Equilibrium of forces

AS · 15 min

A bridge carrying traffic, a shop sign on a bracket and a ladder leaning on a wall all stay still because the forces on them balance in two ways at once: they do not make the object accelerate, and they do not make it turn. This note covers the principle of moments, the two conditions for equilibrium, and the vector triangle for three forces. Every Paper 2 has a structured question on a beam, rod or hinged boom, and Paper 1 tests the vector triangle and the conditions for equilibrium.

The principle of moments

In Moments and couples a see-saw turned because the clockwise and anticlockwise moments were unequal. When they are equal, there is no resultant moment, and the object does not start to rotate.

Definition

Principle of moments: for an object in equilibrium, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about that same point.

The words "about any point" are the most useful part of the statement. If an object is in equilibrium the moments balance about every point, whether or not there is a real pivot there. So you may choose the point, and the best choice is a point where an unknown force acts: that force then has zero moment and drops out of the equation.

Conditions for equilibrium

Definition

An object is in equilibrium when there is no resultant force and no resultant torque (moment) acting on it.

Each condition does a different job:

No resultant force means no linear acceleration (Newton's first law). In practice: the forces upwards equal the forces downwards, and the forces to the left equal the forces to the right. Equivalently, the vectors form a closed polygon.
No resultant torque means no angular acceleration: the object does not start to rotate. In practice: the principle of moments, about any point.

Neither condition is enough on its own. A couple has zero resultant force but makes the object rotate. A single force through the centre of mass has no moment about that point but makes the object accelerate.

Key result

For an object in equilibrium:

\sum F_{\text{up}} = \sum F_{\text{down}} \qquad \sum F_{\text{left}} = \sum F_{\text{right}}

\sum(\text{clockwise moments}) = \sum(\text{anticlockwise moments}) \quad \text{about any point}

Method

Solving a beam, rod or ladder problem

Draw the object and mark every force on it: its weight at the centre of gravity, every support or contact force, every tension, every load.
If there is an unknown force at a hinge or support, take moments about that point so the unknown has no moment.
Write clockwise moments $=$ anticlockwise moments, with each perpendicular distance visible. Solve for one unknown.
Resolve vertically (and horizontally, if needed): total up $=$ total down. Solve for the remaining unknowns.
Check: take moments about a different point. It should balance too.

A beam on two supports

A uniform beam of length $4.0\ \text{m}$ and weight $500\ \text{N}$ rests horizontally on supports A and B at its two ends. A load of $800\ \text{N}$ is placed $1.0\ \text{m}$ from A. Calculate the forces exerted by the supports.

Solution

The beam's weight acts at its midpoint, $2.0\ \text{m}$ from A. Take moments about A, so the unknown force $R_A$ has no moment.

Clockwise moments about A (weight and load): $500 \times 2.0 + 800 \times 1.0 = 1800\ \text{N m}$ .

Anticlockwise moment about A (support B): $R_B \times 4.0$ .

4.0R_B = 1800 \quad\Rightarrow\quad R_B = 450\ \text{N}

Resolve vertically: $R_A + R_B = 500 + 800$ , so

R_A = 1300 - 450 = 850\ \text{N}

Check about B: clockwise $R_A \times 4.0 = 3400\ \text{N m}$ ; anticlockwise $500 \times 2.0 + 800 \times 3.0 = 3400\ \text{N m}$ . Balanced. The support nearer the load takes more of the weight, as you would expect.

Locating a centre of gravity

A non-uniform plank of length $2.00\ \text{m}$ rests horizontally on two bathroom scales, one at each end. The left-hand scale reads $60\ \text{N}$ and the right-hand scale reads $90\ \text{N}$ . Find the weight of the plank and the position of its centre of gravity.

Solution

Resolving vertically, the weight is $W = 60 + 90 = 150\ \text{N}$ .

Let the centre of gravity be a distance $x$ from the left end. Take moments about the left end, so the $60\ \text{N}$ force has no moment:

150 \times x = 90 \times 2.00 \quad\Rightarrow\quad x = 1.20\ \text{m}

The centre of gravity is $1.20\ \text{m}$ from the left end, nearer the scale with the larger reading.

Hinges and cables at an angle

A hinge can push or pull in any direction, so its force has both a horizontal and a vertical component, and usually you do not know either. Taking moments about the hinge removes both at once. Only the perpendicular component of a cable's tension has a moment about the hinge.

A uniform beam of weight 80 N hinged to a wall and held horizontal by a cable at 35° to the beam. A sign of weight 120 N hangs from the free end, 1.5 m from the hinge.

Shop sign on a hinged beam

The diagram shows a uniform horizontal beam of length $1.5\ \text{m}$ and weight $80\ \text{N}$ , hinged to a wall. A sign of weight $120\ \text{N}$ hangs from its free end. A cable from the free end to the wall makes an angle of $35^\circ$ with the beam.

(a) Calculate the tension $T$ in the cable.

(b) Calculate the magnitude and direction of the force exerted by the hinge on the beam.

Solution

(a) Take moments about the hinge. The cable acts at $1.5\ \text{m}$ ; its component perpendicular to the beam is $T\sin 35^\circ$ .

Clockwise moments: $80 \times 0.75 + 120 \times 1.5 = 60 + 180 = 240\ \text{N m}$ .

Anticlockwise moment: $T\sin 35^\circ \times 1.5$ .

T = \frac{240}{1.5\sin 35^\circ} = \frac{240}{0.860} = 279\ \text{N}

(b) Let the hinge force have horizontal component $H$ (away from the wall) and vertical component $V$ (upwards).

Horizontally: the cable pulls the beam towards the wall with $T\cos 35^\circ$ , so

H = 279 \times \cos 35^\circ = 229\ \text{N}

Vertically: $V + T\sin 35^\circ = 80 + 120$ . Since $T\sin 35^\circ = 240/1.5 = 160\ \text{N}$ ,

V = 200 - 160 = 40\ \text{N}

Combine the components:

F = \sqrt{229^2 + 40^2} = 232\ \text{N}, \qquad \tan\phi = \frac{40}{229} \Rightarrow \phi = 9.9^\circ

The hinge pushes on the beam with $232\ \text{N}$ at $9.9^\circ$ above the horizontal, directed away from the wall.

A ladder against a smooth wall

A uniform ladder of length $5.0\ \text{m}$ and weight $200\ \text{N}$ leans against a smooth vertical wall, making an angle of $60^\circ$ with the rough horizontal ground. Calculate (a) the force exerted by the wall on the ladder, (b) the normal and frictional forces exerted by the ground.

Solution

Forces on the ladder: weight $200\ \text{N}$ down at the midpoint; the wall's force $R$ , horizontal (a smooth wall exerts no friction, so its force is perpendicular to the wall); at the foot, a normal force $N$ upwards and friction $F$ horizontally towards the wall.

(a) Take moments about the foot, removing $N$ and $F$ .

The weight's line of action is a horizontal distance $2.5\cos 60^\circ = 1.25\ \text{m}$ from the foot. The wall's horizontal force acts at the top, a vertical height $5.0\sin 60^\circ = 4.33\ \text{m}$ above the foot.

R \times 4.33 = 200 \times 1.25 \quad\Rightarrow\quad R = \frac{250}{4.33} = 57.7\ \text{N}

(b) Resolve vertically: $N = 200\ \text{N}$ . Resolve horizontally: $F = R = 57.7\ \text{N}$ , towards the wall.

If the ground were also smooth, there would be nothing to balance $R$ , and the ladder would slide. This is why ladders are footed or placed on rough ground.

Three forces: the vector triangle

When exactly three non-parallel forces act on an object in equilibrium, two useful facts follow.

The forces form a closed triangle. Draw the three force vectors to scale, head to tail, in any order. Since the resultant is zero, the end of the third arrow lands back on the start of the first. Conversely, if three forces drawn head to tail form a closed triangle, they have no resultant.

Their lines of action pass through one point. If the lines of action of two of the forces meet at a point P, both have zero moment about P. For the moments to balance about P, the third force must also have zero moment about P, so its line of action passes through P too. The forces are concurrent.

Key result

Three coplanar forces in equilibrium can be represented by the three sides of a closed triangle, drawn head to tail, with each side parallel to its force and of length proportional to its magnitude.

The triangle can be solved by scale drawing, by right-angled trigonometry, or by the sine rule:

\frac{F_1}{\sin \alpha_1} = \frac{F_2}{\sin \alpha_2} = \frac{F_3}{\sin \alpha_3}

where $\alpha_1$ is the angle of the triangle opposite the side representing $F_1$ , and so on.

An alternative to the triangle is to resolve: choose two perpendicular directions and set the sum of the components in each direction to zero (see Scalars and vectors). This works for any number of forces. The vector triangle is quicker for three forces and is what Paper 1 often expects you to recognise.

A lamp hanging from two cables

A lamp of weight $50\ \text{N}$ hangs at rest from two cables. One cable makes $30^\circ$ with the horizontal and the other $50^\circ$ with the horizontal, on the opposite side. Find the tension in each cable.

Solution

Three forces act at the knot: the weight $W = 50\ \text{N}$ downwards, $T_1$ (at $30^\circ$ to the horizontal) and $T_2$ (at $50^\circ$ to the horizontal). Draw them head to tail: $W$ straight down, then $T_2$ , then $T_1$ closing the triangle.

The closed vector triangle for the lamp. Each side is parallel to one force; the arrows run head to tail.

Angles in the triangle: $T_2$ is $40^\circ$ from the vertical, $T_1$ is $60^\circ$ from the vertical, so the third angle is $180^\circ - 40^\circ - 60^\circ = 80^\circ$ . The angle opposite $W$ is $80^\circ$ , opposite $T_1$ is $40^\circ$ and opposite $T_2$ is $60^\circ$ .

Sine rule:

\frac{T_1}{\sin 40^\circ} = \frac{T_2}{\sin 60^\circ} = \frac{50}{\sin 80^\circ} = 50.8\ \text{N}

T_1 = 50.8 \times \sin 40^\circ = 32.6\ \text{N}, \qquad T_2 = 50.8 \times \sin 60^\circ = 44.0\ \text{N}

Check by resolving. Horizontally: $T_1\cos 30^\circ = 28.3\ \text{N}$ and $T_2\cos 50^\circ = 28.3\ \text{N}$ , equal. Vertically: $T_1\sin 30^\circ + T_2\sin 50^\circ = 16.3 + 33.7 = 50.0\ \text{N}$ . Correct. The steeper cable carries the larger tension.

Watch out

Leaving out the object's own weight, or putting it at the wrong place. For a uniform object it acts at the midpoint; for a non-uniform one you must be told or find it.
Taking moments about a point and then also including the force at that point. A force through the chosen point has zero moment. Leave it out.
Thinking a hinge force is vertical. A hinge force can have any direction; find both components. Only a smooth surface gives a force known to be perpendicular to it.
Drawing a vector triangle with arrows that do not run head to tail. In a closed triangle for equilibrium, the arrows follow each other round the triangle. If two arrows meet head to head, the triangle shows a resultant, not equilibrium.

Finding the mass of a metre rule by balancing

Aim. Use the principle of moments to find the mass $m$ of a metre rule.

Apparatus. A metre rule; a knife-edge pivot (or a triangular prism); slotted masses of $50\ \text{g}$ to $250\ \text{g}$ on a loop of thread; a balance to check the masses.

Method.

Find the balance point of the rule alone; this is its centre of gravity, close to the $50.0\ \text{cm}$ mark. Record its position $G$ .
Hang a mass $M$ from the thread loop at the $10.0\ \text{cm}$ mark, a distance $0.40\ \text{m}$ from $G$ .
Slide the rule over the pivot until it balances horizontally. Record the distance $x$ from the mass to the pivot.
Repeat for at least five values of $M$ , repeating each balance and averaging $x$ .

Analysis. About the pivot, $Mx = m(0.40 - x)$ . Rearranging,

\frac{1}{x} = \frac{1}{0.40\,m}M + \frac{1}{0.40}

A graph of $1/x$ against $M$ is a straight line with gradient $1/(0.40\,m)$ and intercept $2.5\ \text{m}^{-1}$ . So $m = 1/(0.40 \times \text{gradient})$ , and the intercept is a check. For a typical rule of mass $0.112\ \text{kg}$ , $x$ falls from $27.7\ \text{cm}$ at $M = 50\ \text{g}$ to $12.4\ \text{cm}$ at $M = 250\ \text{g}$ .

Variables. Independent: $M$ . Dependent: $x$ . Controlled: the position of the thread loop; the same rule.

Sources of uncertainty and improvements.

Judging the balance point: the rule tips over a range of positions. Find the positions where it just tips each way and take the midpoint; use half the range as the uncertainty.
The thread loop has width: use a fine thread and read its centre.
The masses may not be exactly as marked: weigh them on a balance.
Small values of $x$ have a large percentage uncertainty: avoid very large $M$ .

Exam tip

"State the principle of moments" needs: for an object in equilibrium; sum of clockwise moments equals sum of anticlockwise moments; about the same point. Missing "in equilibrium" or "about the same point" loses a mark.
"State the two conditions for equilibrium": resultant force is zero and resultant moment (torque) is zero. "The forces are balanced" is too vague.
Examiners repeatedly report candidates taking moments about a point and then including the force acting at that point, or forgetting the object's weight. Make a list of forces before writing the moments equation.
When asked to "show that" a tension is a given value, write the full moments equation with the numbers substituted before rearranging; quote your answer to one more significant figure than the value given.
For hinge or support forces in two dimensions, give the magnitude and the direction (an angle to a stated reference), not just one component.
In Paper 1, "which vector triangle represents the forces?" questions turn on the arrows running head to tail and each side being parallel to its force.

Summary

Principle of moments: for an object in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.
Equilibrium requires both no resultant force and no resultant torque.
Take moments about the point where an unknown force acts, so that force drops out.
Resolve vertically and horizontally for the remaining forces.
A hinge force can act in any direction; find its horizontal and vertical components.
Three forces in equilibrium form a closed vector triangle, head to tail, and their lines of action meet at one point.
Solve a force triangle by scale drawing, right-angled trigonometry or the sine rule, and check by resolving.

Practice

Question

State the principle of moments.
State the two conditions for an object to be in equilibrium, and explain why each one is needed.
A uniform bridge of length $12\ \text{m}$ and weight $4.0 \times 10^{4}\ \text{N}$ rests on supports at its ends. A lorry of weight $1.5 \times 10^{4}\ \text{N}$ is $3.0\ \text{m}$ from one end. Calculate the force on each support.
A uniform metre rule balances horizontally on a pivot at the $60.0\ \text{cm}$ mark when a $0.50\ \text{N}$ weight hangs at the $90.0\ \text{cm}$ mark. Calculate the weight of the rule.
A picture of mass $2.0\ \text{kg}$ hangs from a nail by a string. Each half of the string makes $25^\circ$ with the horizontal. Calculate the tension in the string, and explain why it would be dangerous to make the string tighter so that it is almost horizontal.
A block of weight $40\ \text{N}$ is held at rest on a smooth slope at $25^\circ$ to the horizontal by a force $P$ parallel to the slope. Draw the vector triangle and use it to find $P$ and the normal contact force.
A uniform horizontal boom of length $3.0\ \text{m}$ and weight $400\ \text{N}$ is hinged to a wall. A cable from its free end to the wall makes $30^\circ$ with the boom. A load of $600\ \text{N}$ hangs $2.0\ \text{m}$ from the hinge. Calculate (a) the tension in the cable, (b) the magnitude and direction of the force from the hinge.
A uniform ladder of length $6.0\ \text{m}$ and weight $250\ \text{N}$ rests against a smooth vertical wall at $65^\circ$ to the horizontal ground. A person of weight $700\ \text{N}$ stands $4.0\ \text{m}$ up the ladder. Calculate the force exerted by the wall, and the magnitude of the total force exerted by the ground.
Explain why, if three non-parallel forces keep an object in equilibrium, their lines of action must pass through a single point.

Answers

For an object in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point.
The resultant force must be zero, so that there is no linear acceleration; the resultant torque (moment) about any point must be zero, so that there is no angular acceleration. Both are needed: a couple has zero resultant force but turns the object, and a single force through a point has no moment about it but accelerates the object.
Let the lorry be $3.0\ \text{m}$ from end A. Moments about A: $R_B \times 12 = 4.0 \times 10^{4} \times 6.0 + 1.5 \times 10^{4} \times 3.0 = 2.85 \times 10^{5}$ , so $R_B = 2.4 \times 10^{4}\ \text{N}$ ( $2.375 \times 10^{4}$ ). Vertically: $R_A = 5.5 \times 10^{4} - 2.375 \times 10^{4} = 3.1 \times 10^{4}\ \text{N}$ ( $3.125 \times 10^{4}$ ).
The rule's weight $W$ acts at the $50.0\ \text{cm}$ mark, $0.100\ \text{m}$ from the pivot (anticlockwise, say). The $0.50\ \text{N}$ weight is $0.300\ \text{m}$ on the other side. $W \times 0.100 = 0.50 \times 0.300$ , so $W = 1.5\ \text{N}$ .
Vertically: $2T\sin 25^\circ = 2.0 \times 9.81 = 19.6\ \text{N}$ , so $T = 19.6/(2 \times 0.423) = 23\ \text{N}$ ( $23.2\ \text{N}$ ). As the string approaches horizontal, $\sin\theta \to 0$ and the vertical components can only support the weight if $T$ becomes very large, so the string or nail could break.
The three forces are the weight (vertical), the normal force (perpendicular to the slope) and $P$ (along the slope). The normal force and $P$ are perpendicular, so the triangle is right-angled with the weight as hypotenuse and the angle between the weight and the normal force equal to the slope angle, $25^\circ$ . $P = 40\sin 25^\circ = 16.9\ \text{N}$ ; normal force $= 40\cos 25^\circ = 36.3\ \text{N}$ .
(a) Moments about the hinge: $T\sin 30^\circ \times 3.0 = 400 \times 1.5 + 600 \times 2.0 = 1800\ \text{N m}$ , so $T = 1800/1.5 = 1200\ \text{N}$ . (b) Horizontal: $H = T\cos 30^\circ = 1039\ \text{N}$ away from the wall. Vertical: $V = 1000 - T\sin 30^\circ = 1000 - 600 = 400\ \text{N}$ upwards. Magnitude $= \sqrt{1039^2 + 400^2} = 1110\ \text{N}$ , at $\tan^{-1}(400/1039) = 21^\circ$ above the horizontal.
Moments about the foot. Horizontal distances: weight of ladder $3.0\cos 65^\circ = 1.268\ \text{m}$ ; person $4.0\cos 65^\circ = 1.690\ \text{m}$ . Height of top: $6.0\sin 65^\circ = 5.438\ \text{m}$ . $R \times 5.438 = 250 \times 1.268 + 700 \times 1.690 = 317.0 + 1183.2 = 1500.2$ , so $R = 276\ \text{N}$ . Ground: normal force $= 950\ \text{N}$ , friction $= 276\ \text{N}$ ; total $= \sqrt{950^2 + 276^2} = 989\ \text{N}$ .
Two of the lines of action meet at some point P (they are not parallel). Both of those forces have zero moment about P. The object is in equilibrium, so the resultant moment about P is zero; therefore the third force must also have zero moment about P, which means its line of action passes through P.

The principle of moments

Conditions for equilibrium

Hinges and cables at an angle

Three forces: the vector triangle

Summary

Practice

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