Question

A cylinder of radius $r=0.1m$ and mass $M=2kg$ is placed such that it is in contact with a vertical wall and a horizontal surface as shown in the figure. The coefficient of static friction $\mu$ is (1/3) for both the surfaces. The distance $d=k\times {10}^{-2}m$ from the centre of the cylinder at which a force $F=40N$ should be applied so that the cylinder just starts rotating in the anticlockwise direction. Take $g=10m/s^2$

Answer 1

For equilibrium of the cylinder$\sum F_x=0\Rightarrow N_1=\mu N_2$$\sum F_y=0\Rightarrow N_2+\mu N_1=F+mg$Putting $F=40N,M=2kg$ and $\mu =\frac{1}{3},N_1=18$ and $N_2=54N$

Further,

$F.d=\mu (N_1+N_2)r$$\Rightarrow d=\frac{\mu (N_1+N_2)r}{F}=\frac{\frac{1}{3}(18+54)\left(0.1\right)}{40}=0.06m$