Mechanics: Moments and Statics Depth
As part of A-level Mechanics, moments and statics at depth combines taking moments about a strategically chosen point with resolving forces, often for a rigid body such as a ladder, rod or plank that is non-uniform, loaded unevenly, or on the point of sliding or toppling. The marks reward a complete force diagram, a moment found correctly as force multiplied by perpendicular distance (using sine of the angle where a force is not perpendicular to the body), and choosing a pivot that eliminates an unknown force from the equation. A common source of lost marks is treating a non-uniform body's weight as acting at its midpoint, or missing a force's perpendicular-distance component entirely.
Before you start
Make sure you're comfortable with these topics first:
Method
- Draw a clear diagram showing every force acting on the rigid body (weight at the centre of mass, which may not be the midpoint for a non-uniform body, plus every normal reaction, tension and friction force), labelling all given distances and angles.
- Choose the point to take moments about strategically: a point where an unknown force acts (such as a hinge, or the foot of a ladder) makes that force's moment zero, removing it from the equation.
- For each force not at the pivot, find its moment as magnitude multiplied by perpendicular distance from the pivot; for a force acting at an angle to the body, use magnitude times distance along the body times the sine of the angle between them.
- Set the sum of clockwise moments equal to the sum of anticlockwise moments (equivalently, resultant moment = 0) for a body in equilibrium.
- Combine the moments equation with resolving horizontally and vertically (or parallel and perpendicular to a relevant surface) to find any remaining unknowns - a typical rigid-body problem needs the moments equation plus two resolving equations for three unknown forces.
- For a ladder or similar body in contact with two surfaces, resolve horizontally and vertically for the whole body, take moments (usually about the end in contact with the rougher surface), and if the body is on the point of slipping, add F = mu R at that contact.
- For a toppling (rather than sliding) problem, take moments about the edge the body would rotate about, since the normal reaction and friction both act there at the point of toppling and so drop out of the equation.
Worked example
A uniform ladder AB has length 5 m and weight 200 N. The ladder rests with end A on rough horizontal ground and end B against a smooth vertical wall, at 60 degrees to the horizontal. The coefficient of friction between the ladder and the ground is 0.4. A person of weight 700 N stands on the ladder at point P. Take g = 9.8 m/s^2. Given that the ladder is on the point of slipping, find the distance AP.
- Draw a diagram: the ladder's weight (200 N) acts at its midpoint, 2.5 m from A; the person's weight (700 N) acts at P, distance d from A (to be found); normal reaction R and friction Fr act at A (ground); normal reaction S acts at B (wall, smooth, so horizontal only).
- Resolve vertically: since the wall is smooth (no vertical force at B), R = 200 + 700 = 900 N.
- Since the ladder is on the point of slipping, friction is limiting: Fr = mu R = 0.4 x 900 = 360 N. Resolve horizontally: S = Fr = 360 N.
- Take moments about A (this eliminates both R and Fr, since both act at A): the clockwise moments from the two weights total 200(2.5cos60) + 700d(cos60); the anticlockwise moment from S is 360(5sin60).
- Set clockwise equal to anticlockwise: 200(2.5)(0.5) + 700d(0.5) = 360(5)(0.8660254), giving 250 + 350d = 1558.85 (2 d.p.).
- Solve for d: 350d = 1308.85, so d = 3.74 m (3 s.f.). The person can climb 3.74 m up the ladder before it is on the point of slipping.
Practice questions
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Q1State the two conditions required for a rigid body to be in equilibrium.Show answer
Answer: The resultant force acting on it is zero (in every direction), and the resultant moment about any point is zero.
Q2A non-uniform rod AB has length 3 m and weight 45 N. The rod is suspended horizontally in equilibrium by two vertical strings, one at A and one at B. The tension in the string at A is 30 N. Find the distance of the centre of mass of the rod from A.Show answer
Answer: 1 m from A. (Vertical equilibrium gives the tension at B as 15 N; taking moments about A, 15 x 3 = 45 x x, so x = 1.)
Q3A uniform plank AB has length 6 m and weight 80 N, and balances horizontally on a smooth pivot at its midpoint M. A child of weight 200 N sits at A. Find the distance from M at which a child of weight 250 N must sit on the other side for the plank to balance.Show answer
Answer: 2.4 m from M. (The plank's own weight acts at the pivot, so it has no moment. Taking moments about M: 200 x 3 = 250 x x, so x = 2.4.)
Q4Explain why, when solving a rigid-body equilibrium problem, it is often more efficient to take moments about the point where an unknown reaction acts, rather than about the centre of mass.Show answer
Answer: A force has zero moment about the point where it acts, since its line of action passes through the pivot (perpendicular distance zero). Choosing that point as the pivot removes the unknown reaction from the moments equation entirely, so the equation can be solved for a different unknown without first needing to know that reaction.
Q5A uniform ladder of weight 250 N rests with one end on rough horizontal ground and the other end against a smooth vertical wall, with nobody standing on it. The coefficient of friction between the ladder and the ground is 0.3. Find the minimum angle the ladder can make with the ground without slipping, giving your answer to 1 decimal place.Show answer
Answer: 59.0 degrees. (Taking moments shows tan(theta) = 1/(2mu), so theta = arctan(1/0.6) = arctan(1.667).)
Q6A uniform rectangular block has width 0.4 m and mass 15 kg, and stands on rough horizontal ground that prevents it sliding. A horizontal force P is applied to the block at a height of 0.9 m above the ground and gradually increased. Take g = 9.8 m/s^2. Find the value of P at which the block is on the point of tipping about its lower edge.Show answer
Answer: 32.7 N (3 s.f.). Taking moments about the tipping edge: P(0.9) = 15g(0.2), so P = 15(9.8)(0.2)/0.9.
Q7A steering wheel of radius 0.18 m is turned by a couple: two hands apply equal and opposite forces of 15 N tangentially, on opposite sides of the wheel. Find the total moment of the couple.Show answer
Answer: 5.4 N m. A couple's moment is one force multiplied by the perpendicular distance between the two forces' lines of action, here the diameter: 15 x (2 x 0.18).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A non-uniform plank AB has length 8 m and weight 160 N. The centre of mass of the plank is at G, where AG = 3 m. The plank rests horizontally in equilibrium on two supports, one at C, where AC = 1 m, and one at D, where AD = 7 m. A box of weight 100 N is placed on the plank at the point E, where AE = 5 m. (a) By taking moments about C, find the reaction at D. (4) (b) Find the reaction at C. (3)
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A uniform ladder AB has length 6 m and weight 240 N. The ladder rests with end A on rough horizontal ground and end B against a smooth vertical wall, inclined at 65 degrees to the horizontal. A decorator of weight 560 N stands on the ladder at the point C, where BC = 1.5 m. Take g = 9.8 m/s^2. Given that the ladder is on the point of slipping, find the coefficient of friction between the ladder and the ground, giving your answer to 3 significant figures.
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A uniform cube of side 0.6 m and mass 20 kg rests on rough horizontal ground. The coefficient of friction between the cube and the ground is 0.3. A horizontal force P is applied to a vertical face of the cube, at a height h metres above the ground, and is gradually increased until the cube either slides or topples (about the bottom edge furthest from where P is applied), whichever happens first. Take g = 9.8 m/s^2. (a) Find the value of P at which the cube would be on the point of sliding. (2) (b) Find, in terms of h, the value of P at which the cube would be on the point of toppling. (3) (c) Hence find the value of h above which the cube topples before it slides. (2)
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