M1.5 Statics of a particle
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Mechanics 1 (AS). Friction on a moving particle and connected particles are in M1.3; turning effects on rods and planks are in M1.6; rigid bodies under non-parallel forces, such as ladders, are in M2.
What this page covers
7 points
- Forces as vectors: components, resultants, magnitude and direction, forces written with \(\mathbf{i}\) and \(\mathbf{j}\)
- The forces met in M1 statics: weight, normal reaction, tension, thrust and friction
- Equilibrium of a particle by resolving in two perpendicular directions
- The triangle of forces for three forces
- Particles suspended by two strings, and particles on inclined planes held by horizontal forces or strings
- Friction in equilibrium: \(F \le \mu R\), limiting equilibrium, about to slide up or down a rough plane
- The least force needed to move a particle on a rough surface
In one minute
6 points
- A particle is in equilibrium when the resultant force on it is zero, so every statics question is solved by drawing the forces, resolving in two perpendicular directions and setting each sum to zero.
- A force \(F\) at angle \(\theta\) to a direction has component \(F\cos\theta\) along it and \(F\sin\theta\) perpendicular to it.
- Two directions give two equations, so at most two unknowns can be found.
- On a plane, resolve along and perpendicular to the plane. The normal reaction is not \(mg\cos\alpha\) if another force has a component perpendicular to the plane.
- Friction opposes the motion the particle would have: \(F \le \mu R\). In limiting equilibrium (about to move), \(F = \mu R\).
- With friction there is usually a range of forces that keep a particle at rest: one end with friction acting down the plane, the other with it acting up.
Contents
14 sections
- Which forces act on a particle at rest?
- How do you resolve forces and find a resultant?
- How do you solve an equilibrium problem by resolving?
- When is a triangle of forces quicker?
- How do you deal with a particle on a smooth inclined plane?
- How does friction behave when a particle is at rest?
- What happens on a rough inclined plane?
- What is the least force needed to move a particle?
- Exam tips for M1.5
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
Which forces act on a particle at rest?
Before you can add forces you must find them all. In M1 statics there are only five kinds, and each has a fixed direction rule.
| Force | Where it comes from | Direction |
|---|---|---|
| Weight \(mg\) | Gravity, with \(g = 9.8\) m s⁻² | Vertically downwards |
| Normal reaction \(R\) | A surface the particle rests on | Perpendicular to the surface, away from it |
| Tension \(T\) | A taut string, rope or cable | Along the string, away from the particle (a pull) |
| Thrust | A light rod pushing on the particle | Along the rod, towards the particle (a push) |
| Friction \(F\) | A rough surface | Along the surface, opposing the motion that would happen without it |
The modelling words matter. A particle has all its forces acting at one point, so there is no turning and only resolving is needed. A light string or rod has no weight, so its tension or thrust is the same all along it. Inextensible means it does not stretch. A smooth surface gives no friction at all; a rough surface can give friction up to a maximum \(\mu R\). A rod can pull (tension) or push (thrust), but a string can only pull: a negative tension in a string means the set-up is impossible.
Drawing a vertical normal reaction on an inclined plane, or a horizontal friction force. Both act relative to the surface: the normal reaction at right angles to it, friction along it.
How do you resolve forces and find a resultant?
Any force can be replaced by two perpendicular components. If a force \(F\) makes angle \(\theta\) with a chosen direction, its component along that direction is \(F\cos\theta\) and its component perpendicular to it is \(F\sin\theta\). To add several forces, add their components separately: \(X\) in one direction and \(Y\) in the other. Then the resultant has magnitude \(\sqrt{X^2 + Y^2}\) and makes angle \(\varphi\) with the \(X\)-direction, where \(\tan\varphi = Y/X\).
A force written as \((p\mathbf{i} + q\mathbf{j})\) N already shows its components, and its magnitude is \(\sqrt{p^2 + q^2}\). Use a sketch to fix the sign of each component (a force pointing left has a negative \(\mathbf{i}\)-component), and always give a direction as an angle to a stated line, such as "49.1° above the direction of \(\mathbf{i}\)".
Three horizontal forces act on a particle: 8 N in the direction of \(\mathbf{i}\), 6 N at 60° to \(\mathbf{i}\), and 5 N at 150° to \(\mathbf{i}\), each angle measured anticlockwise from \(\mathbf{i}\). Find the magnitude of the resultant force and the angle it makes with \(\mathbf{i}\).
Components parallel to \(\mathbf{i}\): \(X = 8 + 6\cos 60^\circ + 5\cos 150^\circ = 8 + 3 - 4.330 = 6.670\) N.
Components parallel to \(\mathbf{j}\): \(Y = 0 + 6\sin 60^\circ + 5\sin 150^\circ = 5.196 + 2.5 = 7.696\) N.
Magnitude \(\sqrt{6.670^2 + 7.696^2} = 10.18\) N, and \(\tan\varphi = 7.696 \div 6.670\) gives \(\varphi = 49.1^\circ\). Both components are positive, so the resultant points into the first quadrant. 10.2 N at 49.1° to i
How do you solve an equilibrium problem by resolving?
The method is the same every time:
- Draw a clear diagram with every force on the particle, labelled, and the given angles marked.
- Choose two perpendicular directions, usually horizontal and vertical, or along and perpendicular to a plane. Pick them so that the unknown forces have as few components as possible.
- Resolve in each direction, writing "R(→)" or "R(↑)" so the direction of each equation is clear.
- Solve the two equations simultaneously.
For forces given as vectors, equilibrium means the vector sum is \(\mathbf{0}\): the \(\mathbf{i}\)-components and the \(\mathbf{j}\)-components each add to zero. A useful consequence: if one force is removed from a set in equilibrium, the rest add up to a force equal and opposite to the one removed.
A particle is in equilibrium under three forces: \(\mathbf{F}_1 = (5\mathbf{i} + 2\mathbf{j})\) N, \(\mathbf{F}_2 = (-3\mathbf{i} + 6\mathbf{j})\) N and \(\mathbf{F}_3\). (a) Find \(\mathbf{F}_3\), its magnitude, and the angle it makes with \(\mathbf{i}\). (b) \(\mathbf{F}_2\) is removed. Find the magnitude of the resultant of the two forces that remain.
(a) The forces add to zero: \(\mathbf{F}_3 = -(\mathbf{F}_1 + \mathbf{F}_2) = -(2\mathbf{i} + 8\mathbf{j})\). Magnitude \(\sqrt{2^2 + 8^2} = \sqrt{68} = 2\sqrt{17} = 8.25\) N. The force points left and down, at \(\tan^{-1}(8 \div 2) = 76.0^\circ\) below \(-\mathbf{i}\), so the angle with \(\mathbf{i}\) is \(180^\circ - 76.0^\circ\). F₃ = (−2i − 8j) N, 8.25 N, 104.0° to i (clockwise)
(b) \(\mathbf{F}_1 + \mathbf{F}_3 = -\mathbf{F}_2 = (3\mathbf{i} - 6\mathbf{j})\) N, magnitude \(\sqrt{45} = 3\sqrt{5}\). 6.71 N
A particle P of mass 6 kg hangs in equilibrium from two light inextensible strings AP and BP. The ends A and B are fixed to a horizontal ceiling. AP makes an angle of 50° with the horizontal and BP makes an angle of 30° with the horizontal. Find the tension in each string.
R(→), with tension \(T_1\) in AP and \(T_2\) in BP: \(T_1\cos 50^\circ = T_2\cos 30^\circ\), so \(T_1 = 1.3473\,T_2\).
R(↑): \(T_1\sin 50^\circ + T_2\sin 30^\circ = 6 \times 9.8 = 58.8\).
Substitute: \((1.3473\sin 50^\circ + \sin 30^\circ)T_2 = 58.8\), so \(T_2 = 38.38\) N and \(T_1 = 1.3473 \times 38.38 = 51.71\) N. T₁ = 51.7 N, T₂ = 38.4 N
Keep unrounded values between steps; rounding \(T_2\) to 38 first would move \(T_1\) to 51.2 N.
When is a triangle of forces quicker?
When exactly three forces keep a particle in equilibrium, drawing them tip to tail gives a closed triangle of forces. The sides are proportional to the forces, so the sine rule gives the unknown magnitudes without simultaneous equations. The angle inside the triangle between two sides is 180° minus the angle between those two forces on the force diagram, so sketch carefully. Resolving twice always works too; the triangle is just faster when the angles are awkward. (Lami's theorem is the same sine rule in another form; it is not needed.)
A particle of weight 30 N is held in equilibrium by a light string and a force of magnitude \(P\) newtons. The string makes an angle of 40° with the upward vertical. The force \(P\) acts on the other side of the vertical, at 15° above the horizontal. Find the tension in the string and the value of \(P\).
Angles in the triangle (180° minus the angle between each pair of forces): between T and P the angle is \(180^\circ - 50^\circ - 15^\circ = 115^\circ\), so 65° is opposite W; between W and P it is 105°, so 75° is opposite T; between W and T it is 140°, so 40° is opposite P.
Sine rule: \(\dfrac{30}{\sin 65^\circ} = \dfrac{T}{\sin 75^\circ} = \dfrac{P}{\sin 40^\circ} = 33.10\). So \(T = 33.10\sin 75^\circ = 31.97\) and \(P = 33.10\sin 40^\circ = 21.28\). T = 32.0 N, P = 21.3 N
Check by resolving horizontally: \(T\sin 40^\circ = 20.55\) and \(P\cos 15^\circ = 20.55\).
How do you deal with a particle on a smooth inclined plane?
Resolve parallel and perpendicular to the plane: then the normal reaction (and friction, if any) need no resolving. For a plane at angle \(\alpha\) to the horizontal, the weight has components \(mg\sin\alpha\) down the plane and \(mg\cos\alpha\) into the plane. A horizontal force \(P\) pushing towards the plane makes angle \(\alpha\) with the plane: \(P\cos\alpha\) up the plane and \(P\sin\alpha\) into it, so it increases the normal reaction. A string at angle \(\beta\) to the plane, pulling up the slope, gives \(T\cos\beta\) up the plane and \(T\sin\beta\) away from it, so it reduces the normal reaction.
A particle of mass 3 kg rests on a smooth plane inclined at 25° to the horizontal. It is held in equilibrium by a horizontal force of magnitude \(P\) newtons, acting in the vertical plane containing a line of greatest slope. (a) Find \(P\). (b) Find the normal reaction. (c) The horizontal force is replaced by a force acting up the line of greatest slope. Find the magnitude of this force and the new normal reaction.
(a) R(parallel to the plane): \(P\cos 25^\circ = 3 \times 9.8 \times \sin 25^\circ\), so \(P = 29.4\tan 25^\circ = 13.71\). P = 13.7 N
(b) R(perpendicular to the plane): \(R = 29.4\cos 25^\circ + P\sin 25^\circ = 32.44\). R = 32.4 N
(c) A force along the plane has no perpendicular component: force \(= 29.4\sin 25^\circ = 12.4\) N and \(R = 29.4\cos 25^\circ = 26.6\) N. 12.4 N; R = 26.6 N
Check for (b): \(P\), the weight and \(R\) form a right-angled triangle, so \(R = 29.4 \div \cos 25^\circ = 32.44\) N.
That is only true when no force other than the weight has a component perpendicular to the plane. Always resolve perpendicular to the plane in full.
How does friction behave when a particle is at rest?
Friction is only as large as it needs to be. While the particle stays at rest, \(F\) is found by resolving, not from \(\mu R\). The maximum available friction is \(\mu R\); when \(F\) reaches it, the particle is in limiting equilibrium. "On the point of sliding", "about to move" and "just" all mean \(F = \mu R\). Friction acts along the surface, opposite to the direction the particle would move if the surface were smooth. If the friction needed is greater than \(\mu R\), the particle cannot stay at rest.
A crate of mass 20 kg, modelled as a particle, rests on rough horizontal ground. A rope attached to the crate makes an angle of 30° above the horizontal. When the tension in the rope is 70 N, the crate is on the point of sliding. (a) Find the coefficient of friction. (b) Find the friction force when the tension is 40 N, and show that the crate does not move.
(a) R(↑): \(R + 70\sin 30^\circ = 20 \times 9.8\), so \(R = 196 - 35 = 161\) N. R(→), limiting so \(F = \mu R\): \(F = 70\cos 30^\circ = 60.62\) N, and \(\mu = 60.62 \div 161 = 0.3765\ldots\) μ = 0.38 (2 s.f.)
(b) At 40 N, friction balances the horizontal pull: \(F = 40\cos 30^\circ = 34.6\) N. Now \(R = 196 - 40\sin 30^\circ = 176\) N, so \(\mu R = 0.3765 \times 176 = 66.3\) N \(> 34.6\) N. F = 34.6 N; the crate stays at rest
In (b) the friction is not \(\mu R\): the crate is not on the point of moving.
What happens on a rough inclined plane?
With no applied force, a particle on a rough plane at angle \(\alpha\) stays at rest only if friction can hold the weight component: \(mg\sin\alpha \le \mu mg\cos\alpha\), that is \(\mu \ge \tan\alpha\), whatever the mass.
With a holding force there are two limiting cases. About to slide down: friction acts up the plane and \(F = \mu R\); this gives the smallest force that holds the particle. About to slide up: friction acts down the plane and \(F = \mu R\); this gives the largest force before it moves up. Any force between the two keeps the particle at rest. Re-resolve for each case: the normal reaction usually changes too.
A particle of mass 5 kg is placed on a rough plane inclined at 30° to the horizontal. The coefficient of friction between the particle and the plane is 0.2. A horizontal force of magnitude \(P\) newtons acts on the particle, in the vertical plane containing a line of greatest slope, pushing towards the plane. (a) Show that without the force the particle would slide down the plane. (b) Find \(P\) when the particle is on the point of sliding up the plane. (c) Find \(P\) when the particle is on the point of sliding down the plane. (d) State the range of values of \(P\) for which the particle stays at rest.
(a) Down the plane: \(49\sin 30^\circ = 24.5\) N. Maximum friction: \(\mu R = 0.2 \times 49\cos 30^\circ = 8.49\) N \(< 24.5\) N, so it slides (equivalently \(\tan 30^\circ = 0.577 > 0.2\)).
(b) About to slide up, friction \(\mu R\) down the plane. R(perpendicular): \(R = 49\cos 30^\circ + P\sin 30^\circ\). R(parallel): \(P\cos 30^\circ = 49\sin 30^\circ + 0.2R\). So \(P(\cos 30^\circ - 0.2\sin 30^\circ) = 49(\sin 30^\circ + 0.2\cos 30^\circ)\), and \(P = 49 \times 0.6732 \div 0.766 = 43.06\). P = 43.1 N
(c) About to slide down, friction \(\mu R\) up the plane: \(P\cos 30^\circ + 0.2R = 49\sin 30^\circ\), with the same \(R\). So \(P(\cos 30^\circ + 0.2\sin 30^\circ) = 49(\sin 30^\circ - 0.2\cos 30^\circ)\), and \(P = 49 \times 0.3268 \div 0.966 = 16.58\). P = 16.6 N
(d) 16.6 ≤ P ≤ 43.1
The normal reaction is 64.0 N in (b) and 50.7 N in (c): a single value of \(R\) for both cases would be wrong.
Before resolving on a rough plane, ask "which way would it move if the surface were smooth?" and draw friction the other way. In a range question, state the direction of friction and "\(F = \mu R\)" in words for each case.
What is the least force needed to move a particle?
On rough horizontal ground, pulling at angle \(\theta\) above the horizontal has two effects. The horizontal part, \(P\cos\theta\), is smaller; but the vertical part, \(P\sin\theta\), lifts some of the weight, so \(R\) and the maximum friction \(\mu R\) are smaller too. In limiting equilibrium, \(P\cos\theta = \mu(mg - P\sin\theta)\), so
\[P = \frac{\mu mg}{\cos\theta + \mu\sin\theta}.\]
A neat way to find the minimum: combine \(R\) and \(F = \mu R\) into one total contact force \(S\), which makes angle \(\lambda\) with the normal, where \(\tan\lambda = \mu\) (\(\lambda\) is the angle of friction). The particle is then held by three forces, \(mg\), \(S\) and \(P\). In the triangle of forces \(mg\) is fixed and \(S\) has a fixed direction, so \(P\) is shortest when it is perpendicular to \(S\): at angle \(\lambda\) above the horizontal, with \(P_{\min} = mg\sin\lambda\).
A particle of mass 12 kg rests on rough horizontal ground. The coefficient of friction is 0.35. A force of magnitude \(P\) newtons is applied at an angle \(\theta\) above the horizontal, and the particle is on the point of moving. (a) Show that \(P = \dfrac{41.16}{\cos\theta + 0.35\sin\theta}\). (b) Find \(P\) when the force is horizontal and when \(\theta = 40^\circ\). (c) Find the least possible value of \(P\) and the angle at which it acts.
(a) R(↑): \(R = 117.6 - P\sin\theta\). R(→) with \(F = \mu R\): \(P\cos\theta = 0.35(117.6 - P\sin\theta)\), so \(P(\cos\theta + 0.35\sin\theta) = 41.16\), as required.
(b) \(\theta = 0\): \(P = 41.16\). \(\theta = 40^\circ\): \(P = 41.16 \div (\cos 40^\circ + 0.35\sin 40^\circ) = 41.53\). 41.2 N; 41.5 N
(c) The least force acts at the angle of friction: \(\tan\lambda = 0.35\) gives \(\theta = 19.3^\circ\), and \(P_{\min} = 117.6\sin 19.3^\circ = 117.6 \times 0.330 = 38.85\). 38.8 N at 19.3° above the horizontal
Check in the formula from (a): \(41.16 \div (\cos 19.3^\circ + 0.35\sin 19.3^\circ) = 38.85\) N.
A slightly upward pull reduces the normal reaction, and so the friction, by more than it loses from its horizontal part, up to the angle where \(\tan\theta = \mu\).
Exam tips for M1.5
Draw your own diagram even when one is printed, with every force and every angle. A force left out, or drawn the wrong way, spoils every equation that follows.
Start each equation with its direction, e.g. "R(parallel to plane):", and write it as one complete equation before simplifying. An equation with the right terms, each with the correct sin or cos, shows a sound method, and a sign slip is easier to spot.
When the answer is given, show the resolved equations, the substitution and a final line that matches the printed result. Then use the printed value in later parts.
Give final answers to 2 or 3 significant figures, keeping full calculator values between steps. 16.58 N is better written as 16.6 N.
"On the point of", "about to", "just" and "limiting" mean \(F = \mu R\). "Rests in equilibrium" with no such phrase means only \(F \le \mu R\): find \(F\) by resolving. "Smooth" means no friction at all.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Friction always equals \(\mu R\). | \(\mu R\) is the maximum; friction only provides what is needed. | \(F \le \mu R\); \(F = \mu R\) only in limiting equilibrium (or when sliding). |
| On a plane, \(R = mg\cos\alpha\) always. | Other forces can push into or pull away from the plane. | Resolve perpendicular to the plane with every force included. |
| Friction always acts up a slope. | It opposes the motion that would happen. | About to slide up → friction down the plane, and vice versa. |
| The tension is the same in both strings holding a particle. | Only true for a symmetrical arrangement. | Resolve horizontally: the steeper string carries more. |
| A heavier particle needs a larger \(\mu\) to stay at rest on a slope. | Both the weight component and \(\mu R\) are proportional to \(m\). | With no other force, it rests if \(\mu \ge \tan\alpha\), whatever its mass. |
| Pulling horizontally is always the easiest way to move a particle. | An upward component reduces \(R\) and so the friction. | The least force acts at angle \(\lambda\) above the horizontal, \(\tan\lambda = \mu\). |
How this topic connects
M1.1 Vectors in mechanics
Forces in \(\mathbf{i}\), \(\mathbf{j}\) form, magnitudes and angles with \(\mathbf{i}\) are the same tools as for displacement and velocity.
M1.3 Dynamics
When the resultant is not zero, the same resolving gives \(\sum F = ma\); friction on a sliding particle is \(\mu R\).
M1.6 Moments, and rigid bodies in M2
An extended body also needs zero total moment; ladders and rods with friction combine both conditions. The sine rule from P1 trigonometry drives the triangle of forces.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(F\cos\theta\), \(F\sin\theta\) | Components of a force along and perpendicular to a direction | No — learn it |
| \(R = \sqrt{X^2 + Y^2}\), \(\tan\varphi = Y/X\) | Magnitude and direction of a resultant | No — learn it |
| \(\sum F_x = 0\), \(\sum F_y = 0\) | Equilibrium of a particle (any two perpendicular directions) | No — learn it |
| \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\) | Triangle of forces for three forces | No — learn it |
| \(F \le \mu R\); \(F = \mu R\) when limiting | Friction in equilibrium | No — learn it |
| \(mg\sin\alpha\), \(mg\cos\alpha\) | Components of weight on a plane at angle \(\alpha\) | No — derive it |
| \(\mu \ge \tan\alpha\) | Particle at rest on a rough plane with no other force | No — derive it |
| \(P_{\min} = mg\sin\lambda\), \(\tan\lambda = \mu\) | Least force to move a particle on rough horizontal ground | No — derive it |
Check your understanding
1. Forces of 5 N and 12 N act on a particle at right angles to each other. Find the magnitude of the resultant and its angle with the 5 N force.
\(\sqrt{5^2 + 12^2} = 13\) N; \(\tan^{-1}(12 \div 5) = 67.4^\circ\). 13 N at 67.4° to the 5 N force
2. A 2 kg particle rests on rough horizontal ground (\(\mu = 0.5\)). A horizontal force of 6 N acts on it and it does not move. What is the friction force?
The maximum available is \(\mu R = 0.5 \times 19.6 = 9.8\) N, but friction only balances the 6 N applied. 6 N
3. A particle rests on a rough plane inclined at 20° with no other force applied. What is the least possible value of \(\mu\)?
\(\mu \ge \tan 20^\circ\). 0.364
4. A particle is in equilibrium under \((2\mathbf{i} - 3\mathbf{j})\) N, \((p\mathbf{i} + \mathbf{j})\) N and \((-5\mathbf{i} + q\mathbf{j})\) N. Find \(p\) and \(q\).
\(\mathbf{i}\): \(2 + p - 5 = 0\). \(\mathbf{j}\): \(-3 + 1 + q = 0\). p = 3, q = 2
5. A particle of weight \(W\) hangs on a string and is pulled aside by a horizontal force of size \(W\). Find the angle of the string to the vertical and the tension.
\(\tan\theta = W/W = 1\), so 45°; \(T = \sqrt{W^2 + W^2} = W\sqrt{2}\). 45°, T ≈ 1.41W
Specification checklist
- (M1 5.1) I can resolve a force into two perpendicular components and find the magnitude and direction of the resultant of several coplanar forces, including forces written with \(\mathbf{i}\) and \(\mathbf{j}\).
- (M1 5.2) I can identify the weight, normal reaction, tension, thrust and friction acting on a particle and draw them in the correct directions.
- (M1 5.2) I can solve equilibrium problems for a particle by resolving in two perpendicular directions, or with a triangle of forces when there are three forces.
- (M1 5.2) I can handle particles suspended by two strings and particles on inclined planes held by horizontal forces or by strings at an angle.
- (M1 5.3) I know that friction satisfies \(F \le \mu R\), with \(F = \mu R\) in limiting equilibrium, and I can decide the direction in which friction acts.
- (M1 5.3) I can find the range of forces that keep a particle at rest on a rough plane, and the least force needed to move a particle on a rough surface.