HookWhy the Leaning Tower of Pisa has not fallen over
Building work on the campanile at Pisa began in 1173, and the tower started to lean almost as soon as the third storey went up, because the soft clay beneath one side compressed more than the other. By 1990 the tilt had reached about \(5.5^{\circ}\), the authorities closed the tower to the public, and a committee led by the geotechnical engineer Professor John Burland was asked a single question that is pure A-Level mechanics: is it about to topple? A rigid object topples when the vertical line through its centre of mass falls outside its base. While that line stays inside the footprint, the object's weight produces a restoring moment that rights it; the instant the line crosses the edge, the weight produces an overturning moment instead, and there is nothing left to hold it up. Burland's team calculated that Pisa's line of action was creeping dangerously close to the edge of the foundation.
Their fix was elegant and entirely moment-based: between 1999 and 2001 they slowly extracted soil from underneath the high (north) side, letting the tower settle back the other way. The lean fell from about \(5.5^{\circ}\) to \(3.99^{\circ}\) — roughly half a degree — pulling the weight's line of action back to safety, and the tower reopened in 2001. That whole drama is Chapter 9 of Edexcel Mechanics. A moment is the turning effect of a force, and this section is about when those turning effects cancel — when a rigid body is in equilibrium. You will learn to compute a moment, to write the two conditions a rigid body must satisfy to stay still, to handle non-uniform rods whose weight does not act at the middle, to find the exact instant a plank is on the point of tilting, and finally to tackle non-parallel coplanar forces in the examiner's favourite set-piece: the ladder leaning against a wall.
ModelWhat a moment is — force times perpendicular distance
Push a door near its hinge and it barely moves; push the same door at the handle, far from the hinge, and it swings easily. The force is the same — what changed is the distance from the pivot. That is the whole idea of a moment: the turning effect of a force about a point depends on both how hard you push and how far the push is from the pivot. Formally, the moment of a force \(F\) about a point is
\[\text{moment} = F \times d,\]
where \(d\) is the perpendicular distance from the pivot to the line of action of the force. The unit is the newton metre (\(\text{N m}\)) — never confuse it with the joule, which is also a newton metre but measures energy, not turning. A moment also has a sense: it is either clockwise or anticlockwise, and in a balance we treat one as positive and the other as negative.
The word perpendicular is the whole examination. If a force's line of action passes straight through the pivot, then \(d=0\) and its moment is zero — which is exactly why pulling a spanner directly towards the nut does nothing, and why the hinge forces on a door never help you open it. When a force is applied at an angle, you either use the perpendicular distance to its line of action, or you resolve the force into a component perpendicular to the arm; we return to that in the ladder problem. For now, the everyday version — force at right angles to a rod, distance measured along the rod — is all you need.
Two children balance on a seesaw pivoted at its centre. Aisha has mass \(30\text{ kg}\) and sits \(1.5\text{ m}\) from the pivot. Ben has mass \(45\text{ kg}\). How far from the pivot must Ben sit to balance the seesaw?
Each child's weight is a downward force of \(mg\), acting at right angles to the horizontal plank, so each moment is (weight) \(\times\) (distance from pivot). The seesaw balances when the anticlockwise moment equals the clockwise moment — this is the principle of moments. Taking moments about the pivot:
\[\underbrace{30g \times 1.5}_{\text{Aisha, anticlockwise}} = \underbrace{45g \times d}_{\text{Ben, clockwise}}.\]
The acceleration due to gravity \(g\) appears on both sides, so it cancels — a recurring gift in moments problems where the only forces are weights. Solving, \(d = \dfrac{30 \times 1.5}{45} = \dfrac{45}{45} = 1.0\text{ m}\). Ben, being heavier, sits closer to the pivot, and the heavier-child-sits-closer rule you knew as a six-year-old is precisely \(F_1 d_1 = F_2 d_2\).
ModelEquilibrium of a rigid body — two conditions, and a free choice of pivot
A single particle is in equilibrium when the forces on it balance. A rigid body — a rod, a beam, a ladder, a tower — is larger, so it can also rotate, and stopping the rotation needs a second condition. A rigid body is in equilibrium when both of these hold:
(1) the resultant force is zero — so the forces balance horizontally and vertically; and (2) the resultant moment about any point is zero — equivalently, total clockwise moment = total anticlockwise moment.
The phrase 'about any point' is the single most useful sentence in the chapter. Because the moments balance about every point, you are free to choose the pivot that makes the algebra easiest — and the smart choice is a point through which an unknown force acts, because a force through your chosen pivot has zero moment and vanishes from the equation. Take moments about a support and that support's reaction disappears, leaving one equation in one unknown. This is why an examiner can hand you a beam with two unknown reactions and expect a clean answer: take moments about one support to get the other reaction, then resolve vertically for the first. Model a beam or rod as acting through its centre of mass, and — for a rod described as uniform — that centre of mass sits at the geometric middle.
A uniform beam \(AB\) of length \(4\text{ m}\) and weight \(120\text{ N}\) rests horizontally on two supports, one at each end. A load of \(80\text{ N}\) hangs from the beam \(1\text{ m}\) from \(A\). Find the reactions \(R_A\) and \(R_B\) at the supports.
The beam is uniform, so its \(120\text{ N}\) weight acts at the midpoint, \(2\text{ m}\) from \(A\). Four forces act: \(R_A\) and \(R_B\) up, the \(120\text{ N}\) weight and the \(80\text{ N}\) load down. Take moments about \(A\) to kill \(R_A\):
\[R_B \times 4 = 120 \times 2 + 80 \times 1 = 240 + 80 = 320 \ \Rightarrow\ R_B = 80\text{ N}.\]
Now resolve vertically: \(R_A + R_B = 120 + 80 = 200\), so \(R_A = 200 - 80 = 120\text{ N}\). Always check with a second, independent moment equation — this catches arithmetic slips before the examiner does. Moments about \(B\): \(R_A \times 4 = 120 \times 2 + 80 \times 3 = 240 + 240 = 480\), giving \(R_A = 120\text{ N}\). The two methods agree, so the answer stands.
MechanismNon-uniform rods and the point of tilting
Real planks are not uniform — a scaffold board is worn at one end, a rod may be thicker at one side — so its weight acts at a centre of mass that is not at the middle. You cannot assume the midpoint; you must be told where the centre of mass is, or be given enough information to find it. And moments find it: if a non-uniform rod balances on a single support, or you know the two reactions, one moment equation locates the centre of mass. For a rod of weight \(W\) on supports at \(A\) and \(B\) with the centre of mass a distance \(\bar{x}\) from \(A\), taking moments about \(A\) gives \(R_B \times (\text{length}) = W \times \bar{x}\), and \(\bar{x}\) drops out.
The highest-value idea in the whole section is tilting. Picture a plank resting on two supports with a load moving along it. As the load moves towards one end, more and more of the total weight is carried by the near support and less by the far one — until the far support carries nothing at all. At that instant the plank is on the point of tilting, and it will pivot about the near support if the load moves a hair further. The exam-ready fact is short and decisive: on the point of tilting about a support, the reaction at the other support is zero. Set that reaction to zero, take moments about the pivot support, and the tipping condition falls straight out.
A uniform plank \(AB\) of length \(5\text{ m}\) and mass \(30\text{ kg}\) rests horizontally on two supports at \(C\) and \(D\), where \(AC = 1\text{ m}\) and \(AD = 3\text{ m}\). A child of mass \(20\text{ kg}\) starts at \(A\) and walks towards \(B\). How far from \(A\) can the child walk before the plank tips?
The plank tips about the far support \(D\) (it lifts off at \(C\)), so on the point of tilting the reaction at \(C\) is zero. The only forces left with a moment about \(D\) are the plank's weight and the child's weight. The plank is uniform, so its \(30g\) weight acts at the midpoint, \(2.5\text{ m}\) from \(A\) — that is \(0.5\text{ m}\) on the \(A\)-side of \(D\), giving a restoring (anticlockwise) moment. Let the child be \(x\) metres from \(A\); to tip the plank the child must be on the \(B\)-side of \(D\), a distance \((x-3)\) beyond \(D\), giving a clockwise moment. Taking moments about \(D\) with \(R_C = 0\):
\[20g \times (x-3) = 30g \times 0.5.\]
Again \(g\) cancels: \(20(x-3) = 15\), so \(x - 3 = 0.75\) and \(x = 3.75\text{ m}\). The child can walk to \(3.75\text{ m}\) from \(A\) — that is \(0.75\text{ m}\) beyond support \(D\) — before the plank begins to tip. Notice the lighter the child relative to the plank, the further past \(D\) they may safely go.
CaseNon-parallel coplanar forces — the ladder against a wall
So far every force has been vertical (weights and reactions) — parallel coplanar forces. The final skill is non-parallel coplanar forces: forces pointing in different directions in the same plane, the classic being a ladder. Here two extra ideas matter. First, a smooth contact can only push at right angles to the surface — it exerts a normal reaction and no friction; a rough contact also exerts friction \(F\), up to the limit \(F \le \mu R\) where \(\mu\) is the coefficient of friction and \(R\) the normal reaction. Second, to take the moment of a force that is not perpendicular to the rod, use the perpendicular distance from the pivot to the force's line of action — for a horizontal force acting at the top of the ladder, that distance is simply the ladder's vertical height.
The method never changes. Draw the ladder, mark every force — weight at the centre of mass, the wall's reaction, the ground's normal reaction and friction — then write three equations: resolve horizontally, resolve vertically, and take moments about a well-chosen point (the foot of the ladder is ideal, because two unknowns act there and vanish). Three equations, three unknowns. The examiner's favourite twist asks for the least coefficient of friction that stops the ladder slipping, which is found by setting friction to its limiting value \(F = \mu R\) at the point of slipping.
A uniform ladder \(AB\) of length \(6\text{ m}\) and weight \(200\text{ N}\) rests with its foot \(A\) on rough horizontal ground and its top \(B\) against a smooth vertical wall. The ladder makes an angle of \(60^{\circ}\) with the horizontal. Find the least coefficient of friction between the ladder and the ground for which the ladder does not slip.
Mark the forces. The wall is smooth, so it pushes horizontally with reaction \(S\) at \(B\). The ground gives a vertical normal reaction \(N\) and a horizontal friction \(F\) at \(A\), pointing towards the wall to stop the foot sliding out. The weight \(200\text{ N}\) acts down at the midpoint, \(3\text{ m}\) along the ladder.
Resolve vertically: \(N = 200\text{ N}\). Resolve horizontally: \(F = S\). Now take moments about the foot \(A\), which removes \(N\) and \(F\) at a stroke. The weight's line of action is a horizontal distance \(3\cos 60^{\circ}\) from \(A\); the wall reaction \(S\) is horizontal, so its perpendicular distance from \(A\) is the vertical height of \(B\), namely \(6\sin 60^{\circ}\):
\[S \times 6\sin 60^{\circ} = 200 \times 3\cos 60^{\circ}.\]
So \(S \times 6(0.8660) = 200 \times 3(0.5) = 300\), giving \(S = \dfrac{300}{5.196} = 57.7\text{ N}\). Then \(F = S = 57.7\text{ N}\). For no slipping we need \(F \le \mu N\), i.e. \(\mu \ge \dfrac{F}{N} = \dfrac{57.7}{200} = 0.289\). The least coefficient of friction is \(\mu \approx 0.29\). As a check, the general result for a uniform ladder on a smooth wall is \(\mu_{\min} = \dfrac{1}{2\tan\theta} = \dfrac{1}{2\tan 60^{\circ}} = \dfrac{1}{2\sqrt{3}} = 0.289\) — the same value, which is why a steeper ladder (larger \(\theta\)) is safer.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Every moments question starts the same way: draw a large, clear diagram and mark every force — the weight acting at the centre of mass, both reactions, and any friction or tension. Missing forces, not algebra, lose most of the marks here. Label the weight at the true centre of mass: the midpoint only if the rod is stated to be uniform.
When you take moments, choose the pivot deliberately — a point through which an unknown force acts, because that force then has zero moment and vanishes. State your equation as 'clockwise moments = anticlockwise moments' or as moments about your named point, and be explicit about the perpendicular distance you are using; for a force at an angle, either resolve it or use the perpendicular distance to its line of action, and say which. For any beam-with-two-reactions problem, get one reaction by taking moments about the other support, then resolve vertically for the second, and confirm with a second moment equation about the opposite end.
For tilting problems, write the sentence 'on the point of tilting, the reaction at [the other support] is zero' before you compute anything — that single line is where the method mark sits. For ladder and other non-parallel-force problems, remember a smooth contact gives only a normal reaction, resolve horizontally and vertically as well as taking moments (three equations for three unknowns), and set \(F = \mu R\) only at the point of slipping. Keep \(g\) consistent — Edexcel uses \(g = 9.8\text{ m s}^{-2}\) unless told otherwise — though in pure weight-versus-weight balances it will cancel. Quote moments in \(\text{N m}\), never joules.