Question

Two forces of magnitude $F$ are acting on a uniform disc kept on a rough horizontal surface as shown in the figure. If the frictional force by the horizontal surface on the disc is $nF$, find the value of $n$.

Answer 1

The force $F$ acting at the centre of the disc will produce zero torque about centre. However, force $F$ acting at the topmost point of the disc will tend to make it rotate in clockwise direction.


Hence, friction will act towards left at the bottom to oppose the tendency of slipping. Let $f$ be the friction force acting on the disc.
Applying Newton's $2^{nd}$ law on the disc,
$\Rightarrow 2F-f=ma...\left(i\right)$
Applying equation of torque about the centre of the disc,
${\tau }_{com}=I_{com}\times \alpha$
where ${\tau }_{com}=$ Net torque about centre of mass of the disc.
$\Rightarrow {\tau }_N+{\tau }_{mg}+({\tau }_F{)}_1+({\tau }_F{)}_2+{\tau }_f=I\alpha$
$\Rightarrow 0+0+FR+0+fR=I\alpha$
[${\tau }_N={\tau }_{mg}=0$ because $N$ and $mg$ pass through COM.
Both $F$ and $f$ are tending to rotate disc in clockwise direction, hence the torque will add.]
$\Rightarrow (F+f)R=I\alpha ...\left(ii\right)$

$a=R\alpha$, condition for pure rolling $\&I=\frac{m{R}^2}{2}$
Substituting in Eq.$\left(ii\right)$,
$(F+f)R=\frac{m{R}^2}{2}\times \frac{a}{R}$
$\therefore F+f=\frac{ma}{2}...\left(iii\right)$
Adding Eq.$\left(i\right)\&\left(iii\right)$, we get $f=0$
Hence according to question,
$f=nF$ or $0=nF$
$\therefore n=0$