10 exam-style questions on static and kinetic friction, limiting equilibrium, inclined planes and the angle of friction. Tap Show answer for full working.
A block of mass \(5\,\mathrm{kg}\) rests on a rough horizontal surface. The coefficient of static friction is \(0.4\). Find the least horizontal force required to move the block. Take \(g=10\,\mathrm{m\,s}^{-2}\).
\(R=mg=50\,\mathrm{N}\). \(F_{\max}=\mu R=0.4\cdot50=20\,\mathrm{N}\). Least force \(=20\,\mathrm{N}\).
A particle of mass \(2\,\mathrm{kg}\) is on a rough horizontal surface. A horizontal force of \(6\,\mathrm{N}\) is applied and the particle accelerates at \(1.5\,\mathrm{m\,s}^{-2}\). Find the coefficient of kinetic friction. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Net force \(=ma=3\,\mathrm{N}\). Friction \(=6-3=3\,\mathrm{N}\). \(R=20\,\mathrm{N}\). \(\mu=\dfrac{3}{20}=0.15\).
Explain the difference between static friction and kinetic friction, and state the relationship between the coefficient of static friction \(\mu_s\) and the coefficient of kinetic friction \(\mu_k\).
Static friction acts when surfaces are at rest relative to each other and can take any value up to \(\mu_s R\). Kinetic (sliding) friction acts when surfaces are sliding and has magnitude \(\mu_k R\).
In general \(\mu_s\ge\mu_k\).
A block of mass \(4\,\mathrm{kg}\) rests on a rough plane inclined at \(20^\circ\) to the horizontal. The coefficient of friction is \(0.3\). Determine whether the block remains at rest. If it moves, find its acceleration. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Component down plane: \(mg\sin20^\circ=40\sin20^\circ\approx13.68\,\mathrm{N}\).
\(R=mg\cos20^\circ=40\cos20^\circ\approx37.59\,\mathrm{N}\). \(\mu R=0.3\cdot37.59\approx11.28\,\mathrm{N}\).
Since \(13.68>11.28\), the block slides. Net force \(=13.68-11.28=2.40\,\mathrm{N}\). \(a=\dfrac{2.40}{4}=0.60\,\mathrm{m\,s}^{-2}\).
A particle rests on a rough horizontal surface. The coefficient of friction is \(\mu\). A force of magnitude \(P\) acts on the particle at an angle \(\alpha\) above the horizontal. Show that the least value of \(P\) that will move the particle is \(\dfrac{\mu mg}{\cos\alpha+\mu\sin\alpha}\).
Resolve vertically: \(R+P\sin\alpha=mg \Rightarrow R=mg-P\sin\alpha\).
At limiting equilibrium horizontally: \(P\cos\alpha=\mu R=\mu(mg-P\sin\alpha)\).
\(P\cos\alpha=\mu mg-\mu P\sin\alpha \Rightarrow P(\cos\alpha+\mu\sin\alpha)=\mu mg \Rightarrow P=\dfrac{\mu mg}{\cos\alpha+\mu\sin\alpha}\).
A rough plane is inclined at an angle \(\alpha\) to the horizontal, where \(\tan\alpha=\dfrac{3}{4}\). A particle of mass \(m\) is projected up the plane with speed \(u\). The coefficient of friction is \(\dfrac{1}{4}\). Find the distance the particle travels up the plane before coming to rest.
\(\sin\alpha=\dfrac{3}{5}\), \(\cos\alpha=\dfrac{4}{5}\).
Acceleration down the plane while going up: \(g\sin\alpha+\mu g\cos\alpha=g\bigl(\dfrac{3}{5}+\dfrac{1}{4}\cdot\dfrac{4}{5}\bigr)=g\bigl(\dfrac{3}{5}+\dfrac{1}{5}\bigr)=\dfrac{4g}{5}\).
\(v=0\): \(0=u^2-2\cdot\dfrac{4g}{5}s \Rightarrow s=\dfrac{5u^2}{8g}\).
The angle of friction between a particle and a rough inclined plane is \(15^\circ\). The plane is inclined at \(30^\circ\) to the horizontal. Find the acceleration of the particle down the plane. Take \(g=9.8\,\mathrm{m\,s}^{-2}\).
Angle of friction \(\lambda=15^\circ\), so \(\mu=\tan\lambda=\tan15^\circ\).
Acceleration \(=g\sin\theta-\mu g\cos\theta=g(\sin30^\circ-\tan15^\circ\cos30^\circ)=g\bigl(\dfrac{1}{2}-\tan15^\circ\cdot\dfrac{\sqrt{3}}{2}\bigr)\).
\(\tan15^\circ=2-\sqrt{3}\). Numerical value \(\approx9.8(0.5-0.268\cdot0.866)\approx9.8(0.5-0.232)\approx2.63\,\mathrm{m\,s}^{-2}\).
A particle of mass \(m\) is on a rough horizontal table. The coefficient of friction is \(\mu\). A light inextensible string attached to the particle passes over a smooth pulley at the edge of the table and carries a particle of mass \(2m\) which hangs freely. The system is released from rest with the string taut. Find the acceleration of the system and the tension in the string.
For the hanging mass: \(2mg-T=2ma\). For the table mass: \(T-\mu mg=ma\).
Add: \(2mg-\mu mg=3ma \Rightarrow a=\dfrac{g(2-\mu)}{3}\). \(T=ma+\mu mg=m\cdot\dfrac{g(2-\mu)}{3}+\mu mg=\dfrac{mg}{3}(2-\mu+3\mu)=\dfrac{mg}{3}(2+2\mu)=\dfrac{2mg(1+\mu)}{3}\).
A rough plane is inclined at angle \(\alpha\) where \(\sin\alpha=\dfrac{3}{5}\). A particle of mass \(m\) is held in equilibrium on the plane by a force of magnitude \(\dfrac{1}{2}mg\) acting up the plane. Find the range of possible values of the coefficient of friction \(\mu\).
\(\cos\alpha=\dfrac{4}{5}\). Component of weight down plane \(=mg\cdot\dfrac{3}{5}\). Applied force up \(=\dfrac{1}{2}mg\).
Net force down the plane without friction \(=\dfrac{3}{5}mg-\dfrac{1}{2}mg=\dfrac{1}{10}mg\).
Friction can act either up or down the plane. For equilibrium, friction must balance the net force, so \(|F|=\dfrac{1}{10}mg\), and \(F\le\mu R=\mu mg\cdot\dfrac{4}{5}\).
Thus \(\dfrac{1}{10}mg\le\mu\cdot\dfrac{4}{5}mg \Rightarrow \mu\ge\dfrac{1}{8}\). (Friction cannot exceed limiting, so \(\mu\ge\dfrac{1}{8}\).)
A particle is projected up a rough inclined plane of angle \(30^\circ\) with initial speed \(10\,\mathrm{m\,s}^{-1}\). The coefficient of friction is \(0.2\). Find the speed of the particle when it returns to the point of projection. Take \(g=10\,\mathrm{m\,s}^{-2}\).
Acceleration while going up: \(g\sin30^\circ+\mu g\cos30^\circ=5+0.2\cdot10\cdot\dfrac{\sqrt{3}}{2}=5+\sqrt{3}\approx6.732\,\mathrm{m\,s}^{-2}\).
Distance up: \(0=100-2\cdot6.732\cdot s \Rightarrow s\approx7.43\,\mathrm{m}\).
Acceleration while coming down: \(g\sin30^\circ-\mu g\cos30^\circ=5-\sqrt{3}\approx3.268\,\mathrm{m\,s}^{-2}\).
Speed at bottom: \(v^2=0+2\cdot3.268\cdot7.43\approx48.6 \Rightarrow v\approx7.0\,\mathrm{m\,s}^{-1}\).
(Energy method: work done against friction \(=2\mu mg\cos\theta\cdot s\), so loss in KE \(=2\mu mg\cos\theta\cdot s\), leading to the same result.)