Forces and Newton's Laws – A Level Maths Year 1

1 Forces and Force Diagrams

Kinematics describes motion; dynamics explains it. A force is a push or pull, measured in newtons (N), and it is a vector (DfE content area R · Forces and Newton's laws).

Force	Symbol	Where it acts and which way
Weight	$W = mg$	Vertically downwards, from the centre of mass.
Normal reaction	$R$	Perpendicular to a surface, pushing the object away from it.
Tension	$T$	Along a taut string or rod, pulling away from the object.
Thrust (compression)	$T$	Along a rod, pushing towards the object.
Friction	$F$	Along a rough surface, opposing motion (or attempted motion).
Driving force / resistance	$D$, $R$	Forward force from an engine; resistance opposes motion.

A force diagram shows the object (as a dot or a box) and every force acting on it, each as a labelled arrow. Show the acceleration separately with a double-headed arrow. Never include forces that the object exerts on other things.

Resolve in a direction: add the components of all forces in that direction, taking one way as positive. For now, every force is horizontal or vertical (or given in $\mathbf{i}$, $\mathbf{j}$ form), so there is no trigonometry — that comes in Year 2.

2 Newton's Three Laws

Newton's laws of motion

1st: A body stays at rest or moves with constant velocity unless a resultant force acts on it.
2nd: Resultant force $=$ mass × acceleration: $\mathbf{F} = m\mathbf{a}$
3rd: If body $A$ exerts a force on body $B$, then $B$ exerts an equal and opposite force on $A$.

Equilibrium means the resultant force is zero. By the first law, a particle in equilibrium is either at rest or moving with constant velocity — both count.

Worked Example 1 — F = ma in one dimension

A box of mass $4$ kg is pushed across a floor by a horizontal force of $30$ N against a resistance of $12$ N. Find its acceleration and the normal reaction.

①Horizontally ($\rightarrow$ positive): $30 - 12 = 4a \Rightarrow a = 4.5$ m s$^{-2}$.

②Vertically, no acceleration: $R - 4g = 0 \Rightarrow R = 39.2$ N.

Worked Example 2 — A particle on a string

A particle of mass $2$ kg hangs on a light string and is pulled upwards with tension $26$ N. Find its acceleration.

①Up positive: $26 - 2g = 2a$.

②$26 - 19.6 = 6.4 = 2a \Rightarrow a = 3.2$ m s$^{-2}$ upwards.

Weight and normal reaction are not a third-law pair. Both act on the same object. The third-law partner of the box's weight is the box's gravitational pull on the Earth; the partner of the normal reaction is the box pushing down on the floor.

3 Forces as Vectors

When forces are given in $\mathbf{i}$, $\mathbf{j}$ form, add them component by component (see Vectors in Two Dimensions). Newton's second law applies to each component separately.

In two dimensions

$\mathbf{F}_1 + \mathbf{F}_2 + \cdots = m\mathbf{a}$; equilibrium $\iff$ $\sum \mathbf{F} = \mathbf{0}$

Worked Example 3 — Acceleration from vector forces

Forces $(3\mathbf{i} + 4\mathbf{j})$ N and $(5\mathbf{i} - 10\mathbf{j})$ N act on a particle of mass $2$ kg. Find its acceleration and the magnitude of the acceleration.

①Resultant $= (8\mathbf{i} - 6\mathbf{j})$ N.

②$\mathbf{a} = \dfrac{1}{2}(8\mathbf{i} - 6\mathbf{j}) = (4\mathbf{i} - 3\mathbf{j})$ m s$^{-2}$.

③$|\mathbf{a}| = \sqrt{16 + 9} = 5$ m s$^{-2}$.

Worked Example 4 — Equilibrium

Forces $(2\mathbf{i} - 3\mathbf{j})$, $(-5\mathbf{i} + q\mathbf{j})$ and $(p\mathbf{i} + 7\mathbf{j})$ N act on a particle in equilibrium. Find $p$ and $q$.

①$\mathbf{i}$: $2 - 5 + p = 0 \Rightarrow p = 3$.

②$\mathbf{j}$: $-3 + q + 7 = 0 \Rightarrow q = -4$.

4 Connected Particles: Lifts

For connected objects that move together, you can apply $F = ma$ to the whole system (internal forces cancel) or to each part separately (internal forces appear). Use whichever gives the unknown you want most directly — and use the other as a check.

Worked Example 5 — Person in a lift

A lift of mass $430$ kg carries a man of mass $70$ kg. It accelerates upwards at $0.5$ m s$^{-2}$. Find the tension in the lift cable and the normal reaction of the floor on the man.

①Whole system ($500$ kg), up positive: $T - 500g = 500(0.5)$.

②$T = 4900 + 250 = 5150$ N.

③Man alone: $R - 70g = 70(0.5) \Rightarrow R = 686 + 35 = 721$ N.

$R$ is greater than his weight ($686$ N): he feels heavier while the lift accelerates upwards. When it decelerates on the way up, $R$ is less than $686$ N.

5 Connected Particles: Cars and Trailers

A car and trailer are joined by a light, rigid towbar. When the car pulls, the towbar is in tension; when the car brakes, the trailer pushes on it and the towbar is in thrust.

Worked Example 6 — Tension in a towbar

A car of mass $1200$ kg tows a trailer of mass $400$ kg along a straight horizontal road. The engine's driving force is $2500$ N; resistances are $300$ N on the car and $200$ N on the trailer. Find the acceleration and the tension in the towbar.

①Whole system: $2500 - 300 - 200 = 1600a \Rightarrow a = 1.25$ m s$^{-2}$.

②Trailer alone: $T - 200 = 400(1.25) \Rightarrow T = 700$ N.

③Check with the car: $2500 - 300 - 700 = 1500 = 1200 \times 1.25$ ✓

6 Connected Particles: Pulleys

For particles joined by a light inextensible string over a smooth pulley: the tension is the same on both sides, and both particles have the same magnitude of acceleration. Write an equation for each particle, in its own direction of motion.

Worked Example 7 — Two particles over a pulley

Particles of mass $5$ kg and $3$ kg hang on a light inextensible string over a smooth pulley and are released from rest. Find the acceleration and the tension.

①$5$ kg (down positive): $5g - T = 5a$.

②$3$ kg (up positive): $T - 3g = 3a$.

③Add: $2g = 8a \Rightarrow a = \dfrac{2g}{8} = 2.45$ m s$^{-2}$.

④$T = 3(g + a) = 3(12.25) = 36.75 \approx 36.8$ N.

The force on the pulley from the string is $2T = 73.5$ N, acting downwards — a common follow-up question.

Worked Example 8 — Table and pulley

A block of mass $4$ kg on a smooth horizontal table is joined by a light inextensible string over a smooth pulley at the table's edge to a particle of mass $1$ kg hanging freely. Find the acceleration and tension.

①Block (horizontal): $T = 4a$.

②Hanging particle (down positive): $g - T = a$.

③Add: $g = 5a \Rightarrow a = 1.96$ m s$^{-2}$; $T = 4(1.96) = 7.84$ N.

Do not treat a pulley system as one body with $F = ma$ in a single direction unless you are careful: the weights act in different directions relative to the motion. Writing one equation per particle is always safe.

7 Quick Reference

Newton's 2nd law

Resultant force $= ma$, in the direction of the acceleration.

Weight

$W = mg$, vertically downwards.

Equilibrium

$\sum \mathbf{F} = \mathbf{0}$: at rest or constant velocity.

Vectors

Add $\mathbf{i}$ and $\mathbf{j}$ components separately; $\mathbf{F} = m\mathbf{a}$.

3rd law

Equal and opposite forces on different bodies.

Lift

$R - mg = ma$ (up positive).

Towbar

Tension when pulling; thrust when braking.

Smooth pulley

Same $T$ both sides; same $|a|$; one equation per particle.

Whole system

Internal forces cancel: use it to find $a$ quickly.

8 Practice Questions

Take $g = 9.8$ m s$^{-2}$. Draw a force diagram for every question.

Question 1

A crate of mass $6$ kg is pulled along a horizontal floor by a horizontal force of $40$ N against a resistance of $16$ N. Find its acceleration.

▶ Show solution

$40 - 16 = 6a \Rightarrow a = 4$ m s$^{-2}$.

Question 2

A book of mass $3$ kg rests on a horizontal table. Draw a force diagram and find the normal reaction. State the third-law pair of the normal reaction.

▶ Show solution

Forces on the book: weight $3g$ down, normal reaction $R$ up. Equilibrium: $R = 3g = 29.4$ N.

Its third-law pair is the force of $29.4$ N exerted by the book on the table, acting downwards.

Question 3

Forces $(2\mathbf{i} + \mathbf{j})$ N and $(\mathbf{i} - 5\mathbf{j})$ N act on a particle of mass $0.5$ kg. Find the acceleration and its magnitude.

▶ Show solution

Resultant $= (3\mathbf{i} - 4\mathbf{j})$ N.

$\mathbf{a} = \dfrac{3\mathbf{i} - 4\mathbf{j}}{0.5} = (6\mathbf{i} - 8\mathbf{j})$ m s$^{-2}$; $|\mathbf{a}| = \sqrt{36 + 64} = 10$ m s$^{-2}$.

Question 4

A particle is in equilibrium under forces $(a\mathbf{i} + 2\mathbf{j})$ N, $(4\mathbf{i} - b\mathbf{j})$ N and $(-7\mathbf{i} + 5\mathbf{j})$ N. Find $a$ and $b$.

▶ Show solution

$\mathbf{i}$: $a + 4 - 7 = 0 \Rightarrow a = 3$.

$\mathbf{j}$: $2 - b + 5 = 0 \Rightarrow b = 7$.

Question 5

A woman of mass $60$ kg stands in a lift that is moving upwards and decelerating at $1.2$ m s$^{-2}$. Find the normal reaction of the floor on her.

▶ Show solution

Up positive; decelerating while moving up means $a = -1.2$.

$R - 60g = 60(-1.2) \Rightarrow R = 588 - 72 = 516$ N.

Question 6

A car of mass $1000$ kg tows a trailer of mass $500$ kg. The car brakes with a braking force of $3000$ N acting on the car only; other resistances are negligible. Find the deceleration and the force in the towbar, stating whether it is a tension or a thrust.

▶ Show solution

Whole system: $-3000 = 1500a \Rightarrow a = -2$, a deceleration of $2$ m s$^{-2}$.

Trailer alone: the only horizontal force is from the towbar, $F = 500(-2) = -1000$ N.

The towbar pushes backwards on the trailer with $1000$ N: it is a thrust.

Question 7

Particles of mass $4$ kg and $6$ kg hang on a light inextensible string over a smooth fixed pulley. Find the acceleration, the tension, and the force exerted on the pulley by the string.

▶ Show solution

$6g - T = 6a$ and $T - 4g = 4a$. Adding: $2g = 10a \Rightarrow a = 1.96$ m s$^{-2}$.

$T = 4(g + a) = 4(11.76) = 47.04 \approx 47.0$ N.

Force on pulley $= 2T = 94.1$ N vertically downwards.

Question 8

A block of mass $3$ kg lies on a rough horizontal table, where a constant resistance of $5$ N opposes its motion. It is connected by a light inextensible string over a smooth pulley at the table's edge to a particle of mass $2$ kg hanging freely. Find the acceleration and the tension.

▶ Show solution

Block: $T - 5 = 3a$. Hanging particle: $2g - T = 2a$.

Add: $2g - 5 = 5a \Rightarrow 14.6 = 5a \Rightarrow a = 2.92$ m s$^{-2}$.

$T = 2(g - a) = 2(6.88) = 13.76 \approx 13.8$ N. Check: $3(2.92) + 5 = 13.76$ ✓

Question 9

A particle of mass $2$ kg, initially at rest, is acted on by a constant force $(6\mathbf{i} + 8\mathbf{j})$ N and no other force. Find its speed after $3$ s.

▶ Show solution

$\mathbf{a} = (3\mathbf{i} + 4\mathbf{j})$ m s$^{-2}$, constant.

$\mathbf{v} = \mathbf{u} + \mathbf{a}t = \mathbf{0} + 3(3\mathbf{i} + 4\mathbf{j}) = (9\mathbf{i} + 12\mathbf{j})$ m s$^{-1}$.

Speed $= \sqrt{81 + 144} = 15$ m s$^{-1}$.

Question 10

Particles $A$ (mass $3$ kg) and $B$ (mass $2$ kg) are connected by a light inextensible string over a smooth fixed pulley. They are released from rest with the string taut, $A$ at a height of $1.5$ m above the floor and $B$ well below the pulley.

(a) Find the acceleration and the tension. (b) Find the speed of $A$ when it hits the floor. (c) $A$ does not rebound. Find the further distance $B$ rises before coming instantaneously to rest. (d) State where in your solution you used the fact that the string is inextensible.

▶ Show solution

(a) $A$: $3g - T = 3a$. $B$: $T - 2g = 2a$. Adding: $g = 5a \Rightarrow a = 1.96$ m s$^{-2}$.

$T = 2(g + a) = 2(11.76) = 23.52 \approx 23.5$ N.

(b) $v^2 = 0 + 2(1.96)(1.5) = 5.88 \Rightarrow v = 2.42$ m s$^{-1}$ (3 s.f.).

(c) When $A$ lands the string goes slack, so $B$ moves freely under gravity with $u = 2.42$ upwards: $0 = 5.88 - 2(9.8)s \Rightarrow s = 0.3$ m.

(d) In (a), taking the same magnitude of acceleration for $A$ and $B$ (and in (b)–(c), the same speed for both when $A$ lands).

Forces and Newton's Laws · A Level Maths Year 1 · Created with MathJax