3.4 Mechanics and materials — printable coverage pack

3.4.1.1 · Scalars and vectors

A scalar has magnitude only; a vector has magnitude and direction. Speed, distance and mass are scalars, whereas velocity, displacement, acceleration, force and weight are vectors.
For two perpendicular components, use Pythagoras for the resultant magnitude and trigonometry for its direction.
Resolve a force $F$ at angle $\theta$ into perpendicular components, taking care to match $F\sin\theta$ and $F\cos\theta$ to the stated angle.
Equilibrium means zero resultant force, so an object is either at rest or moving with constant velocity; it does not necessarily mean the object is stationary.
State vector directions unambiguously, for example $39^{\circ}$ south of west, rather than giving an unsupported angle.

Tier 1 · Easy

1. A boat has velocity components $3.0\,\text{m s}^{-1}$ east and $4.0\,\text{m s}^{-1}$ north. Determine the magnitude and direction of its velocity.[2 marks]
Answer
- $5.0\,\text{m s}^{-1}$ at $53^{\circ}$ north of east
Method: The magnitude is $v=\sqrt{3.0^2+4.0^2}=5.0\,\text{m s}^{-1}$ . The direction from east is $\theta=\tan^{-1}(4.0/3.0)=53.1^{\circ}$ , so it is $53^{\circ}$ north of east.

Tier 2 · Standard

1. A $12.0\,\text{kg}$ crate rests on a smooth plane inclined at $30.0^{\circ}$ to the horizontal. Resolve its weight into components parallel and perpendicular to the plane.[3 marks]
Answer
- $58.9\,\text{N}$ down the plane and $102\,\text{N}$ into the plane
Method: The weight is $W=mg=12.0\times9.81=117.72\,\text{N}$ . The parallel component is $W\sin30.0^{\circ}=58.9\,\text{N}$ down the plane. The perpendicular component is $W\cos30.0^{\circ}=102\,\text{N}$ into the plane.

Tier 3 · Hard

1. Two forces act at a point. One is $180\,\text{N}$ due east. The other is $120\,\text{N}$ at $110^{\circ}$ anticlockwise from east. Determine the magnitude and direction of the single force that would produce equilibrium.[5 marks]
Answer
- $179\,\text{N}$ at $39.0^{\circ}$ south of west
Method: Resolve the second force: $F_x=120\cos110^{\circ}=-41.0\,\text{N}$ and $F_y=120\sin110^{\circ}=112.8\,\text{N}$ . The resultant of the given forces is $(180-41.0,112.8)=(139.0,112.8)\,\text{N}$ . Its magnitude is $\sqrt{139.0^2+112.8^2}=179\,\text{N}$ and its angle is $\tan^{-1}(112.8/139.0)=39.0^{\circ}$ north of east. The equilibrant is equal and opposite, so it is $179\,\text{N}$ at $39.0^{\circ}$ south of west.

3.4.1.2 · Moments

The moment of a force about a point is $F d_{\perp}$ , where $d_{\perp}$ is the perpendicular distance from the point to the force's line of action.
For rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments; translational equilibrium also requires zero resultant force.
A couple is a pair of equal and opposite coplanar forces whose lines of action differ. Its moment is one force multiplied by the perpendicular separation of the lines of action.
The weight of a uniform regular solid acts through its centre of mass. Include the object's own weight unless the question says it is negligible.
Take moments about a point through unknown forces when possible, and keep distances perpendicular to each force rather than simply using beam lengths.

Tier 1 · Easy

1. A $25\,\text{N}$ force acts perpendicular to a spanner $0.40\,\text{m}$ from its pivot. Calculate the moment of the force about the pivot.[1 mark]
Answer
- $10\,\text{N m}$
Method: The perpendicular distance is $0.40\,\text{m}$ , so the moment is $Fd=25\times0.40=10\,\text{N m}$ .

Tier 2 · Standard

1. A uniform horizontal beam of length $3.0\,\text{m}$ and weight $180\,\text{N}$ is hinged at its left end and supported vertically at its right end. Determine the vertical support force and the vertical force exerted by the hinge.[3 marks]
Answer
- Support force $90\,\text{N}$ upward; hinge force $90\,\text{N}$ upward
Method: The beam's weight acts $1.5\,\text{m}$ from the hinge. Taking moments about the hinge gives $R(3.0)=180(1.5)$ , so $R=90\,\text{N}$ . Vertical equilibrium then gives $H+90-180=0$ , so the hinge force is also $H=90\,\text{N}$ upward.

Tier 3 · Hard

1. A uniform horizontal beam is $5.0\,\text{m}$ long and weighs $240\,\text{N}$ . It is hinged to a wall at one end. A $360\,\text{N}$ load hangs $3.8\,\text{m}$ from the hinge, and a cable attached to the free end makes $35^{\circ}$ above the beam. Determine the cable tension and the magnitude and direction of the force exerted by the hinge on the beam.[6 marks]
Answer
- Cable tension $686\,\text{N}$ ; hinge force $599\,\text{N}$ at $20.2^{\circ}$ above the horizontal, away from the wall
Method: Only the cable's vertical component produces a moment about the hinge. Thus $(T\sin35^{\circ})(5.0)=240(2.5)+360(3.8)=1968\,\text{N m}$ . Hence $T\sin35^{\circ}=393.6\,\text{N}$ and $T=686\,\text{N}$ . The cable pulls horizontally towards the wall, so equilibrium requires the hinge force $H_x=T\cos35^{\circ}=562\,\text{N}$ away from the wall. Vertically, $H_y+393.6-240-360=0$ , giving $H_y=206.4\,\text{N}$ upward. Therefore $H=\sqrt{562^2+206.4^2}=599\,\text{N}$ and $\theta=\tan^{-1}(206.4/562)=20.2^{\circ}$ above the horizontal.

3.4.1.3 · Motion along a straight line

Velocity is the rate of change of displacement and acceleration is the rate of change of velocity; both require a stated positive direction when signs matter.
On a displacement-time graph, gradient is velocity. On a velocity-time graph, gradient is acceleration and signed area is displacement.
On an acceleration-time graph, signed area is change in velocity. A negative velocity does not necessarily mean the object is slowing down.
The SUVAT equations apply only over intervals of constant acceleration. Do not use one equation across a non-uniform section of a graph.
Near Earth's surface, free-fall acceleration has magnitude $g=9.81\,\text{m s}^{-2}$ ; choose its sign from the declared positive direction.

Tier 1 · Easy

1. An object moves at constant velocity $7.5\,\text{m s}^{-1}$ for $4.0\,\text{s}$ . Calculate its displacement.[1 mark]
Answer
- $30\,\text{m}$
Method: The displacement is the area under the velocity-time graph: $s=vt=7.5\times4.0=30\,\text{m}$ .

Tier 2 · Standard

1. A car accelerates uniformly from $5.0\,\text{m s}^{-1}$ to $17.0\,\text{m s}^{-1}$ in $6.0\,\text{s}$ . Determine its acceleration and displacement during this interval.[3 marks]
Answer
- $2.0\,\text{m s}^{-2}$ and $66\,\text{m}$
Method: The acceleration is $a=(v-u)/t=(17.0-5.0)/6.0=2.0\,\text{m s}^{-2}$ . With uniform acceleration, the mean velocity is $(u+v)/2=11.0\,\text{m s}^{-1}$ , so $s=11.0\times6.0=66\,\text{m}$ .

Tier 3 · Hard

1. A velocity-time graph consists of three straight sections: velocity rises from $0$ to $18\,\text{m s}^{-1}$ in $5.0\,\text{s}$ , remains at $18\,\text{m s}^{-1}$ for $4.0\,\text{s}$ , then falls uniformly to $-6.0\,\text{m s}^{-1}$ in $3.0\,\text{s}$ . Determine the final acceleration, total displacement and total distance travelled.[5 marks]
Answer
- $-8.0\,\text{m s}^{-2}$ ; displacement $135\,\text{m}$ ; distance $140\,\text{m}$
Method: For the final section, $a=(-6-18)/3.0=-8.0\,\text{m s}^{-2}$ . Signed area gives displacement: first triangle $=\tfrac12(5.0)(18)=45\,\text{m}$ , rectangle $=(4.0)(18)=72\,\text{m}$ , and final trapezium $=\tfrac12(18-6)(3.0)=18\,\text{m}$ . Total displacement is $45+72+18=135\,\text{m}$ . In the last section the velocity reaches zero after $18/8=2.25\,\text{s}$ . Its distance is $\tfrac12(2.25)(18)+\tfrac12(0.75)(6)=22.5\,\text{m}$ . Total distance is $45+72+22.5=139.5\,\text{m}$ , or $140\,\text{m}$ to two significant figures.

3.4.1.4 · Projectile motion

When air resistance is neglected, horizontal and vertical motions are independent: horizontal velocity is constant and vertical acceleration is $g$ downward.
Resolve the initial velocity before applying SUVAT separately in each direction; time is the quantity shared by the two motions.
At the highest point, vertical velocity is zero but horizontal velocity is not, so the projectile is still moving.
With air resistance, drag increases with speed and changes both components; a projectile has a shorter range and a non-symmetric trajectory than the no-drag model.
Always state the model used. Do not insert $g$ into the horizontal calculation when air resistance is neglected.

Tier 1 · Easy

1. Air resistance is neglected. State the horizontal and vertical accelerations of a projectile after it has been released.[2 marks]
Answer
- Horizontal acceleration $0$ ; vertical acceleration $9.81\,\text{m s}^{-2}$ downward.
Method: With no air resistance, gravity is the only force. It has no horizontal component, so horizontal acceleration is zero, and it produces vertical acceleration $g=9.81\,\text{m s}^{-2}$ downward.

Tier 2 · Standard

1. A ball leaves a horizontal platform at $12.0\,\text{m s}^{-1}$ . The platform is $20.0\,\text{m}$ above level ground. Air resistance is neglected. Determine the time to reach the ground and the horizontal range.[3 marks]
Answer
- $2.02\,\text{s}$ and $24.2\,\text{m}$
Method: Vertically, $u_y=0$ and $s=20.0\,\text{m}$ downward, so $s=\tfrac12gt^2$ gives $t=\sqrt{2(20.0)/9.81}=2.02\,\text{s}$ . Horizontal speed is constant, so the range is $x=u_xt=12.0\times2.02=24.2\,\text{m}$ .

Tier 3 · Hard

1. A projectile is launched from level ground at $22.0\,\text{m s}^{-1}$ and $35.0^{\circ}$ above the horizontal. It lands at the launch height. Air resistance is neglected. Determine its time of flight, horizontal range and maximum height.[6 marks]
Answer
- $2.57\,\text{s}$ ; $46.4\,\text{m}$ ; $8.12\,\text{m}$
Method: Resolve the initial velocity: $u_x=22.0\cos35.0^{\circ}=18.0\,\text{m s}^{-1}$ and $u_y=22.0\sin35.0^{\circ}=12.6\,\text{m s}^{-1}$ . Returning to the launch height gives $t=2u_y/g=2(12.6)/9.81=2.57\,\text{s}$ . The range is $u_xt=18.0\times2.57=46.4\,\text{m}$ . At maximum height $v_y=0$ , so $0=u_y^2-2gh$ and $h=u_y^2/(2g)=12.6^2/[2(9.81)]=8.12\,\text{m}$ .

3.4.1.5 · Newton's laws of motion

Newton's first law states that an object remains at rest or at constant velocity unless acted on by a resultant external force.
For constant mass, Newton's second law is $\sum F=ma$ . Apply it to one chosen object or to a clearly defined system after drawing a free-body diagram.
Newton's third-law forces are equal and opposite, act on different objects and are the same type of interaction; they therefore do not cancel on one free-body diagram.
Weight is $mg$ , whereas a normal contact force or tension must be determined from the situation and is not automatically equal to the weight.
Choose a positive direction and retain force signs. A zero resultant implies zero acceleration, not zero velocity.

Tier 1 · Easy

1. A resultant force of $12\,\text{N}$ acts on a $4.0\,\text{kg}$ object. Calculate its acceleration.[1 mark]
Answer
- $3.0\,\text{m s}^{-2}$
Method: Newton's second law gives $a=F/m=12/4.0=3.0\,\text{m s}^{-2}$ .

Tier 2 · Standard

1. A lift of total mass $750\,\text{kg}$ accelerates vertically upward at $1.20\,\text{m s}^{-2}$ . Determine the tension in its supporting cable.[3 marks]
Answer
- $8.26\,\text{kN}$
Method: Taking upward as positive, $T-mg=ma$ . Therefore $T=m(g+a)=750(9.81+1.20)=8257.5\,\text{N}=8.26\,\text{kN}$ .

Tier 3 · Hard

1. Blocks A and B, of masses $4.0\,\text{kg}$ and $6.0\,\text{kg}$ , are joined by a light horizontal cord. A horizontal force of $38\,\text{N}$ pulls B. Friction on A is $5.0\,\text{N}$ and friction on B is $7.0\,\text{N}$ , both opposing motion. Determine the acceleration and cord tension. Identify the Newton's-third-law partner of the cord's pull on A.[5 marks]
Answer
- $2.6\,\text{m s}^{-2}$ ; $15.4\,\text{N}$ ; A exerts an equal and opposite pull on the cord.
Method: For the two-block system, the external resultant is $38-5.0-7.0=26\,\text{N}$ and the mass is $10.0\,\text{kg}$ , so $a=26/10.0=2.6\,\text{m s}^{-2}$ . For A alone, $T-5.0=4.0(2.6)$ , hence $T=15.4\,\text{N}$ . The third-law partner of the cord's force on A is the force of A on the cord; it is not another force acting on A.

3.4.1.6 · Momentum

Linear momentum is $p=mv$ and is a vector. Total momentum is conserved in an isolated system even when kinetic energy is not.
Impulse is change in momentum: $J=\Delta p=F\Delta t$ for a constant force, and it equals the signed area under a force-time graph for a varying force.
An elastic collision conserves both total momentum and total kinetic energy. An inelastic collision conserves momentum but not kinetic energy; objects joining is perfectly inelastic.
Choose one direction as positive before substituting velocities. Rebound velocities then carry the opposite sign.
For impact-force questions, longer contact time for the same momentum change reduces the mean force; quote force direction as well as magnitude.

Tier 1 · Easy

1. A $0.80\,\text{kg}$ trolley moving at $3.0\,\text{m s}^{-1}$ collides with a stationary $0.40\,\text{kg}$ trolley. They join, so the collision is perfectly inelastic. Calculate their common velocity.[2 marks]
Answer
- $2.0\,\text{m s}^{-1}$ in the original direction
Method: Momentum is conserved: initial momentum $=0.80(3.0)=2.4\,\text{kg m s}^{-1}$ . The joined mass is $1.20\,\text{kg}$ , so $v=2.4/1.20=2.0\,\text{m s}^{-1}$ .

Tier 2 · Standard

1. A $0.40\,\text{kg}$ trolley moving at $5.0\,\text{m s}^{-1}$ collides with a stationary $0.60\,\text{kg}$ trolley. Afterwards their velocities are $-1.0\,\text{m s}^{-1}$ and $4.0\,\text{m s}^{-1}$ respectively. Show that the collision is elastic.[4 marks]
Answer
- Both momentum and kinetic energy are conserved, so the collision is elastic.
Method: Initial momentum is $0.40(5.0)=2.0\,\text{kg m s}^{-1}$ . Final momentum is $0.40(-1.0)+0.60(4.0)=-0.40+2.40=2.0\,\text{kg m s}^{-1}$ . Initial kinetic energy is $\tfrac12(0.40)(5.0)^2=5.0\,\text{J}$ . Final kinetic energy is $\tfrac12(0.40)(1.0)^2+\tfrac12(0.60)(4.0)^2=0.20+4.80=5.0\,\text{J}$ . Since both totals are unchanged, the collision is elastic.

Tier 3 · Hard

1. A $1200\,\text{kg}$ car moving east at $18\,\text{m s}^{-1}$ collides head-on with an $800\,\text{kg}$ car moving west at $6.0\,\text{m s}^{-1}$ . They lock together, so the collision is perfectly inelastic. Determine their velocity, the kinetic energy dissipated, and the mean force on the $1200\,\text{kg}$ car if contact lasts $0.18\,\text{s}$ .[6 marks]
Answer
- $8.4\,\text{m s}^{-1}$ east; $1.38\times10^5\,\text{J}$ ; $6.4\times10^4\,\text{N}$ west
Method: Take east as positive. Conservation of momentum gives $(1200)(18)+(800)(-6.0)=(2000)v$ , so $v=16800/2000=8.4\,\text{m s}^{-1}$ east. Initial kinetic energy is $\tfrac12(1200)(18)^2+\tfrac12(800)(6.0)^2=208800\,\text{J}$ . Final kinetic energy is $\tfrac12(2000)(8.4)^2=70560\,\text{J}$ , so $138240\,\text{J}=1.38\times10^5\,\text{J}$ is dissipated. For the heavier car, $\Delta p=1200(8.4-18)=-11520\,\text{kg m s}^{-1}$ . Thus $F_{\rm mean}=\Delta p/\Delta t=-11520/0.18=-6.4\times10^4\,\text{N}$ , meaning west.

3.4.1.7 · Work, energy and power

For a constant force, energy transferred is $W=Fs\cos\theta$ , where $\theta$ is the angle between force and displacement.
For a variable force, work is the area under the force-displacement graph; do not use one endpoint force multiplied by the full distance.
Power is the rate of energy transfer, $P=W/t$ , and for steady motion with force parallel to velocity it is also $P=Fv$ .
Efficiency is useful output energy divided by total input energy, or useful output power divided by total input power, and may be expressed as a percentage.
Use joules for energy and watts for power. A common error is to mix an energy numerator with a power denominator.

Tier 1 · Easy

1. A constant force of $75\,\text{N}$ acts in the direction of motion through $12\,\text{m}$ . Calculate the work done.[1 mark]
Answer
- $900\,\text{J}$
Method: Because force and displacement are parallel, $W=Fs=75\times12=900\,\text{J}$ .

Tier 2 · Standard

1. A motor raises a $25\,\text{kg}$ load vertically through $8.0\,\text{m}$ in $6.0\,\text{s}$ . Its electrical input power is $420\,\text{W}$ . Determine its useful output power and efficiency.[3 marks]
Answer
- $327\,\text{W}$ and $77.9\%$
Method: The useful energy transferred is $mgh=25(9.81)(8.0)=1962\,\text{J}$ . Useful power is $P=1962/6.0=327\,\text{W}$ . Efficiency $=327/420=0.7786$ , or $77.9\%$ .

Tier 3 · Hard

1. A force acting on a $2.0\,\text{kg}$ trolley increases linearly from $10\,\text{N}$ to $50\,\text{N}$ over the first $4.0\,\text{m}$ , then remains at $50\,\text{N}$ for another $2.0\,\text{m}$ . The trolley starts from rest, resistive forces are negligible, and the motion lasts $3.0\,\text{s}$ . Determine the work done, final speed and average power.[5 marks]
Answer
- $220\,\text{J}$ ; $14.8\,\text{m s}^{-1}$ ; $73.3\,\text{W}$
Method: Work is the area under the force-displacement graph. Over the first $4.0\,\text{m}$ , $W_1=\tfrac12(10+50)(4.0)=120\,\text{J}$ . Over the next $2.0\,\text{m}$ , $W_2=50(2.0)=100\,\text{J}$ , so total $W=220\,\text{J}$ . With negligible resistance, $W=\Delta E_k=\tfrac12mv^2$ , hence $v=\sqrt{2(220)/2.0}=14.8\,\text{m s}^{-1}$ . Average power is $W/t=220/3.0=73.3\,\text{W}$ .

3.4.1.8 · Conservation of energy

Energy cannot be created or destroyed; in a closed system the total energy remains constant while it is transferred between stores.
Use $E_k=\tfrac12mv^2$ and $\Delta E_p=mg\Delta h$ near Earth's surface, with a consistent reference height.
Work done against resistive forces transfers mechanical energy to internal energy of the object and surroundings; it is not 'lost'.
State assumptions such as negligible air resistance before equating a decrease in gravitational potential energy directly to an increase in kinetic energy.
Retain any initial kinetic or elastic energy in the balance rather than assuming the object starts from rest unless stated.

Tier 1 · Easy

1. A stone is released from rest and falls through $1.80\,\text{m}$ . Air resistance is neglected. Use energy conservation to calculate its speed.[2 marks]
Answer
- $5.94\,\text{m s}^{-1}$
Method: The loss of gravitational potential energy becomes kinetic energy: $mgh=\tfrac12mv^2$ . Mass cancels, so $v=\sqrt{2gh}=\sqrt{2(9.81)(1.80)}=5.94\,\text{m s}^{-1}$ .

Tier 2 · Standard

1. A roller-coaster car starts from rest $12.0\,\text{m}$ above a reference level. Determine its speed when it is $3.0\,\text{m}$ above that level. Resistive forces are neglected.[3 marks]
Answer
- $13.3\,\text{m s}^{-1}$
Method: The vertical drop is $12.0-3.0=9.0\,\text{m}$ . With no resistive work, $mg(9.0)=\tfrac12mv^2$ . Thus $v=\sqrt{2(9.81)(9.0)}=13.3\,\text{m s}^{-1}$ .

Tier 3 · Hard

1. A $70\,\text{kg}$ skier initially travels at $4.0\,\text{m s}^{-1}$ and descends through a vertical height of $25\,\text{m}$ . During the descent, $8.0\,\text{kJ}$ is transferred to internal energy by resistive forces. Determine the skier's final speed.[5 marks]
Answer
- $16.7\,\text{m s}^{-1}$
Method: Initial kinetic energy is $\tfrac12(70)(4.0)^2=560\,\text{J}$ . The gravitational energy decrease is $mgh=70(9.81)(25)=17167.5\,\text{J}$ . Subtracting the $8000\,\text{J}$ resistive transfer gives final kinetic energy $560+17167.5-8000=9727.5\,\text{J}$ . Hence $v=\sqrt{2E_k/m}=\sqrt{2(9727.5)/70}=16.7\,\text{m s}^{-1}$ .

3.4.2.1 · Bulk properties of solids

Density is $\rho=m/V$ in $\text{kg m}^{-3}$ . Estimate or calculate the complete volume before dividing mass by it.
Within the limit of proportionality, Hooke's law gives $F=k\Delta L$ , where stiffness $k$ is in $\text{N m}^{-1}$ ; the elastic limit concerns whether the material returns to its original dimensions.
Elastic strain energy is the area under a force-extension graph and equals $\tfrac12F\Delta L=\tfrac12k(\Delta L)^2$ only for a straight line through the origin.
Tensile stress is $F/A$ and tensile strain is $\Delta L/L$ . Breaking stress is the stress at fracture, not simply the breaking force.
Plastic deformation leaves a permanent extension; brittle materials fracture with little plastic deformation. Do not treat every curved graph as proof that the elastic limit has been exceeded.

Tier 1 · Easy

1. A solid block has mass $2.16\,\text{kg}$ and volume $8.00\times10^{-4}\,\text{m}^3$ . Calculate its density.[1 mark]
Answer
- $2.70\times10^3\,\text{kg m}^{-3}$
Method: Use $\rho=m/V=2.16/(8.00\times10^{-4})=2.70\times10^3\,\text{kg m}^{-3}$ .

Tier 2 · Standard

1. A spring of stiffness $320\,\text{N m}^{-1}$ is compressed by $75\,\text{mm}$ . It launches a $0.150\,\text{kg}$ cart horizontally. Determine the elastic strain energy and the cart's speed if all this energy becomes kinetic energy.[3 marks]
Answer
- $0.900\,\text{J}$ and $3.46\,\text{m s}^{-1}$
Method: Convert the compression: $x=0.075\,\text{m}$ . The elastic strain energy is $E=\tfrac12kx^2=\tfrac12(320)(0.075)^2=0.900\,\text{J}$ . With $E=\tfrac12mv^2$ , $v=\sqrt{2E/m}=\sqrt{1.800/0.150}=3.46\,\text{m s}^{-1}$ .

Tier 3 · Hard

1. A metal wire's force-extension graph is straight from the origin to its elastic limit at $720\,\text{N}$ and $0.015\,\text{m}$ . The force then rises linearly to $900\,\text{N}$ at fracture, where the extension is $0.027\,\text{m}$ . Determine the work done in stretching the wire to fracture and explain why unloading after the elastic limit would not return all this energy.[5 marks]
Answer
- $15.1\,\text{J}$ ; plastic deformation leaves permanent extension and transfers some energy to internal energy.
Method: Work is the area under the graph. Before the elastic limit, $W_1=\tfrac12(720)(0.015)=5.40\,\text{J}$ . From $0.015$ to $0.027\,\text{m}$ , the trapezium area is $W_2=\tfrac12(720+900)(0.027-0.015)=9.72\,\text{J}$ . Total work is $15.12\,\text{J}=15.1\,\text{J}$ . Beyond the elastic limit the wire deforms plastically, so it retains a permanent extension and some work is dissipated as internal energy rather than being recoverable elastic energy.

3.4.2.2 · The Young modulus

Tensile stress is $\sigma=F/A$ in pascals, while tensile strain is $\varepsilon=\Delta L/L$ and is dimensionless.
The Young modulus is $E=\sigma/\varepsilon=FL/(A\Delta L)$ within the linear elastic region; it is the gradient of a stress-strain graph.
Convert $\text{mm}^2$ to $\text{m}^2$ using $1\,\text{mm}^2=10^{-6}\,\text{m}^2$ . Keep $\text{MPa}$ , $\text{GPa}$ and $\text{Pa}$ distinct.
A simple measurement uses a long wire, measures its original length and diameter, adds known loads, and obtains the gradient of force against extension below the limit of proportionality.
Measure diameter with a micrometer at several positions and orientations because area depends on diameter squared; correct for any initial sag or reference-wire movement.

Tier 1 · Easy

1. A tensile force of $120\,\text{N}$ acts on a wire of cross-sectional area $2.0\,\text{mm}^2$ . Calculate the tensile stress in pascals.[2 marks]
Answer
- $6.0\times10^7\,\text{Pa}$
Method: The area is $2.0\,\text{mm}^2=2.0\times10^{-6}\,\text{m}^2$ . Therefore $\sigma=F/A=120/(2.0\times10^{-6})=6.0\times10^7\,\text{Pa}$ , equal to $60\,\text{MPa}$ .

Tier 2 · Standard

1. A wire is $2.00\,\text{m}$ long and $0.800\,\text{mm}$ in diameter. A tensile force of $55.3\,\text{N}$ produces an extension of $2.00\,\text{mm}$ . Determine the tensile stress, tensile strain and Young modulus.[4 marks]
Answer
- $1.10\times10^8\,\text{Pa}$ ; $1.00\times10^{-3}$ ; $1.10\times10^{11}\,\text{Pa}$
Method: The radius is $0.400\,\text{mm}=4.00\times10^{-4}\,\text{m}$ , so $A=\pi r^2=5.03\times10^{-7}\,\text{m}^2$ . Stress is $\sigma=F/A=55.3/(5.03\times10^{-7})=1.10\times10^8\,\text{Pa}$ . Strain is $\varepsilon=\Delta L/L=(2.00\times10^{-3})/2.00=1.00\times10^{-3}$ . Hence $E=\sigma/\varepsilon=(1.10\times10^8)/(1.00\times10^{-3})=1.10\times10^{11}\,\text{Pa}=110\,\text{GPa}$ .

Tier 3 · Hard

1. In a Young-modulus experiment, a wire has original length $2.50\,\text{m}$ and mean diameter $0.420\,\text{mm}$ . Below the limit of proportionality, a force-extension graph has gradient $6.20\times10^3\,\text{N m}^{-1}$ . Starting from the definitions of stress and strain, determine the Young modulus. Explain why the diameter should be measured at several positions and in two perpendicular directions.[6 marks]
Answer
- $1.12\times10^{11}\,\text{Pa}$ ; repeated perpendicular diameter readings account for variation and non-circularity and reduce uncertainty in area.
Method: From $E=\sigma/\varepsilon$ , with $\sigma=F/A$ and $\varepsilon=\Delta L/L$ , $E=(F/A)/(\Delta L/L)=(F/\Delta L)(L/A)$ . The radius is $0.210\,\text{mm}=2.10\times10^{-4}\,\text{m}$ , so $A=\pi r^2=1.385\times10^{-7}\,\text{m}^2$ . Since $F/\Delta L$ is the graph gradient, $E=(6.20\times10^3)(2.50)/(1.385\times10^{-7})=1.12\times10^{11}\,\text{Pa}=112\,\text{GPa}$ . Diameter can vary along the wire and the cross-section may not be perfectly circular; readings at several positions and orientations give a representative mean and reduce the large area uncertainty caused by squaring the diameter.