FrictionAQA A-Level Maths: Revision notes

Friction is a contact force that acts parallel to the surface and opposes motion, or the tendency to move. The model for a rough surface is $F\le\mu R,$ where $R$ is the normal reaction and $\mu$ is the coefficient of friction, a number depending on the two surfaces (it has no unit). Friction can take any value from $0$ up to its maximum $\mu R$: it adjusts to prevent motion. A smooth surface has $\mu=0$, so only the normal reaction acts. A rough surface has friction.

If a body is at rest, friction equals whatever is needed to keep it in equilibrium, as long as $F\le\mu R$. A box of mass $10$ kg on horizontal ground with $\mu=0.4$ has $R=98$ N and $F_{\max}=39.2$ N. A horizontal push of $30$ N gives $F=30$ N, so the box stays at rest. Limiting friction is the state where friction has reached its maximum, $F=\mu R$, and the body is on the point of moving (limiting equilibrium). Phrases such as 'about to slip' or 'just moves' signal $F=\mu R$ with zero acceleration. To decide whether a body moves: find the friction needed for equilibrium and compare it with $\mu R$. If the needed friction is larger, the body moves.

When a body slides, friction takes its maximum value, $F=\mu R$, acting opposite to the motion. Apply Newton's second law along the direction of motion and use $R$ from the perpendicular direction. Example: a $20$ kg crate is pulled by a horizontal $90$ N rope, $\mu=0.3$. $R=196$ N, $F=58.8$ N, so $90-58.8=20a$ and $a=1.56\ \text{m s}^{-2}$. If the rope breaks while the crate moves at $4\ \text{m s}^{-1}$, friction alone decelerates it: $a=-\mu g=-2.94$, so it stops after $s=\frac{16}{2(2.94)}=2.72$ m. On horizontal ground the deceleration is $\mu g$, independent of mass.

On a plane at angle $\alpha$, resolve the weight: $mg\cos\alpha$ perpendicular and $mg\sin\alpha$ down the plane. Then $R=mg\cos\alpha$ and the maximum friction is $\mu mg\cos\alpha$. A block at rest needs friction $mg\sin\alpha$ up the plane. It stays at rest if $mg\sin\alpha\le\mu mg\cos\alpha$, that is $\tan\alpha\le\mu$. In limiting equilibrium, $\mu=\tan\alpha$. Example: $5$ kg block, $\alpha=20^\circ$, $\mu=0.5$: $R=46.0$ N, needed friction $16.8$ N, maximum $23.0$ N. The block stays at rest. If it slides down: $mg\sin\alpha-\mu mg\cos\alpha=ma$.

If a pulling force acts at an angle $\theta$ above the horizontal, its vertical component reduces the normal reaction: $R=mg-T\sin\theta$, so friction is smaller. If it pushes downwards at an angle, $R=mg+P\sin\theta$ and friction increases. Example: $15$ kg box, $\mu=0.35$, rope at $30^\circ$. For limiting equilibrium: $R=147-0.5T$ and $T\cos30^\circ=0.35R$, so $T=49.4$ N. A horizontal rope would need $0.35(147)=51.45$ N, so lifting the rope slightly helps. With $T=70$ N: $R=112$, $F=39.2$, $a=\frac{70\cos30^\circ-39.2}{15}=1.43\ \text{m s}^{-2}$. Modelling: friction assumes a constant $\mu$, surfaces that do not change, and a body treated as a particle.