Pearson Edexcel International Advanced Level Mathematics: Centres of mass and rigid body equilibrium (YMA01) – Revision Notes

These notes condense sections 2 and 5 of unit M2 in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics specification, Issue 3, April 2019 (centres of mass 2.1 to 2.3, moments and statics of rigid bodies 5.1 and 5.2). The specification lists M2 as an optional unit for IAS Further Mathematics and for IAL Mathematics and Further Mathematics.

For full explanations and longer worked examples, read the study guide. When you are ready to test yourself, use the practice questions. Course hub: /boards/edexcel/a-level/mathematics/. Checklist: /checklists/edexcel/a-level/mathematics/. To find weak spots, try the free 10-minute diagnostics.

Key results

Topic	Result
Moment (5.1)	Force × perpendicular distance, in N m. For force F at angle θ to a rod of length d: F d sin θ
Equilibrium (5.2)	Resultant force zero and sum of moments about any point zero
Particles (2.1)	x̄ = Σ(m x)/Σm, ȳ = Σ(m y)/Σm
Composite lamina (2.2)	x̄ = Σ(A x)/ΣA, with area A as the mass for a uniform lamina; a hole has negative area
Friction	F ≤ μR; F = μR only when slipping is about to happen
Weight	W = m g, acting at the centre of mass

The specification allows results from the formulae book to be quoted without proof. Look at your own copy of the booklet to see which are printed, and learn the ones in the next table.

Standard centres of mass (uniform)

Shape	Position
Rectangle, circle	Centre of symmetry
Triangle	On a median, one third of the way from the side to the opposite vertex
Semicircle, radius r	4r/(3π) from the diameter, on the symmetry axis
Sector, radius r, half-angle α	2r sin α/(3α) from the centre, on the symmetry axis

Methods in steps

Centre of mass of a composite lamina

1. Split into rectangles, triangles, circles, semicircles or sectors. Treat holes as negative.
2. Fix axes. Use symmetry to write down one coordinate if possible.
3. Make a table: area, x, y for each part.
4. Divide the sum of area × coordinate by the total area.
5. Give exact values (with π) and then decimals.

Lamina hanging from a point

1. Find G relative to the point of suspension.
2. G is directly below the point.
3. Use a right-angled triangle between the point, G and the side or edge you need to refer to.

Lamina on a rough inclined plane

1. Topple: G is above the lowest contact point, so tan α = (half the base)/(height of G above the plane).
2. Slip: tan α = μ.
3. The smaller angle happens first as the slope rises.

Rod or ladder

1. Draw all forces: weights at centres of mass, normal reactions, friction, tensions.
2. Resolve horizontally and vertically.
3. Take moments about the point where the most unknowns meet, usually the foot.
4. Add friction conditions last: F = μR if limiting, otherwise F ≤ μR.

Beam about to tilt: the reaction at the support it is not tilting about is zero. Say that in words.

Worked reminders

L-shaped lamina. A uniform lamina is made from a rectangle 8 cm by 3 cm, with a second rectangle 3 cm by 4 cm standing on one end of its long side, flush with the left edge. Find the centre of mass from the outer corner O, then the angle the 8 cm edge makes with the vertical when the lamina hangs from O.

Part     Area   x     y
Lower    24     4     1.5
Upper    12     1.5   5
x̄ = (96 + 18)/36 = 19/6 = 3.17 cm      ȳ = (36 + 60)/36 = 8/3 = 2.67 cm
Hanging from O, OG is vertical: tan θ = ȳ/x̄ = (8/3)/(19/6) = 16/19
θ = 40.1° between the 8 cm edge and the vertical

Slope: slip or topple? A uniform rectangular lamina has base 0.8 m on a rough plane and height 0.3 m, and μ = 0.9.

Topple: tan α = 0.4/0.15 = 8/3,  α = 69.4°
Slip:   tan α = 0.9,             α = 42.0°

It slips first, at 42.0°. A wide, low shape slips before it topples unless friction is large.

Tilting beam. A uniform beam AB has length 3 m and mass 12 kg. Supports are at C and D, with AC = 0.4 m and AD = 2.4 m. Find the greatest mass that can be placed at B without tilting.

About to tilt about D, so the reaction at C is zero.
Moments about D:  m g (0.6) = 12 g (0.9)
m = 18 kg

Ladder, smooth wall. A uniform ladder rests at 50° to the ground against a smooth wall, with nobody on it. Find the least coefficient of friction.

Horizontal: S = F.   Vertical: R = W.
Moments about the foot: S L sin 50° = W (L/2) cos 50°
S = W/(2 tan 50°) = 0.420 W
μ ≥ F/R = 0.420  (3 s.f.)

Must-know distinctions

• Light, uniform, non-uniform. A light rod has no weight. A uniform rod’s weight acts at its midpoint. For a non-uniform rod the position of the weight is unknown, and finding it may be the question.
• Equilibrium versus limiting equilibrium. Equilibrium needs F ≤ μR. Only limiting equilibrium gives F = μR.
• Slip versus topple. Slipping depends on μ; toppling depends on the shape.
• Smooth versus rough wall. A smooth wall gives a single horizontal reaction. A rough wall adds friction along the wall.
• Mass versus area. For a uniform lamina, areas work as masses because the lamina has the same mass per unit area everywhere.
• Centre of mass versus point of suspension. In a hanging lamina these two points lie on the same vertical line.

Quick self-test

1. A force of 12 N acts at a perpendicular distance 0.35 m from a pivot. Find its moment.
2. A force of 20 N acts on the end of a 0.9 m rod at 30° to the rod. Find its moment about the other end.
3. Particles of 1 kg and 3 kg are at x = 2 and x = 6. Find x̄.
4. A uniform triangular lamina has height 9 cm. How far is its centre of mass from the base?
5. Find the distance from the diameter of the centre of mass of a uniform semicircular lamina of radius 6 cm.
6. A lamina hangs freely from a point P. Where is G?
7. A uniform rectangle has base 0.4 m and height 0.6 m. At what angle of slope, with enough friction, does it begin to topple?
8. A uniform 4 m beam of mass 20 kg lies on supports at its two ends, A and B. A 10 kg load sits 1 m from A. Find both reactions (g = 9.8).
9. A uniform disc of radius 5 cm has a circular hole of radius 1 cm whose centre is 2 cm from the disc’s centre O. How far is the centre of mass from O?
10. Particles of mass 2 kg at (0, 3) and 6 kg at (4, −1). Find the centre of mass.
11. Find the distance of the centre of mass from the centre of a uniform sector of radius 6 cm and half-angle π/3.
12. Why is “F = μR” wrong for a ladder that is resting steadily?

Answers

1. 12 × 0.35 = 4.2 N m
2. 20 × 0.9 × sin 30° = 9 N m
3. (1×2 + 3×6)/4 = 5
4. 3 cm (a third of the height)
5. 4×6/(3π) = 8/π = 2.55 cm
6. Vertically below P
7. tan α = 0.2/0.3 = 2/3, so 33.7°
8. Moments about B: 4 R_A = 20g(2) + 10g(3) = 70g, so R_A = 17.5g = 171.5 N; R_B = 30g − 171.5 = 122.5 N
9. Hole area π, disc area 25π. Distance = π(2)/(24π) = 1/12 cm = 0.0833 cm, on the side away from the hole
10. x̄ = 24/8 = 3, ȳ = (6 − 6)/8 = 0, so (3, 0)
11. 2(6) sin(π/3)/(3 × π/3) = 6√3/π = 3.31 cm
12. Friction is only as large as needed, so F ≤ μR. F = μR holds only at the point of slipping.

Where marks are usually lost

• Using a distance measured along the rod instead of the perpendicular distance.
• Leaving out a rod’s weight, or putting the weight of a non-uniform rod at its midpoint.
• Using side lengths as masses for a lamina, or forgetting to subtract a hole.
• Mixing origins: one part measured from the left, another from the right.
• Using 4r/(3π) from the centre of a semicircle instead of from the diameter.
• Writing F = μR when the question says only that the body is in equilibrium.
• Forgetting the friction at a rough wall in the vertical equation.
• Stating a tilting condition without saying which reaction is zero.
• Rounding a reaction early and then using the rounded value in the next equation.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, unit M2: sections 2 (centres of mass) and 5 (statics of rigid bodies).