Forces and Newton’s laws – Year 12 A level maths revision notes to download as PDF

Worked example 1

An 800 kg car has driving force 2400 N and resistance 800 N. Find its acceleration and its speed and distance after 10 s from rest.

Resultant \(=2400-800=1600\) N, so \(a=\dfrac{1600}{800}=2\) m/s\(^2\). After 10 s \(v=20\) m/s and the distance is \(\tfrac12(2)(100)=100\) m.

Worked example 2

Forces of 6 N and 8 N act at right angles. Find the resultant.

\(\sqrt{36+64}=10\) N, at \(\tan^{-1}\dfrac86=53.1^\circ\) to the 6 N force.

Worked example 3

Forces \(3\mathbf i+4\mathbf j\) and \(-5\mathbf i+2\mathbf j\) act on a particle in equilibrium with a third force \(\mathbf F\). Find \(\mathbf F\).

The forces sum to zero: \(\mathbf F=-(3\mathbf i+4\mathbf j)-(-5\mathbf i+2\mathbf j)=2\mathbf i-6\mathbf j\).

Worked example 4

A 10 kg crate is pulled along a smooth floor by a rope with tension 40 N at \(30^\circ\) above the horizontal. Find \(a\) and \(R\).

Horizontal: \(40\cos30^\circ=10a\), so \(a=3.46\) m/s\(^2\). Vertical (no vertical acceleration): \(R+40\sin30^\circ=98\), so \(R=78\) N.

Worked example 5

A 5 kg block rests on a smooth \(30^\circ\) slope. Find \(a\) and \(R\).

\(a=9.8\sin30^\circ=4.9\) m/s\(^2\) and \(R=49\cos30^\circ=42.4\) N.

Worked example 6

Particles of 5 kg and 3 kg are connected by a string over a smooth pulley. Find \(a\) and \(T\).

Heavier particle (down): \(5g-T=5a\). Lighter (up): \(T-3g=3a\). Adding: \(2g=8a\), so \(a=\dfrac{2\times9.8}8=2.45\) m/s\(^2\). Then \(T=3(g+a)=36.75\) N.

Worked example 7

A 1200 kg car tows a 600 kg caravan. Driving force 3600 N, resistances 500 N (car) and 300 N (caravan). Find \(a\) and the tension.

Whole system: \(3600-800=1800a\), so \(a=1.56\) m/s\(^2\). Caravan alone: \(T-300=600a\), giving \(T=1233\) N.

Worked example 8

A 60 kg person stands in a lift. Find the reaction when it accelerates upwards at 1.2 m/s\(^2\) and downwards at 1.2 m/s\(^2\).

Up: \(R-588=60\times1.2\), \(R=660\) N (heavier). Down: \(588-R=72\), \(R=516\) N.

Worked example 9: A force at an angle on a rough surface

A box of mass 10 kg is pulled along rough horizontal ground by a rope at \(25^\circ\) above the horizontal with tension 60 N. The coefficient of friction is 0.2. Find the acceleration.

Vertically: \(R+60\sin25^\circ=10g\), so \(R=98-25.36=72.64\) N. Friction \(F=0.2\times72.64=14.53\) N. Horizontally: \(60\cos25^\circ-14.53=10a\), so \(54.38-14.53=39.85=10a\) and \(a=3.99\) m/s\(^2\). The rope lifts the box a little, which reduces the normal reaction and therefore the friction. Always resolve vertically first to find \(R\), then use \(F=\mu R\).

Worked example 10: Resolving on a slope

A 12 kg crate is held at rest on a smooth slope of \(35^\circ\) by a force \(P\) acting parallel to the slope up the slope. Find \(P\) and the normal reaction.

Parallel to the slope: \(P=mg\sin35^\circ=12\times9.8\times0.5736=67.5\) N. Perpendicular: \(R=mg\cos35^\circ=12\times9.8\times0.8192=96.3\) N. If \(P\) is removed the crate accelerates down the slope at \(g\sin35^\circ=5.62\) m/s\(^2\) and \(R\) is unchanged. Notice that the mass cancels in the acceleration, so every object slides down a smooth slope at the same rate.

Worked example 11: Newton’s third law

A 70 kg skater pushes off a wall with a force of 140 N for 0.5 s. Find the speed gained and the force of the skater on the wall.

By Newton\'s third law the wall pushes the skater with 140 N, and the skater pushes the wall with an equal and opposite force of 140 N. The skater\'s acceleration is \(\dfrac{140}{70}=2\) m/s\(^2\), so after 0.5 s the speed is \(2\times0.5=1\) m/s. The two forces in a third-law pair act on different bodies, which is why they do not cancel each other in the equation of motion of the skater.

Worked example 12: A pulley with a particle on a smooth slope

A 5 kg particle on a smooth slope of \(30^\circ\) is attached by a string over a pulley at the top to a 3 kg particle hanging freely. Find the acceleration.

Take the direction of motion with the 5 kg particle moving down the slope if \(5g\sin30^\circ>3g\). Here \(5g\sin30^\circ=24.5\) N and \(3g=29.4\) N, so the hanging particle goes down and the particle on the slope goes up it. Equations: \(3g-T=3a\) and \(T-5g\sin30^\circ=5a\). Add: \(29.4-24.5=8a\), so \(a=0.6125\) m/s\(^2\). Then \(T=3(9.8-0.6125)=27.56\) N. Check the tension in the other equation: \(27.56-24.5=3.06=5\times0.6125\) ✓.

Worked example 13: Tension in a rope hauling a load

A crane lifts a 250 kg load with an acceleration of 0.4 m/s\(^2\) upwards. Find the tension in the cable and the cable tension when the load moves at constant speed.

\(T-mg=ma\), so \(T=250(9.8+0.4)=2550\) N. At constant speed \(a=0\), so \(T=mg=2450\) N. The tension is larger than the weight when the load speeds up and smaller than the weight when it slows down while rising: for a deceleration of 0.4 m/s\(^2\) the tension is \(250(9.8-0.4)=2350\) N.