The moment of a force about a point, using the perpendicular distance or by splitting the force into components

The two conditions for a rigid body to be in equilibrium, for parallel and for non-parallel coplanar forces

Beams on supports, including the point at which a beam is about to tilt

Rods hinged to a wall and held by a string: the tension, and the magnitude and direction of the force at the hinge

Ladders against smooth or rough walls on rough ground: limiting equilibrium, the least coefficient of friction, and how far a person can climb

Rods resting on smooth pegs with one end on rough ground

1. Every statics question in M2 is solved by one method: draw the forces, resolve in two directions and take moments about one well-chosen point, giving three equations for at most three unknowns.

2. The moment of a force about a point is force × perpendicular distance from the point to the line of action, in N m, with a sense: clockwise or anticlockwise.

3. For a force at an angle, either find the perpendicular distance with trigonometry, or split the force into components and add the moments of the components.

4. A hinge can push or pull in any direction: give it two unknown components X and Y, then combine them for the magnitude and direction.

5. A smooth surface or peg gives only a normal reaction (perpendicular to the surface, or to the rod at a peg). A rough surface adds friction, with F≤μR, and F=μR when slipping is about to happen.

6. Friction opposes the way the contact point would slide. For a ladder, the foot tends to slide away from the wall and the top tends to slide down.

Here F is the magnitude of the force in N and d the perpendicular distance from the point to the line of action in m. In the second form, l is the distance from the point to where the force acts and θ is the angle between the force and that line. This is not in the formulae booklet: learn it.

A moment has a sense: clockwise or anticlockwise. A force whose line of action passes through the point has zero moment about that point. The line of action is the infinite straight line through the force, and you may slide a force along its line of action without changing its moment.

Why sin θ? Only the part of the force perpendicular to the rod turns it.

The part along the rod pushes or pulls along a line through the pivot, so its moment is zero. Choose one sense as positive (say anticlockwise) and stick to it for every force in the same equation.

WORKED EXAMPLE · QUESTION

A light rod AB of length 0.80 m is smoothly pivoted at B. A force of magnitude 24 N acts at A, in the plane of the rod, making an angle of 35° with AB produced. Find the moment of the force about B.

SOLUTION

Method 1 Perpendicular distance from B to the line of action: d=0.80sin 35°=0.4589 m, so moment =24×0.4589= 11.0 N m

Method 2 Components at A: perpendicular to AB, 24sin 35°=13.766 N; along AB, 24cos 35°=19.660 N (moment 0). Moment =13.766×0.80= 11.0 N m (11.013 N m)

Both methods are the same calculation, 24×0.80×sin 35°, arranged differently.

Split the force into horizontal and vertical components (or i and j components). Each component has its own moment; add them with signs. For a force Xi+Yj acting at a point whose position relative to the pivot is xi+yj, the moment, anticlockwise positive, is

You do not need to learn this as a formula: picture each component and decide whether it turns the body clockwise or anticlockwise. This is the quickest method when a force acts at a corner of a frame or lamina, or when the perpendicular distance is awkward to find.

WORKED EXAMPLE · QUESTION

A force (3i+4j) N acts at the point P with position vector (2i-j) m. The unit vectors i and j are horizontal and vertical. Find the moment of the force about (a) the origin O and (b) the point Q with position vector (5i+2j) m, stating the sense in each case.

SOLUTION

(a) The vertical component, 4 N, acts 2 m to the right of O: 4×2=8 N m anticlockwise. The horizontal component, 3 N to the right, acts 1 m below O: 3×1=3 N m anticlockwise (a force to the right, below the pivot, turns anticlockwise). Total =8+3= 11 N m anticlockwise

(b) Position of P relative to Q: (2-5)i+(-1-2)j=-3i-3j. Moment, anticlockwise positive: (-3)(4)-(-3)(3)=-12+9=-3, so 3 N m clockwise

The same force has different moments about different points — always say which point.

In two dimensions this gives three independent equations: resolve in two directions and take moments about one point. So at most three unknowns can be found. You may replace a resolving equation by a second moments equation about a different point, which is useful when it removes an unknown.

Take moments about the point where the most unknown forces act: their moments are zero there.

Parallel forces (all vertical): the resolving equation and one moments equation are enough.

About to tilt about one support means the beam is just losing contact with the other: set the reaction there to zero.

For a uniform rod the weight acts at the midpoint. For a non-uniform rod the centre of mass is wherever the question puts it, and often it is the unknown.

WORKED EXAMPLE · QUESTION

A non-uniform plank AB has length 6 m and mass 30 kg. Its centre of mass is 2.5 m from A. The plank rests horizontally on two supports at C and D, where AC = 1 m and AD = 4 m. A child of mass 40 kg stands on the plank. Model the plank as a rod and the child as a particle. (a) The child stands 3.5 m from A. Find the reactions at C and D. (b) The child walks towards B. Find the distance from A at which the plank is about to tilt.

SOLUTION

(a) Moments about D (removes R_D): R_C×3=30g×1.5+40g×0.5=65g, so R_C=65g÷3= 212 N (212.3 N). Resolve vertically: R_D=70g-212.3= 474 N (473.7 N)

(b) About to tilt about D, so R_C=0. Moments about D with the child x m from A: 30g×(4-2.5)=40g×(x-4), so 45=40x-160 and x=5.125 m from A (so 0.875 m from B)

g cancels in part (b): keep it as a symbol until the end and the answer is exact.

A smooth hinge exerts a single force whose direction is unknown. Model it as a horizontal component X and a vertical component Y, each with an assumed direction. Then:

1. Take moments about the hinge first: X and Y both vanish, leaving one equation for the tension.

2. Resolve horizontally and vertically for X and Y. A negative answer just means that component acts the other way.

3. Magnitude R=√(X²+Y²); direction tan⁻¹(Y/X) above or below the horizontal — say which.

The force at the hinge is not along the rod in general. It is along the rod only for a light rod with forces at its two ends. A useful check: if exactly three non-parallel forces act on a body in equilibrium, their lines of action pass through one point.

WORKED EXAMPLE · QUESTION

A uniform rod AB has length 2.4 m and mass 8 kg. The end A is freely hinged to a vertical wall. The rod is held horizontal by a light inextensible string joining B to the point C on the wall, 1.8 m vertically above A. A particle of mass 5 kg hangs from B. The rod and the string lie in a vertical plane perpendicular to the wall. Find (a) the tension in the string and (b) the magnitude and direction of the force exerted on the rod at A.

SOLUTION

Geometry BC=√(2.4²+1.8²)=3.0 m, so the angle θ between the string and the rod has sin θ=1.8/3.0=0.6 and cos θ=2.4/3.0=0.8.

(a) Moments about A: T×0.6×2.4=8g×1.2+5g×2.4=21.6g, so T=21.6g÷1.44=15g= 147 N

(b) Resolve horizontally (X away from the wall) and vertically (Y upwards): X=Tcos θ=12g=117.6 N; Y+Tsin θ=13g, so Y=4g=39.2 N.

Then R=g√(12²+4²)=4√10 g= 124 N (123.96 N), at tan⁻¹(4/12)= 18.4° above the horizontal, away from the wall

A ladder is modelled as a rod; uniform means its weight acts at the midpoint. People are particles at a stated distance along it.

At a smooth vertical wall: one force, the normal reaction S, horizontal and away from the wall.

On rough horizontal ground: the normal reaction R upwards and friction F horizontally towards the wall (the foot tends to slide away).

The standard set of equations: resolve vertically, R = total weight; resolve horizontally, F=S; take moments about the foot A, which removes both R and F; and in limiting equilibrium F=μR (in general F≤μR). For a ladder of length l at angle α to the horizontal, the top is lsin α above A and the midpoint is l/2cos α from A horizontally.

WORKED EXAMPLE · QUESTION

A uniform ladder of mass 20 kg and length l rests with its foot on rough horizontal ground and its top against a smooth vertical wall. The ladder makes an angle of 65° with the ground and is in equilibrium. Find the least possible value of the coefficient of friction between the ladder and the ground.

SOLUTION

Step 1 Moments about the foot: S×lsin 65°=20g×l/2cos 65°, so S=20g/2tan 65°=45.70 N.

Step 2 Resolve: F=S=45.70 N and R=20g=196 N.

Step 3 F≤μR, so μ≥45.70÷196=1/2tan 65°=0.2332. The least μ is 0.233

The mass and the length both cancel: for a uniform ladder on its own, the least μ depends only on the angle.

WORKED EXAMPLE · QUESTION

A uniform ladder AB has length 6 m and mass 15 kg. The end A rests on rough horizontal ground and the end B rests against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall and makes an angle of 60° with the ground. The coefficient of friction between the ladder and the ground is 0.35. A decorator of mass 75 kg climbs slowly up the ladder. Model the decorator as a particle. (a) Find the distance the decorator can climb along the ladder from A before it slips. (b) Find the least coefficient of friction that would allow her to reach the top.

SOLUTION

(a) Resolve vertically: R=90g=882 N. Limiting friction: F=0.35×90g=31.5g=308.7 N. Resolve horizontally: S=F=31.5g. Moments about A, decorator x m from A: S×6sin 60°=15g×3cos 60°+75g×xcos 60°. Dividing by gcos 60°: 189tan 60°=45+75x, so 75x=282.36 and x=3.76 m (3.7648 m)

(b) At the top, x=6. Moments about A: S×6sin 60°=(15×3+75×6)gcos 60°, so S=495g/6tan 60°=47.63g. With F=S and R=90g: μ≥47.63÷90= 0.529

Climbing higher increases the moment about A, so the wall pushes harder and more friction is needed; R does not change.

The wall now gives a normal reaction S and a friction force. As the foot slides away, the top slides down the wall, so the wall's friction on the ladder acts upwards. "About to slip" for a ladder means slipping at both ends at once (the ladder cannot slide at one end only), so both frictions are limiting: F₁=μ₁ R and F₂=μ₂ S.

Wall friction now appears in the vertical equation, so R is less than the total weight. Moments about the foot now include the wall friction too: its moment arm is the horizontal distance from A to the wall.

WORKED EXAMPLE · QUESTION

A uniform ladder of mass 12 kg rests with its foot on rough horizontal ground and its top against a rough vertical wall. The coefficient of friction is 0.4 at the ground and 0.25 at the wall. The ladder is about to slip when it makes an angle θ with the ground. (a) Find the normal reactions at the ground and at the wall. (b) Find θ.

SOLUTION

(a) Limiting at both ends. Resolve horizontally: S=0.4R. Resolve vertically, with wall friction 0.25S upwards: R+0.25S=12g, so R+0.1R=12g. Then R=12g÷1.1= 107 N (106.91 N) and S=0.4R= 42.8 N (42.76 N)

(b) Moments about the foot, ladder length 2a: S×2asin θ+0.25S×2acos θ=12g×acos θ. Divide by 2acos θ: S(tan θ+0.25)=6g, so tan θ=(6×9.8÷42.764)-0.25=1.375-0.25=1.125 and θ=48.4°

In general, for a uniform ladder limiting at both ends, tan θ=(1-μ₁μ₂)÷(2μ₁) — derive it each time; do not quote it.

A smooth peg (or a smooth rail or edge touching the rod along its length) gives a normal reaction perpendicular to the rod — not vertical. If the rod is at angle α to the horizontal, the peg reaction N makes angle α with the vertical, with components Ncos α up and Nsin α horizontally.

Take moments about the foot of the rod: only N and the weight have moments, and N acts perpendicular to the rod, so its moment is simply N × (distance along the rod). Friction at the foot opposes the slide: the horizontal part of N pushes the foot away from the peg, so friction acts towards the peg.

WORKED EXAMPLE · QUESTION

A uniform rod AB has length 4 m and mass 10 kg. The end A rests on rough horizontal ground and the rod rests on a smooth peg C, where AC = 3 m. The rod is in a vertical plane and makes an angle of 30° with the ground. The rod is about to slip. (a) Find the force exerted on the rod by the peg. (b) Find the coefficient of friction between the rod and the ground, giving your answer in exact form.

SOLUTION

(a) Moments about A (N is perpendicular to the rod): N×3=10g×2cos 30°=10√3 g, so N=10√3 g/3=10g/√3= 56.6 N, perpendicular to the rod

(b) Resolve vertically: R+Ncos 30°=10g, so R=10g-5g=5g=49 N. Resolve horizontally: F=Nsin 30°=5g/√3=28.29 N. Limiting, F=μR: μ=(5g/√3)/5g=1/√3= √3/3 (≈ 0.577)

Keeping g as a symbol makes the exact answer drop out cleanly.

M1 Moments and M1 Statics of a particle

This page extends M1 moments of parallel forces to forces at any angle, and uses the M1 friction model F≤μR and resolving in two directions.

M2.3 Centres of mass

Non-uniform rods and laminas need the position of the centre of mass before you can take moments; uniform shapes have results in the formulae booklet, and a lamina hanging freely in equilibrium uses the same moments idea.

P1–P3 trigonometry and IAL Physics

Ladder and peg questions lead to equations in tan θ, exact values such as sin 30° and cos 30°, and occasionally identities to tidy an answer. In IAL Physics Unit 1 the principle of moments and the centre of gravity are the same ideas, applied to beams, levers and cranes.

1. A force of 50 N acts at one end of a rod of length 1.2 m, at 40° to the rod. Find its moment about the other end.

2. A uniform rod AB of weight 40 N is hinged at A to a wall and held horizontal by a string attached at B, making 30° with the rod. Find the tension and the force at the hinge.

3. A uniform ladder rests against a smooth wall at 70° to rough ground. Find the least coefficient of friction.

4. A rod rests on a smooth peg. In what direction does the peg push on the rod, and why?

5. A ladder rests against a rough wall on rough ground and is about to slip. In which direction does friction act at each end?

6. A uniform beam AB of length 5 m and mass 20 kg rests horizontally on supports at A and B. A person of mass 30 kg stands 1 m from A. Find the reactions.

1. 50×1.2sin 40°=38.6 N m (38.57 N m).

2. M(A): Tsin 30°×l=40×l/2, so T=40 N. X=40cos 30°=34.6 N, Y=40-20=20 N, so the hinge force is 40 N at 30° above the horizontal — the three forces meet at the point above the midpoint.

3. M(foot): Ssin 70°=½ Wcos 70°, so F=S=W/(2tan 70°) and R=W. μ≥1/(2tan 70°)=0.182.

4. Perpendicular to the rod. A smooth contact cannot exert any force along the surface of the rod, so only the normal reaction remains.

5. At the ground, towards the wall (the foot tends to slide away). At the wall, upwards (the top tends to slide down). Both frictions are limiting.

6. M(A): R_B×5=20g×2.5+30g×1=80g, so R_B=16g=157 N. Resolve: R_A=50g-16g=34g=333 N (3 s.f.).