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This lesson covers AQA Spec 3.4.1.1 (Moments) in full depth, including the principle of moments, centre of mass, toppling versus sliding, and coplanar force problems.
Spec mapping (AQA 7408 §3.4.1.1): This lesson develops moments of a force about a pivot, the couple as a pair of equal-and-opposite parallel forces with non-coincident lines of action, the principle of moments for rotational equilibrium, and the conditions for full static equilibrium (zero resultant force and zero resultant moment about any point). The strand also includes centre of mass / centre of gravity for uniform and composite bodies, and the geometric criteria for sliding versus toppling on an inclined surface. (Refer to the official AQA specification document for exact wording.)
Synoptic links: Moments and equilibrium underpin static analysis across mechanics. (i) Newton's laws (Lesson 3, §3.4.1.4) provide the translational equilibrium half (ΣF = 0); the moment principle adds the rotational half (ΣM = 0). Both must hold for static equilibrium. (ii) Stress and Young's modulus (Lesson 7, §3.4.2.2) uses the same force-balance machinery to analyse loaded structural members — bridges, beams, cranes — where the load distribution is the engineering input and the moment equation gives reactions at the supports. (iii) Circular motion and rotational dynamics (Year 13) generalises torque from "moment" to "torque vector," and the rotational analogue of F = ma is τ = Iα. The conceptual base is laid here. (iv) Bridge and crane design in engineering follows directly from the principle of moments — every external statics problem is a moment-balance problem at heart.
Key Definition: The moment of a force about a point (or pivot) is the product of the force and the perpendicular distance from the line of action of the force to the point.
Moment = F × d
where d is the perpendicular distance. SI unit: N m.
If the force acts at an angle θ to the line joining the point of application to the pivot:
Moment = F × r × sin θ
where r is the distance from the pivot to the point of application.
A moment can be clockwise or anticlockwise about the pivot.
A spanner is 0.30 m long. A force of 50 N is applied at the end, perpendicular to the spanner. Find the moment about the nut.
Solution:
Moment = F × d = 50 × 0.30 = 15 N m
The same 50 N force is applied at 60° to the spanner handle. Find the moment.
Solution:
Moment = F × r × sin θ = 50 × 0.30 × sin 60° = 50 × 0.30 × 0.866 = 13.0 N m
Exam Tip: Always use the perpendicular distance to the line of action. If the force is at an angle, either resolve the force or use r sin θ.
Key Definition: A couple consists of two equal and opposite parallel forces whose lines of action do not coincide.
A couple produces pure rotation — no resultant translational force.
Torque of a couple = F × d
where d is the perpendicular distance between the lines of action of the two forces.
Note: The torque of a couple is the same about any point. You do not need to specify a pivot.
A steering wheel has a diameter of 0.40 m. A driver applies equal and opposite forces of 12 N at opposite ends of a diameter. Find the torque.
Solution:
Torque = F × d = 12 × 0.40 = 4.8 N m
For a body in rotational equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point.
Σ Clockwise moments = Σ Anticlockwise moments
This applies about ANY chosen pivot point. A wise choice of pivot simplifies the calculation by eliminating unknown forces that pass through the pivot.
A uniform beam of length 4.0 m and weight 200 N is supported at its ends, A and B. A 500 N load is placed 1.0 m from end A. Find the reaction forces at A and B.
Solution: The beam's weight (200 N) acts at its centre, 2.0 m from A.
Taking moments about A (to eliminate the reaction at A):
Clockwise moments = Anticlockwise moments
500 × 1.0 + 200 × 2.0 = R_B × 4.0
500 + 400 = 4.0 R_B
R_B = 900/4.0 = 225 N
Resolving vertically: R_A + R_B = 500 + 200 = 700 N
R_A = 700 − 225 = 475 N
Check: Take moments about B:
R_A × 4.0 = 500 × 3.0 + 200 × 2.0 = 1500 + 400 = 1900
R_A = 1900/4.0 = 475 N ✓
A uniform plank of length 6.0 m and mass 30 kg is supported at points P (1.0 m from the left end) and Q (1.0 m from the right end). Loads of 200 N and 400 N are placed at the left end and right end respectively. Find the reactions at P and Q.
Solution: Weight of plank = 30 × 9.81 = 294.3 N, acting at the centre (3.0 m from left end).
Taking moments about P (to find R_Q):
Clockwise moments: 294.3 × (3.0 − 1.0) + 400 × (6.0 − 1.0) = 294.3 × 2.0 + 400 × 5.0 = 588.6 + 2000 = 2588.6 N m
Anticlockwise moments: 200 × 1.0 + R_Q × (6.0 − 1.0 − 1.0) = 200 + R_Q × 4.0
So: 200 + 4.0 R_Q = 2588.6
R_Q = 2388.6/4.0 = 597 N
Resolving vertically: R_P + R_Q = 200 + 294.3 + 400 = 894.3 N
R_P = 894.3 − 597 = 297 N
The centre of mass (centre of gravity for a uniform gravitational field) is the point where the entire weight of an object can be considered to act.
For a uniform object, the centre of mass is at the geometric centre.
For an L-shaped or composite object, find the centre of mass by taking moments.
A thin uniform L-shaped plate is made by joining two rectangles:
Find the position of the centre of mass.
Solution:
x̄ = (m_A × x_A + m_B × x_B) / (m_A + m_B) = (8.0 × 0.5 + 6.0 × 2.5) / (8.0 + 6.0)
x̄ = (4.0 + 15.0) / 14.0 = 19.0/14.0 = 1.36 m
ȳ = (m_A × y_A + m_B × y_B) / (m_A + m_B) = (8.0 × 2.0 + 6.0 × 0.5) / 14.0
ȳ = (16.0 + 3.0) / 14.0 = 19.0/14.0 = 1.36 m
For a rigid body to be in complete equilibrium (static equilibrium):
For coplanar forces, this gives three independent equations:
An object on a slope will either slide or topple depending on the geometry and friction.
Sliding occurs when the component of weight along the slope exceeds the maximum friction force:
mg sin θ > μmg cos θ → tan θ > μ → θ > tan⁻¹(μ)
Toppling occurs when the vertical line through the centre of mass falls outside the base of the object. For a uniform block of width w and height h on a slope at angle θ:
The block topples when: tan θ > w/h
Which happens first?
A uniform block is 0.40 m wide and 1.20 m tall on a slope. The coefficient of friction is μ = 0.50. Does the block slide or topple first as the slope angle increases?
Solution:
Sliding angle: θ_slide = tan⁻¹(μ) = tan⁻¹(0.50) = 26.6°
Toppling angle: θ_topple = tan⁻¹(w/h) = tan⁻¹(0.40/1.20) = tan⁻¹(0.333) = 18.4°
Since θ_topple < θ_slide, the block topples first (at 18.4°).
Exam Tip: A tall, narrow object with a rough surface is more likely to topple. A short, wide object on a smooth surface is more likely to slide.
If exactly three non-parallel coplanar forces act on a body in equilibrium, their lines of action must pass through a single point (they are concurrent).
This can be proven by taking moments about any point — if two forces intersect at a point, the third must also pass through that point for the net moment to be zero.
A triangle of forces can be drawn: the three force vectors, placed nose-to-tail, form a closed triangle.
A uniform ladder of length 5.0 m and weight 250 N rests against a smooth vertical wall, with its foot 1.5 m from the base of the wall on a rough horizontal floor. The ladder is in equilibrium. Find (a) the reaction force at the wall, (b) the normal reaction at the floor, and (c) the friction force at the floor, and (d) the coefficient of friction required between the ladder and the floor.
Solution:
The ladder's base is at 1.5 m horizontal distance from the wall. Length 5.0 m, so the top of the ladder is at vertical distance √(5.0² − 1.5²) = √(25 − 2.25) = √22.75 = 4.77 m above the ground. The angle of the ladder above the horizontal: cos θ = 1.5/5.0 = 0.30, so θ = 72.5°. (Or sin θ = 4.77/5.0 = 0.954.)
Three forces act on the ladder: weight W = 250 N vertically downward at the centre (2.5 m along the ladder); normal reaction from the wall N_W horizontally (the wall is smooth, so no friction there); and reaction from the floor consisting of normal N_F vertically upward and friction F horizontally toward the wall.
Vertical equilibrium: N_F = W = 250 N.
Take moments about the foot of the ladder (to eliminate N_F and F from the equation):
Clockwise: W × (horizontal distance from foot to centre of gravity) = 250 × (½ × 1.5) = 250 × 0.75 = 187.5 N m.
Anticlockwise: N_W × (vertical height of top) = N_W × 4.77.
For equilibrium: N_W × 4.77 = 187.5 → N_W = 187.5 / 4.77 = 39.3 N.
Horizontal equilibrium: F = N_W = 39.3 N.
Coefficient of friction needed: μ = F/N_F = 39.3/250 = 0.157.
The friction coefficient must be at least 0.157 for the ladder to stay in place. Any μ_s above this is fine; any μ_s below it and the ladder will slide.
This is a classic A-Level statics problem: it ties together translational equilibrium (two equations), rotational equilibrium (one equation about a strategically chosen pivot), and the limiting friction condition. The choice of pivot at the foot of the ladder is the key insight — it eliminates two unknowns (N_F and F) in the moment equation, leaving N_W as the only unknown.
A simplified suspension-bridge model: a horizontal rigid beam of length 20 m and weight 8000 N is supported at its midpoint by a vertical cable that connects upward to a single fixed point. The beam is hinged at one end; the other end is free to rotate about the hinge. A 2500 N load is suspended from the free end. Find the tension in the supporting cable.
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