Mechanics Overview – A Level Maths Year 2

Year 1 mechanics (see the Year 1 Mechanics Overview) dealt with particles moving in straight lines: suvat, motion graphs, variable acceleration with calculus, and $F = ma$ along one direction. Year 2 keeps every one of those tools and lifts the restrictions.

Year 2 mechanics is about two dimensions and extended bodies. Forces act at angles, so you resolve. Surfaces are rough, so you model friction. Objects can turn, so you take moments. And motion is no longer along a line, so position, velocity and acceleration become vectors.

What is new compared with Year 1

Moments and rigid bodies · resolving forces at angles · inclined planes
Friction $F \leqslant \mu R$ · ladders · projectiles
suvat and calculus with $\mathbf{i}$, $\mathbf{j}$ vectors · $\mathbf{F} = m\mathbf{a}$ in vector form

DfE content area	Year 1	Year 2 adds
Q · Kinematics	suvat in 1-D; $v = \dfrac{\mathrm{d}s}{\mathrm{d}t}$, $a = \dfrac{\mathrm{d}v}{\mathrm{d}t}$	Vector suvat and calculus; projectiles
R · Forces and Newton's laws	$F = ma$ along a line; connected particles	Resolving; equilibrium; inclined planes; friction
S · Moments	—	Moments; rigid bodies in equilibrium; ladders

Use $g = 9.8$ m s$^{-2}$ unless told otherwise, and give final answers to 2 or 3 significant figures — answers that depend on $g = 9.8$ should not be given to more accuracy than that.

Mechanics questions are full of modelling words. Each one tells you something precise about what to include or ignore.

Word	Meaning in the model
Rigid body / rod	Has size but does not bend; forces at different points cause turning.
Uniform	Weight acts at the geometric centre (the midpoint of a rod).
Light	Mass is negligible — no weight.
Smooth	No friction.
Rough	Friction acts, with $F \leqslant \mu R$.
Limiting equilibrium	At rest, but on the point of moving: $F = \mu R$.
On the point of tilting	The reaction at the support being lifted off is zero.
Moment	Turning effect: force $\times$ perpendicular distance (N m).
Resolve	Find the component of a force in a given direction.
Line of greatest slope	The direction straight up or down an inclined plane.
Projectile	A particle moving freely under gravity alone (no air resistance).
Trajectory	The path of a projectile — a parabola.

1 — Moments (S)

$M = Fd$, $d$ the perpendicular distance. In equilibrium: clockwise moments $=$ anticlockwise moments about any point

2 — Resolving (R)

$F$ at $\theta$ to a direction has component $F\cos\theta$ along it and $F\sin\theta$ perpendicular to it

3 — Friction (R)

$F \leqslant \mu R$, with equality when moving or in limiting equilibrium

4 — Projectiles (Q)

Horizontal: $x = (u\cos\theta)t$ Vertical: $y = (u\sin\theta)t - \tfrac12gt^2$

5 — Vectors and calculus (Q, R)

$\mathbf{v} = \dot{\mathbf{r}}$, $\mathbf{a} = \dot{\mathbf{v}}$ constant $\mathbf{a}$: $\mathbf{v} = \mathbf{u} + \mathbf{a}t$, $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \tfrac12\mathbf{a}t^2$ $\mathbf{F} = m\mathbf{a}$

Worked Example — Several techniques in one question

A particle of mass $2$ kg is released from rest on a rough plane inclined at $30^\circ$, with $\mu = 0.4$. Show that it slides, and find its speed after it has moved $2$ m.

①Resolve perpendicular: $R = 2g\cos 30^\circ = 16.97$ N.

②Friction: $F_{\max} = 0.4 \times 16.97 = 6.79$ N.

③Component of weight down the slope: $2g\sin 30^\circ = 9.8$ N $> 6.79$ N, so it slides.

④$F = ma$ down the slope: $9.8 - 6.79 = 2a \Rightarrow a = 1.505$ m s$^{-2}$.

⑤suvat: $v^2 = 0 + 2(1.505)(2) = 6.02$, so $v = 2.45$ m s$^{-1}$.

Equivalently: a particle slides on a rough slope when $\tan\alpha > \mu$. Here $\tan 30^\circ = 0.577 > 0.4$ ✓

The Year 2 mechanics content, and the page that covers it. The code is the DfE content area.

SMomentsThe moment of a force, equilibrium of rigid bodies, beams on supports, tilting and non-uniform rods.
RResolving Forces and Inclined PlanesComponents and resultants, equilibrium of a particle, smooth slopes and pulleys.
R
SFriction$F \leqslant \mu R$ on horizontal and inclined planes, connected particles, and ladders.
QProjectilesTime of flight, range, maximum height, the trajectory equation and projection from a height.
Q
RKinematics with Calculus and VectorsVector suvat, differentiating and integrating $\mathbf{i}$, $\mathbf{j}$ components, and $\mathbf{F} = m\mathbf{a}$.

Assuming $R = mg$. On a slope $R = mg\cos\alpha$, and any force pulling at an angle changes $R$ further. Always find $R$ by resolving perpendicular to the surface.

Assuming friction is always $\mu R$. $\mu R$ is the maximum. If the body is at rest and not about to move, friction is whatever balances the other forces.

Using a distance that is not perpendicular. A moment uses the perpendicular distance to the line of action. For a ladder at angle $\theta$, a horizontal force at the top has moment arm $L\sin\theta$ about the foot, not $L$.

Mixing horizontal and vertical in projectiles. Gravity only affects the vertical motion. The horizontal velocity stays at $u\cos\theta$ throughout the flight.

Using level-ground formulae from a height. $R = \dfrac{u^2\sin2\theta}{g}$ is only valid when launch and landing heights are equal.

Forgetting the vector constant of integration. When integrating $\mathbf{a}$ or $\mathbf{v}$, the constant is $c_1\mathbf{i} + c_2\mathbf{j}$, found from the initial conditions.

Moment

$Fd$, perpendicular distance, N m.

Rigid body equilibrium

Resultant force $0$ and resultant moment $0$.

About to tilt

Reaction at the other support $= 0$.

Slope

$mg\sin\alpha$ down, $mg\cos\alpha$ into the plane.

Friction

$F \leqslant \mu R$; about to slip on a slope: $\mu = \tan\alpha$.

Ladder (smooth wall)

$\mu \geqslant \dfrac{1}{2\tan\theta}$ for a uniform ladder alone.

Projectile range

$\dfrac{u^2\sin2\theta}{g}$ on level ground.

Trajectory

$y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$.

Vectors

Speed $= |\mathbf{v}|$; calculus on each component.

Ten questions drawing on the whole of Year 2 mechanics. Use $g = 9.8$ m s$^{-2}$ and give answers to 3 significant figures where they are not exact.

Question 1

Find the moment about a hinge of a $15$ N force acting perpendicular to a door at $0.8$ m from the hinge.

▶ Show solution

$M = 15 \times 0.8 = 12$ N m.

Question 2

Resolve a $25$ N force acting at $30^\circ$ above the horizontal into horizontal and vertical components.

▶ Show solution

Horizontal $25\cos 30^\circ = 21.7$ N; vertical $25\sin 30^\circ = 12.5$ N.

Question 3

A $5$ kg box on rough horizontal ground ($\mu = 0.2$) is pushed by a horizontal force of $20$ N. Find its acceleration.

▶ Show solution

$F_{\max} = 0.2 \times 5g = 9.8$ N $< 20$ N, so it moves.

$20 - 9.8 = 5a \Rightarrow a = 2.04$ m s$^{-2}$.

Question 4

A uniform rod $AB$ of length $3$ m and mass $10$ kg rests on supports at $A$ and $B$. A $5$ kg mass is placed $1$ m from $A$. Find the reactions.

▶ Show solution

Moments about $A$: $3R_B = 10g(1.5) + 5g(1) = 20g \Rightarrow R_B = \dfrac{20g}{3} = 65.3$ N.

$R_A = 15g - \dfrac{20g}{3} = \dfrac{25g}{3} = 81.7$ N.

Question 5

Find the acceleration of a particle sliding down a smooth plane inclined at $20^\circ$.

▶ Show solution

$a = g\sin 20^\circ = 3.35$ m s$^{-2}$.

Question 6

A ball is thrown horizontally at $10$ m s$^{-1}$ from a point $45$ m above level ground. Find the time of flight and the horizontal distance travelled.

▶ Show solution

$45 = 4.9t^2 \Rightarrow t = 3.03$ s.

$x = 10 \times 3.030 = 30.3$ m.

Question 7

A particle has initial velocity $(2\mathbf{i} + 3\mathbf{j})$ m s$^{-1}$ and constant acceleration $(\mathbf{i} - 2\mathbf{j})$ m s$^{-2}$. Find its velocity after $2$ s.

▶ Show solution

$\mathbf{v} = (2\mathbf{i} + 3\mathbf{j}) + 2(\mathbf{i} - 2\mathbf{j}) = 4\mathbf{i} - \mathbf{j}$ m s$^{-1}$.

Question 8

A particle has position $\mathbf{r} = t^3\,\mathbf{i} + (2t^2 - t)\,\mathbf{j}$. Find its speed and direction of motion when $t = 1$.

▶ Show solution

$\mathbf{v} = 3t^2\,\mathbf{i} + (4t - 1)\mathbf{j}$. At $t = 1$: $\mathbf{v} = 3\mathbf{i} + 3\mathbf{j}$.

Speed $= 3\sqrt2 = 4.24$ m s$^{-1}$, moving at $45^\circ$ to $\mathbf{i}$.

Question 9

A uniform ladder rests against a smooth vertical wall with its foot on rough horizontal ground, making $65^\circ$ with the ground. Find the least possible value of $\mu$.

▶ Show solution

Let the ladder have length $2a$ and weight $W$. Resolving: $R = W$, $F = S$.

Moments about the foot: $S \times 2a\sin 65^\circ = W \times a\cos 65^\circ \Rightarrow S = \dfrac{W}{2\tan 65^\circ}$.

$F \leqslant \mu R \Rightarrow \mu \geqslant \dfrac{1}{2\tan 65^\circ} = 0.233$.

Question 10

A ball is kicked from level ground at $24.5$ m s$^{-1}$ at $30^\circ$ above the horizontal.

(a) Find the time of flight. (b) Find the range. (c) Find the greatest height. (d) Find the length of time for which the ball is at least $5$ m above the ground.

▶ Show solution

$u_x = 24.5\cos 30^\circ = 21.22$, $u_y = 24.5\sin 30^\circ = 12.25$.

(a) $0 = 12.25t - 4.9t^2 \Rightarrow t = \dfrac{12.25}{4.9} = 2.5$ s.

(b) $21.22 \times 2.5 = 53.0$ m.

(c) $H = \dfrac{12.25^2}{2 \times 9.8} = 7.66$ m.

(d) $5 = 12.25t - 4.9t^2 \Rightarrow 4.9t^2 - 12.25t + 5 = 0$.

$t = \dfrac{12.25 \pm \sqrt{150.06 - 98}}{9.8} = \dfrac{12.25 \pm 7.215}{9.8}$, so $t = 0.514$ s or $1.99$ s.

Time at least $5$ m up $= 1.986 - 0.514 = 1.47$ s.