the moments produced by each hanging mass and verify that their algebraic sum is zero.

Data Table

Trial	\$d_1\$ (m)	\$d_2\$ (m)	\$m_1\$ (kg)	\$m_2\$ (kg)	Moment left \$M1 = m1 g d_1\$ (N·m)	Moment right \$M2 = m2 g d_2\$ (N·m)	\$\sum M\$ (N·m)
1	0.10	0.20	0.200	0.100	\$0.200 \times 9.8 \times 0.10 = 0.196\$	\$0.100 \times 9.8 \times 0.20 = 0.196\$	0 (within experimental error)
2	0.15	0.25	0.150	0.090	\$0.150 \times 9.8 \times 0.15 = 0.221\$	\$0.090 \times 9.8 \times 0.25 = 0.221\$	0 (within experimental error)
3	0.12	0.30	0.250	0.100	\$0.250 \times 9.8 \times 0.12 = 0.294\$	\$0.100 \times 9.8 \times 0.30 = 0.294\$	0 (within experimental error)

Theoretical Explanation

When the board is in equilibrium, the clockwise moments must balance the anticlockwise moments. Using the definition of moment:

\$M{\text{left}} = m1 g d1,\qquad M{\text{right}} = m2 g d2\$

Setting the algebraic sum to zero gives:

\$m1 g d1 - m2 g d2 = 0 \;\;\Longrightarrow\;\; m1 d1 = m2 d2\$

Thus, for any pair of forces acting at different distances, the product of force and lever arm must be equal for the net moment to vanish.

Sources of Error

Friction at the pivot may introduce a small resisting torque.

Inaccurate measurement of distances \$d1\$ and \$d2\$.

Masses may not be perfectly centred on the strings, altering the effective lever arm.

Air currents or vibrations.

Conclusion

The experiment confirms that an object can be in static equilibrium when the sum of the moments about any axis is zero. By adjusting the masses so that \$m1 d1 = m2 d2\$, the clockwise and anticlockwise torques cancel, resulting in no resultant moment. This demonstrates the principle that for equilibrium the turning effects of all forces must balance.

Describe an experiment to demonstrate that there is no resultant moment on an object in equilibrium