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.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/943143/original_10.png)
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
2nd law on the disc,
⇒2F−f=ma ...(i)
Applying equation of torque about the centre of the disc,
τcom=Icom×α
where
τcom= Net torque about centre of mass of the disc.
⇒τN+τmg+(τF)1+(τF)2+τf=Iα
⇒0+0+FR+0+fR=Iα
[
τN=τmg=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.]
⇒(F+f)R=Iα ...(ii)
a=Rα, condition for pure rolling
& I=mR22
Substituting in Eq.
(ii),
(F+f)R=mR22×aR
∴F+f=ma2...(iii)
From Eq.
(i) & (iii), we get
f=0
Hence according to question,
f=nF or
0=nF
∴n=0