Forces in two dimensions and friction – Year 13 A level maths revision notes to download as PDF

Worked example 1

Forces of 10 N along the \(x\)-axis and 8 N at \(60^\circ\) above it act on a particle. Find the resultant.

Components \(10+8\cos60^\circ=14\) and \(8\sin60^\circ=6.93\). Magnitude \(\sqrt{14^2+6.93^2}=15.6\) N, direction \(\tan^{-1}\dfrac{6.93}{14}=26.3^\circ\) above the \(x\)-axis.

Worked example 2

Forces \((4\mathbf i+3\mathbf j)\) N and \((-\mathbf i+5\mathbf j)\) N act on a 2 kg particle with a further force so the resultant is \((6\mathbf i+8\mathbf j)\) N. Find the resultant of the first two and the acceleration.

The first two give \(3\mathbf i+8\mathbf j\), magnitude \(\sqrt{73}=8.54\) N at \(69.4^\circ\) to \(\mathbf i\). Under a resultant \((6\mathbf i+8\mathbf j)\) N, \(\mathbf a=\dfrac{\mathbf F}{m}=(3\mathbf i+4\mathbf j)\ \mathrm{m\,s^{-2}}\), magnitude \(5\ \mathrm{m\,s^{-2}}\).

Worked example 3

A 50 N weight hangs from two strings at \(30^\circ\) and \(60^\circ\) to the horizontal. Find the tensions.

Horizontally \(T_1\cos30^\circ=T_2\cos60^\circ\), so \(T_2=\sqrt3\,T_1\). Vertically \(T_1\sin30^\circ+T_2\sin60^\circ=50\), giving \(0.5T_1+1.5T_1=50\), so \(T_1=25\) N and \(T_2=43.3\) N.

Worked example 4

A 40 N lamp hangs from two equal cables, each at \(30^\circ\) to the horizontal. Find \(T\). Then find \(\mathbf F_3\) if \(\mathbf F_1=(2\mathbf i-\mathbf j)\) and \(\mathbf F_2=(-5\mathbf i+4\mathbf j)\) are in equilibrium with it.

\(2T\sin30^\circ=40\), so \(T=40\) N: a shallow angle needs a large tension. For vectors set the sum to zero: \(\mathbf F_3=(3\mathbf i-3\mathbf j)\) N.

Worked example 5

A 20 kg crate on a horizontal floor has \(\mu=0.35\). Find the push needed to move it, and the pull \(P\) at \(25^\circ\) above the horizontal that keeps it moving at constant speed.

\(R=196\) N, limiting friction \(68.6\) N, so a push of 68.6 N is just enough. With the rope: \(R=196-P\sin25^\circ\) and \(P\cos25^\circ=0.35R\), giving \(P=65.1\) N. A rope above the horizontal reduces \(R\) and so reduces friction.

Worked example 6

A 4 kg block rests on a rough \(30^\circ\) plane. Find \(F\), \(R\) and the least \(\mu\).

\(F=mg\sin30^\circ=19.6\) N and \(R=33.9\) N, so \(\mu\ge\dfrac{19.6}{33.9}=\tan30^\circ=0.577\).

Worked example 7

(a) On a \(25^\circ\) plane with \(\mu=0.4\) does a block slide? (b) What force parallel to a \(20^\circ\) plane pushes a 6 kg block up at constant speed if \(\mu=0.25\)?

(a) \(\tan25^\circ=0.466>0.4\), so it slides, with \(a=g(\sin25^\circ-0.4\cos25^\circ)=0.589\ \mathrm{m\,s^{-2}}\). (b) \(P=mg\sin20^\circ+\mu mg\cos20^\circ=33.9\) N.

Worked example 8

A 3 kg block on a rough table (\(\mu=0.2\)) is joined by a string over a smooth pulley to a hanging 2 kg block. Find \(a\) and \(T\).

\(2g-T=2a\) and \(T-0.2(3g)=3a\), so \(a=\dfrac{2g-0.6g}{5}=2.74\ \mathrm{m\,s^{-2}}\) and \(T=2(g-a)=14.1\) N. The string is light and inextensible, so tension and acceleration are the same for both particles.

Worked example 9: Resolving at an angle to the vertical

A force of 50 N acts at \(20^\circ\) to the vertical. Find its horizontal and vertical components.

The angle is measured from the vertical, so the vertical component is the adjacent side and the horizontal component is the opposite side. Vertical \(=50\cos20^\circ=47.0\) N. Horizontal \(=50\sin20^\circ=17.1\) N. Sketch the right-angled triangle first: the angle given is next to the component with cosine.

Worked example 10: A resultant using i and j

Forces of \((5\mathbf i-2\mathbf j)\) N, \((-3\mathbf i+7\mathbf j)\) N and \((\mathbf i+\mathbf j)\) N act on a particle of mass 3 kg. Find the acceleration.

Add: \((5-3+1)\mathbf i+(-2+7+1)\mathbf j=3\mathbf i+6\mathbf j\) N. \(\mathbf a=\dfrac13(3\mathbf i+6\mathbf j)=(\mathbf i+2\mathbf j)\) m s\(^{-2}\) with magnitude \(\sqrt5=2.24\) m s\(^{-2}\). The direction is \(\tan^{-1}2=63.4^\circ\) above \(\mathbf i\).

Worked example 11: Limiting friction in equilibrium

A 10 kg box rests on a rough floor, \(\mu=0.5\). A horizontal force of 30 N is applied. Does the box move? What is the friction force?

\(R=98\) N so the maximum friction is \(0.5\times98=49\) N. The applied force 30 N is less than 49 N, so the box stays at rest. Friction is only as large as needed: \(F=30\) N, not 49 N. The formula \(F=\mu R\) applies only when the box is about to move or moving.

Worked example 12: Atwood machine: two hanging masses

Masses of 5 kg and 3 kg hang from the ends of a light string over a smooth pulley. Find the acceleration and the tension.

Heavier mass moves down: \(5g-T=5a\). Lighter mass moves up: \(T-3g=3a\). Add: \(2g=8a\) so \(a=\dfrac{2\times9.8}8=2.45\) m s\(^{-2}\). Then \(T=3(g+a)=3(12.25)=36.75\) N. Check with the other mass: \(T=5(g-a)=5(7.35)=36.75\) N.

Worked example 13: Pushing a block up a smooth slope at constant speed

What force parallel to a smooth \(20^\circ\) slope pushes a 6 kg block up the slope at constant speed? Then repeat if \(\mu=0.25\).

Smooth: constant speed means zero resultant, so \(P=mg\sin20^\circ=6\times9.8\times0.3420=20.1\) N. Rough: friction acts down the slope, \(F=\mu mg\cos20^\circ=0.25\times55.26=13.8\) N. So \(P=20.1+13.8=33.9\) N. The rough slope needs 69% more force.

Worked example 14: Finding the angle at which a body slips

A box rests on a rough ramp whose coefficient of friction with the box is 0.6. Find the steepest angle at which the box can rest, and decide whether a ramp at \(35^\circ\) is safe.

A body rests in equilibrium only while \(\tan\alpha\le\mu\), so the steepest angle is \(\tan^{-1}0.6=31.0^\circ\). A ramp at \(35^\circ\) has \(\tan35^\circ=0.700>0.6\), so friction cannot hold the box and it will slide. The mass of the box is irrelevant, because both the weight component and the maximum friction are proportional to \(m\).