Forces and acceleration
When a single unbalanced force acts on a particle, or when the resultant of several forces acting on a particle is non-zero, the particle accelerates in the direction of that resultant force. This is Newton's Second Law:
\[\mathbf{F}_{\text{resultant}} = m\mathbf{a}\]Here \(m\) is the mass of the particle (in kg), \(\mathbf{a}\) is its acceleration (in m s\(^{-2}\)), and \(\mathbf{F}_{\text{resultant}}\) is the vector sum of all forces acting (in N). If the resultant force is zero, the particle is in equilibrium — it either remains at rest or moves with constant velocity.
Balanced forces — static equilibrium
An object that is stationary has zero acceleration, so the resultant of all forces acting on it must be zero. We consider equilibrium in two directions separately.
A block rests on a horizontal surface. Its weight \(mg\) acts downward; the normal reaction \(R\) acts upward. Since the block is stationary: \(R - mg = 0\), so \(R = mg\).
A horizontal pushing force \(P\) is applied to a stationary block. Friction \(F\) acts in the opposite direction. Horizontal equilibrium gives: \(P - F = 0\), so \(F = P\).
Why do turning effects matter?
The three situations below cannot be fully explained by force balance alone — they involve the turning effect (moment) of a force. Each one raises a natural question for you to think about.
The door handle The hinge is on the left; the handle is on the far right. Why is the handle placed as far as possible from the hinge?
The wheelbarrow The wheel acts as a pivot. The load sits close to the wheel; the handles extend well behind it. Why are the arms of the wheelbarrow so long?
The spanner A force \(F\) is applied at the end of the handle, a distance \(d\) from the nut. Why is it easier to turn the nut when the handle is longer? Why is the pulling force best applied at right angles to the spanner?
All three questions share the same answer: the further a force is applied from the pivot point, the larger the moment (turning effect) it produces. The formal study of moments and their role in equilibrium is the subject of the pages below.
Choose a topic to explore:
What is a particle?
In mechanics we often simplify a real object into a particle: a point mass that has mass but no physical size. All forces on a particle act through a single point, so there is no possibility of rotation — the only condition for equilibrium is that the resultant force is zero.
What is a rigid rod?
A rigid rod (or lamina, plank, beam or ladder) is a body with a definite length and shape that does not change. Unlike a particle, forces can act at different points along the rod. This means the rod can:
- Translate (move as a whole) if the resultant force is non-zero.
- Rotate about a point (pivot) if the resultant moment is non-zero.
For a rigid rod to be in full equilibrium, both conditions must hold:
The moment of a force
The moment of a force about a point (the pivot or fulcrum) measures its turning effect. For a force \(F\) acting perpendicular to the line joining the pivot to the point of application:
\[\text{Moment} = F \times d\]where \(d\) is the perpendicular distance from the pivot to the line of action of the force. The unit is the newton-metre (N m).
A moment is clockwise (CW) or anticlockwise (ACW) according to the direction it would cause rotation. Anticlockwise can be taken as positive or negative, then clockwise has the opposite sign. Once you have chosen for your question, apply the sign consistently to all moments.
For the rod above to be in rotational equilibrium about the pivot: \(F_1 d_1 = F_2 d_2\).
If the force or distance increases on one side of the equation then the moment increases and the rod will be pulled down on the side that has the increased force or increased distance from the pivot.
Worked Example — Seesaw
Two children sit on a uniform seesaw of length 3.2 m pivoted at its midpoint. Child A (mass 40 kg) sits 1.6 m from the pivot. Child B (mass 50 kg) sits at a distance \(x\) m from the pivot on the other side. Find \(x\) for the seesaw to be in equilibrium (take \(g = 9.8\) m s\(^{-2}\)).
Direction of tension
When two people pull a spring apart, they feel the spring trying to pull their hands back together. So tension is pulling both ends of the spring towards the centre.
Modelling a string
In mechanics, a string (or wire, rope, cable) is modelled as light (having negligible mass) and inextensible (it does not stretch). The force exerted by a string is called its tension, and it always acts along the string, away from the object it is attached to — strings can only pull, never push.
| Assumption | What it means |
|---|---|
| Light | The weight of the string itself is ignored |
| Inextensible | The string does not stretch; its length is constant |
| Taut | If a string is taut it transmits tension; if slack, tension is zero |
Simplifying the force diagram
An (ideal) string is inextensible. However, the tension behaves the same in a spring or string, in that the forces pull towards the centre.
Even though the tension from a spring or string acts inwards from both ends, we are normally only interested in the effect the tension has on one object, say a box being pulled. Often the force diagram will only show one tension, acting on the box, but still pulling to the centre of the string. This is to simplify the diagram and concentrate on one object at a time and only the forces acting on that one object.
Slack strings and horizontal tension
A string that is slack (loose, not pulled tight) has zero tension — it exerts no force on the object it is attached to. Only when a string is taut (pulled straight and tight) does it transmit a tension force.
A force applied to a taut (fully pulled and straight) string causes the string to transmit the point of application of the force to the end of the string attached to the object being pulled.
Why do we do this? Because when analyzing a situation where we draw a force diagram, we want to consider each object separately and in turn. We need to consider all, and only, the forces acting directly on the object of interest. We can think of a string as a device for transferring a remote acting force to a force acting directly on the object we are interested in.
Tension in a string supporting a hanging object
If an object of mass \(m\) hangs in equilibrium from a single vertical string, the tension \(T\) in the string must equal the weight of the object:
\[T = mg\]If the object is supported by two strings at different angles, we resolve forces horizontally and vertically and solve the resulting equations simultaneously.
Resolving horizontally: \(\;T_1 \cos\alpha = T_2 \cos\beta\)
Resolving vertically: \(\;T_1 \sin\alpha + T_2 \sin\beta = mg\)
Worked Example — Rod supported by two strings
A uniform rod AB of length 4 m and mass 6 kg is suspended horizontally by two vertical strings. String 1 is attached 0.5 m from A; String 2 is attached 0.5 m from B. Find the tension in each string.
Worked Example — Rod supported by two asymmetric strings
A uniform rod AB of length 4 m and mass 6 kg is suspended horizontally by two vertical strings. String 1 is attached 1 m from A; String 2 is attached 0.5 m from B. Find the tension in each string.
Tension versus thrust
A light rod is a two-force member: it can only exert a force along its own length. The direction of that force depends on whether the rod is being stretched or compressed.
| Condition | Force in rod | Rod is... |
|---|---|---|
| Rod being pulled apart | Tension \(T\) (pulling inward) | in tension |
| Rod being pushed together | Thrust \(C\) (pushing outward) | in compression (thrust) |
A string can only be in tension (it goes slack if compressed). A rod can sustain both tension and thrust (compression), which is why rods are used in structural frameworks.
Rods in frameworks
A practical example of rods in a framework is a three-legged stool. Each leg is a light rod that can carry a compressive force (thrust). The seat is supported by three legs; the weight of the seat (and anyone sitting on it) must be balanced by the upward forces that the legs exert on the seat.
The diagram below shows the forces acting on the seat of the stool. The three legs push upward on the seat with equal forces \(S\), and the weight \(W\) of the seat acts downward through its centre.
For the seat to be in vertical equilibrium: \[3S = W\] so each leg carries one-third of the total weight. Each leg is in thrust (compression) — the seat pushes down on the leg, and the leg pushes back up on the seat.
Worked Example — Strut and tie
A light rod AB is hinged to a vertical wall at A. A light rod BC connects B to a point C on the wall directly above A, with BC making an angle of 30° to the horizontal. A load of 200 N hangs vertically from B. Find the thrust in AB and the tension in BC. Draw a diagram showing the forces acting at B.
The uniform rod
A uniform rod has its mass distributed evenly along its length. This means its centre of gravity (and thus the point through which its weight acts) is at the midpoint of the rod.
A rod of length \(2l\) and mass \(m\) has its weight \(mg\) acting downward through the point at distance \(l\) from each end.
Horizontal uniform bar on two supports
Consider a uniform horizontal bar of mass \(m\) and length \(L\), supported by two vertical strings (or ropes/wires) attached at distances \(a\) and \(b\) from the left-hand end, where \(a < b\).
With the bar in equilibrium, we write two equations:
\[\text{(↑) } \quad T_1 + T_2 = mg\] \[\text{(Moments about A)} \quad T_1 \cdot a + T_2 \cdot b = mg \cdot \tfrac{L}{2}\]It is most efficient to take moments about the point where one tension acts, to find the other directly.
Worked Example — Non-uniform bar
A non-uniform plank AB of length 5 m and mass 8 kg is supported by two strings attached at A and at a point C, where C is 3 m from A. The plank rests horizontally. The tension in the string at A is twice the tension in the string at C. Find the distance of the centre of gravity, CG, from A.
What is a hinge?
A hinge (or pin joint) attaches a rod to a fixed support (such as a wall) in a way that allows free rotation. This means:
- A hinge can exert a reaction force in any direction — not just perpendicular to the rod.
- A hinge cannot exert a moment (couple) — it offers no rotational resistance.
Because the direction of the hinge reaction is unknown, we represent it by two components: a horizontal component \(H\) and a vertical component \(V\). Solving the equilibrium equations determines both \(H\) and \(V\), from which the magnitude and direction of the reaction can be found.
The hinge at the wall provides components \(H\) (horizontal) and \(V\) (vertical). These are found by applying \(\sum F_x = 0\), \(\sum F_y = 0\), and \(\sum M = 0\).
Worked Example — Hinged rod with a string
A uniform horizontal rod AB of length 2 m and mass 5 kg is hinged to a wall at A. The rod is held horizontal by a light string attached at B making an angle of 25° to the rod and directed toward the wall. Find the tension in the string and the reaction at the hinge.
The ladder problem
A ladder leaning against a wall is one of the most common and important applications of moments and equilibrium in mechanics. The typical set-up involves a uniform ladder of length \(2l\) (or \(L\)) and mass \(m\), leaning against a smooth vertical wall and resting on a rough horizontal floor, making an angle \(\theta\) with the floor.
The four forces acting on the ladder are:
| Force | Direction | Point of action |
|---|---|---|
| Weight \(mg\) | Vertically downward | Midpoint of ladder |
| Normal reaction \(R\) | Vertically upward | Base A (floor) |
| Friction \(F\) | Horizontal (toward wall) | Base A (floor) |
| Normal reaction \(S\) | Horizontal (away from wall) | Top B (wall — smooth) |
The wall is smooth, so it can only push the ladder horizontally — no friction at B. The floor is rough, providing both \(R\) (vertical) and \(F\) (horizontal friction) at A.
Equations of equilibrium for a ladder
Taking the length of the ladder as \(2l\) and the angle with the floor as \(\theta\):
\[\text{(↑)} \quad R = mg\] \[\text{(→)} \quad F = S\] \[\text{(Moments about A)} \quad S \cdot 2l\sin\theta = mg \cdot l\cos\theta\]Simplifying the moments equation:
\[S = \frac{mg\cos\theta}{2\sin\theta} = \frac{mg}{2}\cot\theta\]Since at the point of slipping \(F = \mu R\) (where \(\mu\) is the coefficient of friction), and \(F = S\), the minimum angle at which the ladder can rest without slipping satisfies:
\[\mu R = \frac{mg}{2}\cot\theta \implies \mu mg = \frac{mg}{2}\cot\theta \implies \cot\theta = 2\mu \implies \tan\theta = \frac{1}{2\mu}\]Worked Example — Ladder with a person climbing
A uniform ladder of length 6 m and mass 20 kg leans against a smooth vertical wall on rough horizontal ground, making an angle of 70° with the horizontal. A person of mass 60 kg climbs the ladder. Find how far up the ladder the person can climb before the ladder slips, given that the coefficient of friction at the ground is 0.35.