FrictionEdexcel International A Level Maths: Revision notes

Friction is a contact force along the surface that opposes sliding, or the tendency to slide. Its direction is opposite to the (intended) relative motion. The normal reaction $R$ is the perpendicular contact force. A surface described as smooth has no friction; a rough surface does. Friction acts on the particle, parallel to the surface, and is always shown on the force diagram together with the weight, $R$ and any applied forces. The size of the friction force depends on $R$, which is why finding $R$ comes first in every friction problem.

The coefficient of friction $\mu$ depends on the two surfaces. The friction force $F$ satisfies $F\le\mu R.$

• Equilibrium, not on the point of moving: $F<\mu R$; $F$ equals whatever is needed to balance the other forces.
• Limiting equilibrium (on the point of slipping): $F=\mu R$.
• Moving: $F=\mu R$ in the direction opposing the motion. Example: a $12$ kg box on a floor with $\mu=0.35$: $R=117.6$ N and $\mu R=41.2$ N. A $30$ N push leaves the box at rest with $F=30$ N. A $50$ N push gives $50-41.16=12a$, so $a=0.737$ m s⁻².

A force at an angle changes $R$. For a particle of mass $m$ pulled by a rope at $\theta$ above the horizontal with tension $T$: $R=mg-T\sin\theta,\quad\text{horizontal resultant }T\cos\theta-F.$ Pulling upwards reduces $R$ and therefore the friction. Pushing downwards at an angle increases $R$. Example: $m=2$, $\mu=0.4$, $\theta=25^\circ$. On the point of moving: $T\cos25^\circ=0.4(19.6-T\sin25^\circ)$, so $T=7.29$ N. With $T=10$: $R=15.37$ N and $10\cos25^\circ-0.4(15.37)=2a$, so $a=1.46$ m s⁻².

On a plane inclined at $\theta$, $R=mg\cos\theta$ and the weight component down the slope is $mg\sin\theta$. A particle on a rough slope with no other force:

• stays at rest if $mg\sin\theta\le\mu mg\cos\theta$, i.e. $\tan\theta\le\mu$,
• is in limiting equilibrium if $\tan\theta=\mu$. The mass cancels, so whether it slides does not depend on the mass. With an extra force $X$ up the slope, friction can act either way. Find the range of $X$ for equilibrium: the minimum has friction up the slope at its limit ($X+\mu R=mg\sin\theta$) and the maximum has friction down the slope at its limit ($X=mg\sin\theta+\mu R$).

When a particle moves on a rough slope use $F=\mu R$ with $R=mg\cos\theta$. Sliding down: $mg\sin\theta-\mu mg\cos\theta=ma$. Moving up: $-mg\sin\theta-\mu mg\cos\theta=ma$ (friction acts down the slope), a greater deceleration. Example: $m=6$, $\theta=30^\circ$, $\mu=0.5$, projected up at $5$ m s⁻¹: $a=-9.14$ m s⁻², so $s=\frac{25}{2(9.14)}=1.37$ m. At rest it slides back if $mg\sin\theta>\mu mg\cos\theta$ (here $29.4>25.5$, or $\tan30^\circ=0.577>0.5$). It stays at rest if $\mu\ge\tan\theta$.

1. Draw a force diagram: weight, $R$, friction, applied forces.
2. Resolve perpendicular to the surface to find $R$.
3. Decide the case: equilibrium ($F\le\mu R$), limiting ($F=\mu R$) or moving ($F=\mu R$ opposing motion).
4. Resolve along the surface (equilibrium) or use $F=ma$ (moving).
5. For a range of values, solve each limit separately and state an inequality. Use $g=9.8$ and give answers to 2 or 3 significant figures.