Suspended laminas and equilibrium angles

The key fact

A lamina freely suspended from a point P hangs in equilibrium with P and the centre of mass G in a vertical line. (Otherwise the weight would have a moment about P and the lamina would turn.)

Finding the angle

1. Find G relative to convenient axes.

2. Draw the lamina with the line PG vertical.

3. Use a right-angled triangle containing PG and the edge you want: tan θ = opposite/adjacent, where the sides are distances measured from P along and perpendicular to that edge.

Pearson's examiners report candidates using the reciprocal of the correct ratio, so label the triangle carefully on your sketch.

Making it hang a certain way

A particle may be attached so that the lamina hangs with an edge vertical or at a given angle. Write the new centre of mass in terms of the unknown mass m, then use the condition (for example, G lies on a particular line through P).

Worked example

The lamina from the composite lesson (8 by 6 rectangle with a 2 by 2 corner removed) has centre of mass (3.73, 2.82) relative to O. It is hung from O. Find the angle between OA (the x-axis edge) and the vertical.

1. OG is vertical.
2. Measured from O: G is 3.73 along OA and 2.82 perpendicular to it.
3. tan θ = 2.82/3.73, so θ = 37.1°.

Answer: 37.1°

Check yourself

1. A uniform rectangular lamina 6 cm by 4 cm is hung from one corner. Find the angle between the longer side through that corner and the vertical, in degrees to 1 decimal place.
   Show answer

   33.7 °. G is 3 along the long side and 2 perpendicular to it, so tan θ = 2/3, θ = 33.7°.

2. The L-shaped lamina with centre of mass (2.2, 2.2) relative to its corner O is hung from O. Find the angle between the x-axis edge and the vertical.
   Show answer

   45 °. tan θ = 2.2/2.2 = 1, so θ = 45°.

3. A uniform square lamina of side 10 cm and mass 4 kg has corners O(0, 0) and (10, 10). A 2 kg particle is attached at (10, 0). The lamina is hung from O. Find the angle between the edge along the x-axis and the vertical, in degrees to 1 decimal place.
   Show answer

   26.6 °. G = ((4×5 + 2×10)/6, (4×5 + 2×0)/6) = (6.67, 3.33). tan θ = 3.33/6.67 = 0.5, θ = 26.6°.

4. A uniform semicircular lamina of radius 6 cm is hung from one end A of its diameter. Find the angle between the diameter and the vertical, in degrees to 1 decimal place.
   Show answer

   23.0 °. G is 6 cm along the diameter from A and 4r/(3π) = 2.55 cm from it. tan θ = 2.55/6, θ = 23.0°.

5. A uniform rectangular lamina OABC of mass 3 kg has O(0, 0), A(6, 0), B(6, 4), C(0, 4). A particle of mass m kg is attached at C. Hung from O, OA makes 45° with the vertical. Find m.
   Show answer

   0.75 kg. G = (9/(3 + m), (6 + 4m)/(3 + m)). At 45°, x̄ = ȳ: 9 = 6 + 4m, so m = 0.75 kg.

6. A lamina hangs freely in equilibrium from a point P. What must be true?
   • Its centre of mass is vertically below P
   • Its longest edge is vertical
   • Its centre of mass is level with P
   • The tension equals half its weight
   Show answer

   Its centre of mass is vertically below P. Otherwise the weight has a moment about P and the lamina rotates.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy
• Pearson Edexcel IAL Mechanics M2 (WME02) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.