A particle in equilibrium has zero resultant force, so the components in any direction add up to zero. Two resolving equations, two unknowns: that is the whole method.
A particle is in equilibrium when the vector sum of the forces on it is zero. Equivalently, the sum of the components in ANY direction is zero, so resolving in two perpendicular directions gives two equations for two unknowns.
Fig. 1: a 40 N weight hanging from two strings at 30° and 50° to the horizontal (drawn to scale); resolving gives tensions of 26.1 N and 35.2 N
Section 01
Zero resultant, two equations.
In the lab, set the hanging weight to 40 N and the string angles to \(30^\circ\) and \(50^\circ\). Read the two tensions and check them against Fig. 1. Then make one string steeper and watch its tension rise while the other falls: the components must still cancel.
Equilibrium means the particle is at rest (or moving at constant velocity): the resultant force is zero.
Resolve twice, in two perpendicular directions, and set each total to zero.
Choose clever directions: resolving perpendicular to an unknown force removes it from that equation.
The lab uses \(g = 9.81\). In Paper 4 always use \(g = 10\ \text{m s}^{-2}\).
Worked example A weight on two strings
A particle of weight 40 N hangs in equilibrium from two light strings inclined at \(30^\circ\) and \(50^\circ\) to the horizontal (Fig. 1). Find the tensions.
From (1): \(T_1 = \dfrac{T_2\cos 50^\circ}{\cos 30^\circ} = 0.7422\,T_2\).
Substitute in (2): \(0.3711\,T_2 + 0.7660\,T_2 = 40\), so \(T_2 = 35.2\) N.
\(T_1 = 0.7422 \times 35.18 = 26.1\) N.
Answer: \(T_1 = 26.1\) N, \(T_2 = 35.2\) N. The steeper string carries more tension.
Examiner trap
Keep at least 4 significant figures in the middle of the working. Rounding \(T_2\) to 35 before finding \(T_1\) gives 26.0 instead of 26.1 and can cost the accuracy mark.
Section 02
Smooth slopes.
A contact is "smooth" when we model it as having no friction: the only contact force is the normal reaction, perpendicular to the surface. On a slope, resolve parallel and perpendicular to the plane, so the normal reaction appears in only one equation.
Limitation of the smooth model: real surfaces always have some friction, so a smooth plane predicts a particle that slides however gentle the slope. It is a good model only when friction is small (ice, a polished surface, a well-oiled bearing).
Worked example Held by a horizontal force
A particle of mass 2 kg rests on a smooth plane inclined at \(25^\circ\) to the horizontal, held by a horizontal force of \(P\) N. Find \(P\) and the normal reaction.
Weight \(= 20\) N. The horizontal force makes \(25^\circ\) with the plane.
Parallel to the plane: \(P\cos 25^\circ = 20\sin 25^\circ\), so \(P = 20\tan 25^\circ = 9.33\).
Perpendicular to the plane: \(R = 20\cos 25^\circ + P\sin 25^\circ = 18.13 + 3.94\).
\(R = 22.1\) N.
Answer: \(P = 9.33\), \(R = 22.1\) N. The horizontal force pushes into the plane, so \(R\) is MORE than \(20\cos 25^\circ\).
Worked example Held by a force at an angle to the plane
A particle of mass 5 kg rests on a smooth plane inclined at \(30^\circ\), held by a force of \(P\) N acting at \(20^\circ\) above a line of greatest slope, up the plane. Find \(P\) and the normal reaction.
Answer: \(P = 26.6\), \(R = 34.2\) N. This time the force lifts the particle off the plane, so \(R\) is LESS than \(50\cos 30^\circ\).
Examiner trap
The normal reaction is NOT always \(mg\cos\theta\). Any other force with a component perpendicular to the plane changes it. Always write the full perpendicular equation.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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A particle is in equilibrium. What is the resultant force on it?
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A particle of weight 20 N hangs on a string and is held aside by a horizontal force of 15 N. What is the tension?
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What does it mean to model a contact as smooth?
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A particle of mass 4 kg rests on a smooth plane inclined at \(30^\circ\), held by a force \(P\) N parallel to the plane. Find \(P\).
Section 03
Finding an unknown force.
When the size and direction of a force are both unknown, resolve to find its two components, then rebuild it with Pythagoras and inverse tan.
Worked example Magnitude and direction
A particle of weight 30 N is in equilibrium under its weight, a horizontal force of 16 N and a force of \(P\) N at \(\theta^\circ\) to the upward vertical. Find \(P\) and \(\theta\).
Horizontally: \(P\sin\theta = 16\).
Vertically: \(P\cos\theta = 30\).
Square and add: \(P^2 = 16^2 + 30^2 = 1156\), so \(P = 34\).
Divide: \(\tan\theta = \dfrac{16}{30}\), so \(\theta = 28.1\).
Answer: \(P = 34\), \(\theta = 28.1\)
Your turn
A particle of mass 3 kg hangs on a light string. A horizontal force of 12 N holds it in equilibrium with the string inclined to the vertical. Find the tension and the angle of the string to the vertical.
Vertically \(T\cos\alpha = 30\), horizontally \(T\sin\alpha = 12\). So \(T = \sqrt{30^2 + 12^2} = 32.3\) N and \(\alpha = \tan^{-1}\dfrac{12}{30} = 21.8^\circ\).
Examiner trap
Dividing the two equations gives \(\tan\theta\) only if you put the right one on top: \(\dfrac{P\sin\theta}{P\cos\theta} = \tan\theta\). Check which component used sin before you divide.
Stage 2 · Exam
Exam-style questions
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A particle of weight 50 N hangs from two light strings, each inclined at \(20^\circ\) to the horizontal. Find the tension in each string.
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A particle of mass 2 kg rests on a smooth plane inclined at \(25^\circ\) to the horizontal, held by a horizontal force of \(P\) N. Find \(P\).
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A particle of mass 6 kg rests on a smooth plane inclined at \(20^\circ\), held by a force acting at \(30^\circ\) above a line of greatest slope. Find the normal reaction.
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A particle of weight \(W\) N hangs from two strings at right angles to each other, inclined at \(35^\circ\) and \(55^\circ\) to the vertical. The tension in the first string is 20 N. Find \(W\).
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