Notes · PhysicsUK · A-Levels

Mechanics and materials

This chapter builds classical mechanics from vectors and forces up to momentum, work and energy, and then applies it to the mechanical properties of solids. It derives the constant-acceleration equations from motion graphs, states and uses Newton's laws, applies conservation of momentum and energy to collisions and machines, and defines the Young modulus and stress-strain behaviour that engineers use to choose materials.

6 sections~18 min reading time3 competenciesLevel Foundation 1 · Standard 4 · Advanced 1

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Exam profile

AO1 · State Newton's laws, define the mechanical quantities and describe elastic and plastic material behaviourAO2 · Resolve forces, use the suvat equations, apply conservation of momentum and energy, and calculate the Young modulus and strain energyAO3 · Interpret motion graphs and stress-strain curves and evaluate the Young-modulus and terminal-velocity required practicals

Operators:stateresolvecalculatederiveexplaindetermine

basic level

AS-Level requires vectors and equilibrium, moments, the suvat equations and projectiles, Newton's laws, momentum, work-energy-power and the Young modulus.

higher level

The full A-Level extends momentum to impulse from force-time graphs, treats elastic and inelastic collisions quantitatively, and links stress-strain behaviour to strain energy and material selection.

Depth

Text

Contents · 6 sections

Mechanics and materials

Vectors, resolution and equilibrium#

FoundationAQA 7408 3.4.1.1DfE GCE Physics - scalars and vectors

Resolving a vector into components

Fig. 1A force

F

at angle

\theta

to the horizontal has a horizontal component

F\cos\theta

and a vertical component

F\sin\theta

; the components and

F

form a right-angled triangle.

A scalar has magnitude only (mass, time, energy, speed); a vector has magnitude and direction (displacement, velocity, acceleration, force). Vectors add by the triangle or parallelogram rule, placing them nose-to-tail; the single vector from the start of the first to the end of the last is the resultant. Two perpendicular vectors of magnitudes

a

and

b

have a resultant of magnitude

\sqrt{a^2 + b^2}

at an angle

\tan^{-1}(b/a)

to the first.

The reverse process, resolution, splits one vector into two perpendicular components, usually horizontal and vertical. A force

F

at angle

\theta

to the horizontal has a horizontal component

F\cos\theta

and a vertical component

F\sin\theta

. Resolving is the key to nearly every mechanics problem because the components act independently and can be treated separately.

An object is in equilibrium when the resultant force on it is zero (and, for an extended body, the resultant moment is also zero). For forces this means the components in any two perpendicular directions each sum to zero. Equivalently, three forces in equilibrium form a closed triangle when drawn nose-to-tail - a powerful graphical test.

On an inclined plane it is easiest to resolve along and perpendicular to the slope rather than horizontally and vertically. For a block of weight

mg

on a slope at angle

\theta

, the component of weight down the slope is

mg\sin\theta

and the component pressing into the slope is

mg\cos\theta

; the normal reaction balances the latter, and friction (or an applied force) balances the former if the block is in equilibrium.

F_x = F\cos\theta, \qquad F_y = F\sin\theta

Resolving a force

Horizontal and vertical components of a force at angle $\theta$ to the horizontal.

R = \sqrt{F_x^2 + F_y^2}, \qquad \theta = \tan^{-1}\!\dfrac{F_y}{F_x}

Resultant of perpendicular components

Magnitude and direction of the resultant.

Worked example

Resultant of two forces

Two forces act at a point: $6.0\,\text{N}$ due east and $8.0\,\text{N}$ due north. Find the magnitude and direction of the resultant.

The forces are perpendicular
East and north are at right angles, so the resultant is the hypotenuse of a right triangle.
Magnitude
$R = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,\text{N}$ .
Direction
Angle north of east: $\tan^{-1}(8.0/6.0)$ .
$\theta = \tan^{-1}\!\left(\dfrac{8.0}{6.0}\right) = 53^{\circ}$

Result: The resultant is $10\,\text{N}$ at $53^{\circ}$ north of east.

Typical mistakes

Swapping sine and cosine when resolving - the component along the direction of the angle uses cosine, the perpendicular one uses sine.
Adding vector magnitudes arithmetically instead of accounting for their directions.

Active revision

A $50\,\text{N}$ force acts at $30^{\circ}$ above the horizontal. Find its horizontal and vertical components, and the resultant of this force combined with a $20\,\text{N}$ horizontal force in the same direction.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Moments, couples and centre of mass#

StandardAQA 7408 3.4.1.2DfE GCE Physics - moments

The moment of a force about a point is its turning effect, defined as the force multiplied by the perpendicular distance from the point (the pivot) to the line of action of the force: moment

= F d

, in newton metres. Only the perpendicular distance counts, so a force whose line of action passes through the pivot has no moment. Moments are described as clockwise or anticlockwise.

For a body in rotational equilibrium the principle of moments applies: the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point. Combined with the condition that the resultant force is zero, this lets us solve for unknown forces on beams, bridges and levers.

A couple is a pair of equal, antiparallel forces whose lines of action do not coincide; it produces a turning effect but no resultant force. The moment of a couple (its torque) is the magnitude of one force multiplied by the perpendicular separation of the two forces,

\text{torque} = F s

. A couple turns a steering wheel or a tap without pushing it sideways.

The centre of mass (or centre of gravity in a uniform field) is the single point at which the whole weight of a body can be taken to act. For a symmetric, uniform object it lies at the geometric centre. Its position governs stability: an object topples once its centre of mass passes beyond the edge of its base, so a low centre of mass and a wide base give stability - the reasoning behind racing-car and bus design.

\text{moment} = F d

Moment of a force

Force times the perpendicular distance from the pivot to the line of action.

\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}

Principle of moments

The condition for rotational equilibrium about any point.

Worked example

A loaded beam

A uniform beam of length $3.0\,\text{m}$ and weight $200\,\text{N}$ is pivoted at its left end. A $150\,\text{N}$ load hangs $2.0\,\text{m}$ from the pivot. What upward force applied at the right end keeps the beam horizontal?

Take moments about the pivot
The beam's weight acts at its centre, $1.5\,\text{m}$ from the pivot; the load acts at $2.0\,\text{m}$ ; the unknown force $F$ acts at $3.0\,\text{m}$ .
Balance clockwise and anticlockwise moments
Anticlockwise (from $F$ ) equals clockwise (from weight and load).
$F \times 3.0 = 200 \times 1.5 + 150 \times 2.0 = 600\,\text{N m}$
Solve for F
$F = 600/3.0 = 200\,\text{N}$ .

Result: An upward force of $200\,\text{N}$ at the right end holds the beam horizontal.

Typical mistakes

Using the distance along the force rather than the perpendicular distance from the pivot to the line of action.
Forgetting the weight of a beam acting at its centre of mass when taking moments.

Active revision

A uniform $4.0\,\text{m}$ plank of weight $120\,\text{N}$ rests on two supports, one at each end. A child of weight $300\,\text{N}$ stands $1.0\,\text{m}$ from the left support. Find the force on each support.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Kinematics: the suvat equations and projectiles#

StandardAQA 7408 3.4.1.3AQA 7408 3.4.1.4DfE GCE Physics - motion

Velocity-time graph: area gives displacement

Fig. 2For uniform acceleration from rest, the velocity rises linearly; the shaded area under the line is the displacement, and the gradient is the acceleration.

Motion in a straight line is described by displacement

s

, velocity

v

(rate of change of displacement) and acceleration

a

(rate of change of velocity). On a velocity-time graph the gradient is the acceleration and the area beneath the line is the displacement. These two graphical facts are the foundation from which the equations of motion follow.

For constant acceleration, the four suvat equations relate the five quantities

s, u, v, a, t

. They are derived directly from the velocity-time graph: since the graph is a straight line from

u

v

v = u + at

(gradient), and the displacement is the area of the trapezium,

s = \dfrac{(u + v)}{2}\,t

. Eliminating

v

gives

s = ut + \tfrac{1}{2}at^2

, and eliminating

t

gives

v^2 = u^2 + 2as

. Each equation omits one variable, so choose the one that avoids the quantity you neither know nor want.

Motion under gravity is the special case with

a = g = 9.81\,\text{m}\,\text{s}^{-2}

downward. An object in free fall, or thrown vertically, obeys the same suvat equations; taking one direction as positive and being consistent with signs is essential. Air resistance eventually balances weight, giving a constant terminal speed at which the resultant force, and so the acceleration, is zero.

A projectile has independent horizontal and vertical motions. The horizontal velocity is constant (no horizontal force, ignoring air resistance), so horizontal distance is

u_x t

; the vertical motion is free fall under

g

. The two are linked only by the common time. Splitting a launch velocity into

u\cos\theta

(horizontal) and

u\sin\theta

(vertical) and treating each axis with suvat solves range, maximum height and time of flight.

v = u + at

suvat (no s)

From the gradient of the velocity-time graph.

s = ut + \tfrac{1}{2}at^2

suvat (no v)

Displacement from initial velocity and constant acceleration.

v^2 = u^2 + 2as

suvat (no t)

Links velocities and displacement without time.

A horizontally launched projectile

Fig. 3The horizontal velocity stays constant while the vertical velocity grows under gravity; the two independent motions share only the time of flight.

Worked example

Horizontal projectile from a cliff

A stone is thrown horizontally at $15\,\text{m}\,\text{s}^{-1}$ from the top of a $45\,\text{m}$ cliff. Find the time to reach the ground and how far from the base it lands. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Vertical motion gives the time
Vertically $u_y = 0$ , $s = 45\,\text{m}$ , $a = g$ . Use $s = \tfrac{1}{2}gt^2$ , so $t = \sqrt{2s/g} = \sqrt{90/9.81}$ .
$t = \sqrt{\dfrac{2 \times 45}{9.81}} = 3.03\,\text{s}$
Horizontal motion gives the range
Horizontally the velocity is constant: $x = u_x t = 15 \times 3.03$ .
Evaluate
$x = 45.4\,\text{m}$ .

Result: The stone lands after $3.0\,\text{s}$ , about $45\,\text{m}$ from the base of the cliff.

Typical mistakes

Using a suvat equation when the acceleration is not constant (for example after terminal velocity is reached).
Mixing the horizontal and vertical motions of a projectile instead of linking them only through the common time.

Active revision

A ball is thrown horizontally at $12\,\text{m}\,\text{s}^{-1}$ from a cliff $20\,\text{m}$ high. Find the time to land and the horizontal distance travelled. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Newton's laws, momentum and impulse#

StandardAQA 7408 3.4.1.5AQA 7408 3.4.1.6DfE GCE Physics - Newton's laws and momentum

Free-body diagram of a block on an incline

Fig. 4The three forces on a block on a slope: the weight vertically down, the normal reaction perpendicular to the surface, and friction up the slope.

Newton's first law states that a body remains at rest or moves with constant velocity unless acted on by a resultant force - a statement of inertia. The second law relates the resultant force to the rate of change of momentum; for constant mass it reduces to

F = ma

. The third law states that when body A exerts a force on body B, body B exerts an equal and opposite force on A; these paired forces act on different bodies, which is why they never cancel.

The weight of a body is the gravitational force on it,

W = mg

. On a free-body diagram every force acting on the object is drawn as an arrow from the object; the resultant of these determines the acceleration through

F = ma

. Distinguishing the forces on the object from the forces the object exerts on others (the third-law partners) is essential to avoid double counting.

Momentum is

p = mv

, a vector. Newton's second law in its general form is

F = \dfrac{\Delta p}{\Delta t}

, and rearranged it gives impulse:

F\,\Delta t = \Delta p

, the change in momentum. The impulse equals the area under a force-time graph, which is why a longer collision time (a crumple zone, an air bag, bending the knees) reduces the peak force for a given change in momentum.

In any collision or explosion with no external resultant force, the total momentum is conserved:

\sum m u = \sum m v

. In an elastic collision kinetic energy is also conserved; in an inelastic collision some kinetic energy is transferred to other forms (heat, sound, deformation), though momentum is still conserved. A perfectly inelastic collision is one in which the bodies stick together and move off with a common velocity.

F = ma = \dfrac{\Delta p}{\Delta t}

Newton's second law

Resultant force equals mass times acceleration, or the rate of change of momentum.

F\,\Delta t = \Delta p

Impulse

The impulse of a force equals the change in momentum it produces.

\sum m u = \sum m v

Conservation of momentum

Total momentum before equals total momentum after, with no external resultant force.

Worked example

A perfectly inelastic collision

A $2.0\,\text{kg}$ trolley moving at $3.0\,\text{m}\,\text{s}^{-1}$ collides with and sticks to a stationary $1.0\,\text{kg}$ trolley. Find the common velocity afterwards and the kinetic energy lost.

Conserve momentum
Total momentum before $= 2.0 \times 3.0 + 1.0 \times 0 = 6.0\,\text{kg m s}^{-1}$ . After, mass $3.0\,\text{kg}$ moves at $v$ .
$v = \dfrac{6.0}{3.0} = 2.0\,\text{m}\,\text{s}^{-1}$
Kinetic energy before
$\tfrac{1}{2}(2.0)(3.0)^2 = 9.0\,\text{J}$ .
Kinetic energy after
$\tfrac{1}{2}(3.0)(2.0)^2 = 6.0\,\text{J}$ , so $3.0\,\text{J}$ is lost.

Result: They move off together at $2.0\,\text{m}\,\text{s}^{-1}$ ; $3.0\,\text{J}$ of kinetic energy is lost, so the collision is inelastic.

Typical mistakes

Treating a Newton's-third-law pair as forces that cancel - they act on different bodies, so they never cancel on one body.
Forgetting that momentum is a vector: opposite directions must be given opposite signs before adding.

Active revision

A $1200\,\text{kg}$ car travelling at $20\,\text{m}\,\text{s}^{-1}$ collides with and sticks to a stationary $800\,\text{kg}$ car. Find their common velocity and state whether kinetic energy is conserved.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Work, energy, power and conservation of energy#

StandardAQA 7408 3.4.1.7AQA 7408 3.4.1.8DfE GCE Physics - work, energy and power

Work is done when a force moves its point of application, and it equals the force multiplied by the distance moved in the direction of the force:

W = Fs\cos\theta

, where

\theta

is the angle between force and displacement. Work is a transfer of energy, measured in joules; a force perpendicular to the motion (such as the tension in a string swinging a mass in a circle) does no work.

Kinetic energy is the energy of motion,

E_k = \tfrac{1}{2}mv^2

; gravitational potential energy near the Earth's surface is

E_p = mgh

. The principle of conservation of energy states that energy cannot be created or destroyed, only transferred between forms. For a body in free fall with no resistance, the loss of potential energy equals the gain in kinetic energy, so

mgh = \tfrac{1}{2}mv^2

and the speed after falling a height

h

v = \sqrt{2gh}

Power is the rate of doing work or transferring energy,

P = W/t

, in watts. For a constant force moving an object at speed

v

, the power delivered is

P = Fv

. This is why a car engine of fixed power produces a smaller driving force at high speed, and why climbing stairs quickly requires more power than climbing them slowly even though the work done is the same.

No real machine transfers all its input energy usefully; efficiency is the ratio of useful output energy (or power) to total input,

\text{efficiency} = \dfrac{\text{useful output}}{\text{total input}}

, always less than 1. The wasted energy usually ends up as heat through friction and resistance. Tracking energy through a system - a Sankey-style accounting - identifies where it is lost and how a design might be improved.

W = Fs\cos\theta

Work done

Force times distance moved in the direction of the force.

E_k = \tfrac{1}{2}mv^2, \qquad E_p = mgh

Kinetic and potential energy

The energy of motion and the gravitational potential energy near the surface.

P = \dfrac{W}{t} = Fv

Power

Rate of energy transfer; equals force times velocity for a constant force.

Worked example

Speed at the foot of a slope

A $60\,\text{kg}$ skier starts from rest and descends a smooth slope through a vertical height of $25\,\text{m}$ . Using energy conservation, find the speed at the bottom. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Equate energy transfers
With no friction, the loss of potential energy equals the gain in kinetic energy: $mgh = \tfrac{1}{2}mv^2$ .
Cancel the mass and solve
$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 25}$ .
$v = \sqrt{490.5} = 22.1\,\text{m}\,\text{s}^{-1}$

Result: The skier reaches about $22\,\text{m}\,\text{s}^{-1}$ ; the mass cancels, so it does not affect the speed.

Typical mistakes

Forgetting the $\cos\theta$ factor when force and displacement are not parallel.
Confusing power with energy - power is the rate of energy transfer, not the total energy.

Active revision

A pump raises $150\,\text{kg}$ of water through $12\,\text{m}$ every minute. Calculate the useful output power, and the efficiency if the pump draws $400\,\text{W}$ . Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Materials: Hooke's law, the Young modulus and stress-strain#

AdvancedAQA 7408 3.4.2.1AQA 7408 3.4.2.2DfE GCE Physics - bulk properties of solids

Elastic region of a stress-strain graph

Fig. 5Within the elastic region stress is proportional to strain; the gradient of this straight line is the Young modulus. Beyond the limit of proportionality (top of the line) the material begins to yield.

Density is mass per unit volume,

\rho = m/V

, in

\text{kg}\,\text{m}^{-3}

. When a material is stretched, Hooke's law states that the extension is proportional to the applied force up to a limit:

F = k\,\Delta L

, where

k

is the spring constant (stiffness). The limit of proportionality is the point beyond which this linear relationship fails. The work done stretching a spring within this region is stored as elastic strain energy, equal to the area under the force-extension graph,

E = \tfrac{1}{2}F\,\Delta L = \tfrac{1}{2}k\,\Delta L^2

To describe a material rather than a particular sample, we use stress and strain. Tensile stress is force per unit cross-sectional area,

\sigma = F/A

(in pascals); tensile strain is the fractional extension,

\varepsilon = \Delta L / L

(dimensionless). Within the elastic region the two are proportional, and their ratio is the Young modulus

E = \sigma/\varepsilon

, a property of the material that measures its stiffness. Steel has a Young modulus of about

2\times10^{11}\,\text{Pa}

A stress-strain graph reveals a material's behaviour. Up to the limit of proportionality the graph is a straight line whose gradient is the Young modulus. Beyond the elastic limit the material no longer returns to its original length when unloaded - it has entered plastic deformation. The yield point is where it begins to extend rapidly for little extra stress; the ultimate tensile stress is the maximum it can withstand; and the breaking (fracture) stress is where it snaps.

Materials are classified by how they behave. A brittle material (glass, cast iron) obeys Hooke's law up to fracture and breaks without plastic deformation. A ductile material (copper, mild steel) undergoes large plastic deformation before breaking and can be drawn into wires. A material's toughness is the total energy it absorbs before fracture, the area under the whole stress-strain curve; a stiff material has a large Young modulus, and a strong material a high breaking stress - these are different properties.

F = k\,\Delta L

Hooke's law

Force is proportional to extension up to the limit of proportionality.

\sigma = \dfrac{F}{A}, \qquad \varepsilon = \dfrac{\Delta L}{L}

Stress and strain

Force per unit area and fractional extension.

E = \dfrac{\sigma}{\varepsilon} = \dfrac{F L}{A\,\Delta L}

Young modulus

The gradient of the linear region of the stress-strain graph.

Worked example

Young modulus of a wire

A wire of length $1.5\,\text{m}$ and diameter $0.40\,\text{mm}$ extends by $0.90\,\text{mm}$ under a load of $30\,\text{N}$ . Find the Young modulus.

Cross-sectional area
Radius $= 0.20\,\text{mm} = 2.0\times10^{-4}\,\text{m}$ , so $A = \pi r^2 = \pi(2.0\times10^{-4})^2 = 1.257\times10^{-7}\,\text{m}^2$ .
Stress and strain
$\sigma = F/A = 30/1.257\times10^{-7} = 2.39\times10^{8}\,\text{Pa}$ ; $\varepsilon = \Delta L/L = 0.90\times10^{-3}/1.5 = 6.0\times10^{-4}$ .
Young modulus
$E = \sigma/\varepsilon$ .
$E = \dfrac{2.39\times10^{8}}{6.0\times10^{-4}} = 4.0\times10^{11}\,\text{Pa}$

Result: The Young modulus is $4.0\times10^{11}\,\text{Pa}$ (a very stiff material).

Typical mistakes

Confusing stiffness (spring constant $k$ , sample-dependent) with the Young modulus (material property).
Using diameter as radius when finding cross-sectional area $A = \pi r^2$ for a wire.

Active revision

A steel wire of length $2.0\,\text{m}$ and diameter $0.50\,\text{mm}$ stretches by $1.2\,\text{mm}$ under a load of $45\,\text{N}$ . Calculate the Young modulus of the steel.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Mechanics and materials

Key points

Resultant of two forces

Key points

A loaded beam

Key points

Horizontal projectile from a cliff

Key points

A perfectly inelastic collision

Key points

Speed at the foot of a slope

Key points

Young modulus of a wire

Sources