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 `sum F_(x) = 0 rArr N_(1) = mu N_(2)`
`sum F_(y) = 0 rArr N_(2) + mu N_(1) = F + mg`
Putting F = 40 N, M = 2 kg and `mu = (1)/(3), N_(1) = 18 and N_(2) = 54 N`
Furthuer, F.d `= mu(N_(1) + N_(2)) r`
`rArr" "d = (mu(N_(1)+N_(2))r)/(F) = ((1)/(3)(18+54)(0.1))/(40) = 0.06 m`