Equilibrium of a Particle

AS · M1 · 13 min

A particle is in equilibrium when the forces on it balance exactly, so that it stays at rest or keeps moving with constant velocity. Equilibrium questions are a staple of Paper 4: a particle hanging from strings, a ring threaded on a string, a block held on a slope. They all use one principle, that the components of the forces in any direction add to zero, and one technique, resolving. This note builds the method for particles on strings and on horizontal surfaces; slopes and friction follow in the next notes.

The equilibrium condition

Definition

A particle is in equilibrium when the resultant force on it is zero. It is then either at rest or moving in a straight line with constant speed.

The syllabus states the principle like this: when a particle is in equilibrium, the vector sum of the forces acting is zero, or equivalently, the sum of the components in any direction is zero.

Equilibrium of a particle

\sum F_x = 0 \qquad\text{and}\qquad \sum F_y = 0

for any pair of perpendicular directions $x$ and $y$ . Resolving in two perpendicular directions gives two independent equations, so you can find at most two unknowns.

Two points are worth stressing.

Constant velocity is equilibrium too. A car moving at a steady $20\ \text{m s}^{-1}$ along a straight road, or a crate being pushed across a floor at constant speed, has zero resultant force. The phrase "moves at constant speed in a straight line" is an equilibrium signal.
Any direction works. You are free to choose which two perpendicular directions to resolve in. A good choice makes the algebra much easier.

Choosing the directions

The two equations you get are equally valid for any pair of perpendicular directions, but some choices give simpler equations:

Horizontal and vertical for particles hanging on strings or on horizontal surfaces, because the weight and usually several other forces are already vertical or horizontal.
Parallel and perpendicular to a slope for particles on inclined planes, because the normal reaction and friction lie along these.
Perpendicular to an unknown force when you want to eliminate it. A force has no component at right angles to itself, so resolving perpendicular to a force you do not need removes it from the equation entirely.

Solving an equilibrium problem

Draw a clear force diagram showing every force, with all given angles marked.
Choose two perpendicular directions. State them ("resolving vertically", "resolving parallel to the plane").
Resolve every force in the first direction and set the total to zero. Write the equation with every term, before simplifying.
Do the same for the second direction.
Solve the two equations simultaneously. If the unknowns are a magnitude and an angle in the form $P\cos\theta$ and $P\sin\theta$ , square and add, then divide.
Check the answer is sensible: tensions are positive, angles lie in the range the diagram shows.

Three horizontal forces

Three horizontal forces act on a particle on a smooth horizontal table: $12\ \text{N}$ due east, $9\ \text{N}$ due north and a force $P\ \text{N}$ at a bearing chosen so that the particle stays in equilibrium. Find $P$ and its direction.

Solution

Take east as the $x$ -direction and north as the $y$ -direction, and let $P$ have components $P_x$ and $P_y$ .

\text{east: } 12 + P_x = 0 \Rightarrow P_x = -12, \qquad \text{north: } 9 + P_y = 0 \Rightarrow P_y = -9

P = \sqrt{12^2 + 9^2} = 15\ \text{N}

$P$ points south-west of the particle: towards the west at $\tan^{-1}\left(\tfrac{9}{12}\right) = 36.9^\circ$ south of west. In other words, $P$ is equal and opposite to the resultant of the other two forces, which is always true for a particle in equilibrium.

Particles hanging from strings

A particle hanging from strings has weight acting vertically down and a tension along each string, pointing away from the particle. Resolve horizontally and vertically.

A particle of weight 30 N hangs from two light strings attached to a horizontal ceiling. The strings make angles of 35° and 50° with the vertical.

Two strings at different angles

A particle of weight $30\ \text{N}$ hangs in equilibrium from two light inextensible strings attached to a horizontal ceiling. The strings make angles of $35^\circ$ and $50^\circ$ with the vertical. Find the tension in each string.

Solution

Let the tensions be $T_1$ (string at $35^\circ$ ) and $T_2$ (string at $50^\circ$ ). The angles are measured from the vertical, so the vertical components are the cos ones.

Resolving horizontally:

T_1\sin 35^\circ = T_2\sin 50^\circ \tag{1}

Resolving vertically:

T_1\cos 35^\circ + T_2\cos 50^\circ = 30 \tag{2}

From (1), $T_1 = \dfrac{T_2\sin 50^\circ}{\sin 35^\circ} = 1.3356\,T_2$ . Substitute in (2):

1.3356\,T_2\cos 35^\circ + T_2\cos 50^\circ = 30 \quad\Rightarrow\quad (1.0940 + 0.6428)\,T_2 = 30 \quad\Rightarrow\quad T_2 = 17.27

Then $T_1 = 1.3356 \times 17.27 = 23.07$ .

T_1 = 23.1\ \text{N}, \qquad T_2 = 17.3\ \text{N}

The string nearer the vertical carries the larger tension, which is a useful sanity check: a more vertical string does more of the job of holding up the weight.

Tip

Elimination of an unknown by clever resolving: resolving perpendicular to one string removes that tension. Here, resolving perpendicular to the $50^\circ$ string gives an equation in $T_1$ only: $T_1\sin(35^\circ + 50^\circ) = 30\sin 50^\circ$ , so $T_1 = \dfrac{30\sin 50^\circ}{\sin 85^\circ} = 23.1\ \text{N}$ in one line. This is optional; the horizontal-and-vertical method always works.

A particle of weight 5 N hangs on a light string. A horizontal force P holds it in equilibrium with the string at 20° to the vertical.

A string pulled aside by a horizontal force

A particle of weight $5\ \text{N}$ hangs from a light string attached to a fixed point. A horizontal force $P\ \text{N}$ holds the particle in equilibrium with the string at $20^\circ$ to the vertical.

(a) Find the tension in the string and the value of $P$ .

(b) The string will break if its tension exceeds $8\ \text{N}$ . The horizontal force is gradually increased. Find the greatest angle the string can make with the vertical before it breaks, and the corresponding value of $P$ .

Solution

(a) Resolving vertically: $T\cos 20^\circ = 5$ , so

T = \frac{5}{\cos 20^\circ} = 5.32\ \text{N}

Resolving horizontally: $P = T\sin 20^\circ = 5.3209 \times 0.3420 = 1.82\ \text{N}$ .

(b) With the string at angle $\theta$ to the vertical, resolving vertically still gives $T\cos\theta = 5$ . The tension reaches $8\ \text{N}$ when

\cos\theta = \frac{5}{8} \quad\Rightarrow\quad \theta = 51.3^\circ

Then $P = 8\sin\theta$ . Since $\cos\theta = \tfrac58$ , $\sin\theta = \sqrt{1 - \tfrac{25}{64}} = \tfrac{\sqrt{39}}{8}$ , so $P = \sqrt{39} = 6.24\ \text{N}$ .

As a check, $T^2 = P^2 + 5^2$ : $64 = 39 + 25$ .

Smooth rings and pegs

A smooth ring threaded on a string can slide freely along it, so it exerts no friction on the string. The same is true of a string passing over a smooth peg or pulley. The consequence is the most important fact in these problems:

Key result

If a light string passes through a smooth ring, or over a smooth peg or pulley, the tension is the same on both sides.

So a ring on a string gives you two tension forces of the same magnitude $T$ , pointing along the two parts of the string, away from the ring.

A smooth ring R of weight 6 N is threaded on a light string with ends fixed at A and at B, vertically below A. A horizontal force of X N holds the ring with AR at 30° and BR at 60° to the vertical.

Smooth ring on a string

A small smooth ring $R$ of mass $0.6\ \text{kg}$ is threaded on a light inextensible string. One end of the string is attached to a fixed point $A$ and the other to a fixed point $B$ vertically below $A$ . A horizontal force of magnitude $X\ \text{N}$ acts on $R$ , away from $AB$ . In equilibrium, $AR$ makes $30^\circ$ with the vertical and $BR$ makes $60^\circ$ with the vertical. Find the tension in the string and the value of $X$ .

Solution

The ring is smooth, so the tension $T$ is the same in $AR$ and $BR$ . The weight is $6\ \text{N}$ .

$AR$ pulls up and towards $AB$ ; $BR$ pulls down and towards $AB$ .

Resolving vertically:

T\cos 30^\circ - T\cos 60^\circ - 6 = 0 \quad\Rightarrow\quad T(0.8660 - 0.5) = 6 \quad\Rightarrow\quad T = 16.4\ \text{N}

Resolving horizontally:

X = T\sin 30^\circ + T\sin 60^\circ = 16.392(0.5 + 0.8660) = 22.4

So $T = 16.4\ \text{N}$ and $X = 22.4$ . (Exactly, $T = 6(\sqrt3 + 1)$ and $X = 12 + 6\sqrt3$ .)

Watch out

Do not give a smooth ring two different tensions $T_1$ and $T_2$ . If the question says the ring is smooth, or the string passes over a smooth peg, there is one tension. Conversely, if two separate strings are tied to a particle, the tensions are different and you need two unknowns.

Problems with a limit

Many questions give a maximum: a string that breaks above a certain tension, or a particle on the point of lifting off. The method is to work out the forces in terms of the unknown, then apply the limit to whichever force reaches it first.

The greatest mass the strings can support (exam standard)

A particle of mass $m\ \text{kg}$ hangs in equilibrium from two light strings attached to a horizontal ceiling, making angles of $35^\circ$ and $50^\circ$ with the vertical, as in the earlier diagram. Each string will break if its tension exceeds $50\ \text{N}$ . Find the greatest possible value of $m$ , and the tension in the other string when $m$ takes this value.

Solution

Repeating the earlier working with weight $10m$ instead of $30$ :

T_1\sin 35^\circ = T_2\sin 50^\circ, \qquad T_1\cos 35^\circ + T_2\cos 50^\circ = 10m

Solving as before gives

T_1 = 0.76897 \times 10m = 7.6897m, \qquad T_2 = 0.57577 \times 10m = 5.7577m

$T_1 > T_2$ for every $m$ , so the $35^\circ$ string reaches $50\ \text{N}$ first:

7.6897m = 50 \quad\Rightarrow\quad m = 6.50

Then $T_2 = 5.7577 \times 6.5022 = 37.4\ \text{N}$ .

The key step is deciding which string breaks first. Setting both tensions to $50$ at once would be wrong: the equations would be inconsistent.

Other methods (not required)

Triangle of forces and Lami's theorem

For exactly three forces in equilibrium, the forces drawn head to tail form a closed triangle, so the sine rule applies (this is the triangle of forces). Lami's theorem is a packaged version: each force is proportional to the sine of the angle between the other two. The syllabus says these methods are acceptable where suitable but are not required and will not be referred to in questions. Resolving always works and is what mark schemes are written around, so make it your default.

Common mistakes

Equilibrium errors

Sin and cos swapped because the angle was measured from the vertical rather than the horizontal. Mark the angle on the diagram and decide adjacent or opposite each time.
Missing a force. Every string attached gives a tension; every surface gives a normal reaction; the weight is always there.
Two tensions for a smooth ring (should be one), or one tension for two separate strings (should be two).
Applying a breaking limit to both strings at once. Find which reaches the limit first.
Treating constant velocity as needing a resultant force. If the velocity is constant, the forces balance.

Exam technique

Exam tip

Write each resolved equation in full with every force before simplifying: "Resolving vertically: $T_1\cos 35^\circ + T_2\cos 50^\circ = 30$ ". Each correct equation typically earns a method mark even if you never finish solving.
An equation needs the right number of terms. Mark schemes often specify "three terms, allow sin/cos mix" for the method mark, so a missing force loses marks that a sign slip would not.
Leave the final answers to 3 significant figures, but carry more figures through the simultaneous equations.
"Show that" questions give the answer: show every step, including the equations you resolved, and give the result to more figures than stated to show it was not copied.

Summary

Equilibrium means zero resultant force: the particle is at rest or moving with constant velocity.
Resolve in two perpendicular directions and set each total to zero. Two equations give at most two unknowns.
Choose directions that make the algebra easy: horizontal and vertical for strings, along and perpendicular to a slope for planes, perpendicular to an unwanted force to eliminate it.
A smooth ring, peg or pulley has the same tension on both sides of the string.
For limits (breaking strings, losing contact), find each force in terms of the unknown, then apply the limit to the one that reaches it first.
Triangle of forces and Lami's theorem are acceptable but never required.

Practice

Question

Three horizontal forces act on a particle in equilibrium: $5\ \text{N}$ due east, $12\ \text{N}$ due north and a third force $P\ \text{N}$ . Find $P$ and the angle it makes with due south.
A particle of mass $2\ \text{kg}$ hangs from two light strings, each making $30^\circ$ with the horizontal. Find the tension in each string.
A particle of mass $1.2\ \text{kg}$ hangs from a light string. A horizontal force of $5\ \text{N}$ holds it in equilibrium with the string inclined to the vertical. Find the tension and the angle the string makes with the vertical.
A particle of mass $4\ \text{kg}$ hangs from two light strings attached to a ceiling. One string makes $50^\circ$ with the horizontal and the other makes $20^\circ$ with the horizontal, on the opposite side. Find the two tensions.
Four forces act on a particle in equilibrium: $20\ \text{N}$ at angle $\alpha$ above the positive $x$ -axis where $\tan\alpha = \tfrac34$ ; $26\ \text{N}$ at angle $\beta$ above the negative $x$ -axis where $\tan\beta = \tfrac{12}{5}$ ; $P\ \text{N}$ along the negative $y$ -axis; and $Q\ \text{N}$ along the negative $x$ -axis. Find $P$ and $Q$ .
A smooth ring of mass $0.4\ \text{kg}$ is threaded on a light string whose ends are attached to two fixed points. A horizontal force $X\ \text{N}$ holds the ring in equilibrium with the two parts of the string making angles of $30^\circ$ and $50^\circ$ with the horizontal, both above the ring and on opposite sides of it. The force $X$ acts towards the side of the $50^\circ$ string. Find the tension and $X$ .
A particle of weight $W\ \text{N}$ hangs from two strings making $40^\circ$ and $60^\circ$ with the vertical. The string at $40^\circ$ breaks if its tension exceeds $30\ \text{N}$ ; the other breaks if its tension exceeds $25\ \text{N}$ . Find the greatest possible value of $W$ .
A particle of mass $2\ \text{kg}$ hangs from a light string. A force of magnitude $P\ \text{N}$ , acting at angle $\phi$ above the horizontal, holds the particle in equilibrium with the string at $30^\circ$ to the vertical. (a) Find $P$ in terms of $\phi$ by resolving perpendicular to the string. (b) Hence find the least possible value of $P$ , and the value of $\phi$ for which it occurs.

Answers

$P_x = -5$ , $P_y = -12$ , so $P = 13\ \text{N}$ . It points south-west of the particle, at $\tan^{-1}\left(\tfrac{5}{12}\right) = 22.6^\circ$ west of due south.
Horizontal components cancel by symmetry, so both tensions equal $T$ . Vertically: $2T\sin 30^\circ = 20$ , so $T = 20\ \text{N}$ .
Horizontally $T\sin\theta = 5$ ; vertically $T\cos\theta = 12$ . So $T = \sqrt{5^2 + 12^2} = 13\ \text{N}$ and $\theta = \tan^{-1}\left(\tfrac{5}{12}\right) = 22.6^\circ$ .
Horizontally: $T_1\cos 50^\circ = T_2\cos 20^\circ$ . Vertically: $T_1\sin 50^\circ + T_2\sin 20^\circ = 40$ . Substituting $T_2 = T_1\cos 50^\circ/\cos 20^\circ$ :

T_1\left(\sin 50^\circ + \frac{\cos 50^\circ\sin 20^\circ}{\cos 20^\circ}\right) = 40 \Rightarrow T_1 = 40.0\ \text{N}, \quad T_2 = 27.4\ \text{N}

( $T_1$ is exactly $40$ , since the bracket simplifies to $\sin 70^\circ/\cos 20^\circ = 1$ .) 5. $\sin\alpha = 0.6$ , $\cos\alpha = 0.8$ ; $\sin\beta = \tfrac{12}{13}$ , $\cos\beta = \tfrac{5}{13}$ . $x$ : $16 - 10 - Q = 0$ , so $Q = 6$ . $y$ : $12 + 24 - P = 0$ , so $P = 36$ . 6. One tension $T$ . Vertically: $T\sin 30^\circ + T\sin 50^\circ = 4$ , so $T = \dfrac{4}{0.5 + 0.7660} = 3.16\ \text{N}$ . Horizontally the $30^\circ$ part pulls one way with $T\cos 30^\circ$ and the $50^\circ$ part the other way with $T\cos 50^\circ$ , so $X = T(\cos 30^\circ - \cos 50^\circ) = 3.1594 \times 0.2232 = 0.705$ . 7. Horizontally $T_1\sin 40^\circ = T_2\sin 60^\circ$ ; vertically $T_1\cos 40^\circ + T_2\cos 60^\circ = W$ . Solving: $T_1 = 0.8794W$ , $T_2 = 0.6527W$ . Limits: $0.8794W \le 30 \Rightarrow W \le 34.1$ ; $0.6527W \le 25 \Rightarrow W \le 38.3$ . The first string breaks first, so the greatest $W$ is $34.1$ . 8. (a) The weight $20\ \text{N}$ has component $20\sin 30^\circ = 10$ perpendicular to the string. The force $P$ at $\phi$ above the horizontal makes angle $|\phi - 30^\circ|$ with the perpendicular to the string (the string's perpendicular is itself $30^\circ$ above the horizontal). So $P\cos(\phi - 30^\circ) = 10$ and $P = \dfrac{10}{\cos(\phi - 30^\circ)}$ . (b) $P$ is least when $\cos(\phi - 30^\circ) = 1$ , that is $\phi = 30^\circ$ : the force acts perpendicular to the string. Then $P = 10\ \text{N}$ .

The equilibrium condition

Choosing the directions

Particles hanging from strings

Smooth rings and pegs

Problems with a limit

Other methods (not required)

Common mistakes

Exam technique

Summary

Practice

Console