A solid cylinder with `r= 0.1 m` and mass `M = 2 kg` is placed such that it is in contact with the vertical and a horizontal surface as shown in Fig. The coefficient of friction is `mu = (1//3)` for both the surfaces. Find the distance (in `CM`) from the centre of the cylinder at .which a force `F = 40 N` should be applied vertically so that the cylinder just starts rotating in anticlockwise direction.

Answer

6

Explanation

For equilibrium of the cylinder in the horizontal direction
`N_(1)=muN_(2)`……….i

In the vertical direction
`N_(2)+muN_(1)=F+mg`…….ii
Solving these two equations with `F=40N, M=2kg` and `mu=1/3`, we get
`N_(1)=18N`
and `N_(2)=54N`
Now, `Fd=mu(N_(1)+N_(1))r`
`:. d=(M(N_(1)+N_(2))r)/F`
`=((1/3)(18+50)(0.1))/40=0.06m`