FrictionEdexcel A-Level Maths: Revision notes

Friction is a contact force along a surface that opposes sliding (or the tendency to slide). The model for a rough surface is $F\le\mu R,$ where $R$ is the normal reaction and $\mu$ is the coefficient of friction, which depends on the two surfaces. The three cases are:

• At rest: $F$ is just enough to keep equilibrium, so $F<\mu R$ (or $F\le\mu R$).
• Limiting equilibrium: the body is about to slide and $F=\mu R$ (the maximum).
• Sliding: $F=\mu R$, acting opposite to the motion. A smooth surface has $\mu=0$ and no friction.

Resolve vertically first to find $R$, then horizontally. A box of mass 10 kg on a floor with $\mu=0.3$: $R=10g=98$ N, so the maximum friction is $\mu R=29.4$ N.

• Push of 25 N: the box stays at rest and $F=25$ N.
• Push of 35 N: the box moves and $35-29.4=10a$, so $a=0.56\ \text{m s}^{-2}$. If the applied force is at an angle to the floor, it changes $R$. A rope pulling at $30^\circ$ above the horizontal with tension $T$ gives $R=mg-T\sin30^\circ$, and the horizontal pull is $T\cos30^\circ$. A push at an angle below the horizontal increases $R$.

Resolve parallel and perpendicular to the plane. For a particle of mass $m$ on a plane at angle $\alpha$: $R=mg\cos\alpha$ (no other perpendicular forces) and the weight component down the plane is $mg\sin\alpha$. Equilibrium on the plane: friction acts up the plane, $F=mg\sin\alpha\le\mu mg\cos\alpha$, so the particle stays at rest if $\mu\ge\tan\alpha$. Limiting equilibrium (about to slip): $\mu=\tan\alpha$. Sliding down: $mg\sin\alpha-\mu mg\cos\alpha=ma$, so $a=g(\sin\alpha-\mu\cos\alpha)$. Example: $\alpha=20^\circ$, $\mu=0.3$: $a=9.8(0.3420-0.3(0.9397))=0.589\ \text{m s}^{-2}$. If a particle slides up the plane, friction acts down the plane.

For a crate pulled by a rope at $30^\circ$ above the horizontal with tension 80 N, mass 20 kg and $\mu=0.25$:

• vertically $R=20g-80\sin30^\circ=156$ N
• friction (moving) $F=0.25\times156=39$ N
• horizontally $80\cos30^\circ-39=20a$, so $a=1.51\ \text{m s}^{-2}$. Once $a$ is known, use the suvat equations. For a body that is slowing down with no driving force, the deceleration is $\mu g$ on a horizontal surface, independent of the mass.

For a block on a rough table joined to a hanging particle, write one equation for each: friction acts on the block only. With $B$ 6 kg on the table ($\mu=0.4$) and $C$ 4 kg hanging: $R=58.8$ N, $F=23.52$ N, $4g-T=4a$ and $T-23.52=6a$, giving $a=1.568\ \text{m s}^{-2}$ and $T=32.9$ N. When the hanging particle lands the string goes slack. The block now decelerates under friction alone at $\mu g=3.92\ \text{m s}^{-2}$, so the extra distance is $\frac{v^2}{2\mu g}$. Combine the two stages with the suvat equations.