A short, right circular cylinder of weight $W$ rests in a horizontal, $V$ - shaped notch of angle $2\alpha$ . If the coefficient of friction is $\mu$, find the horizontal force parallel to the axis necessary for slipping to occur.

For equilibrium of the cylinder in vertical direction, we have 

                    $2N \sin \alpha = W$

$\therefore N = \frac{W}{2\sin \alpha }$

The force needed to pull the cylinder along horizontal direction,

                                          $F = 2f = 2\mu N$

                                             $$\begin{gathered} = 2\mu \times \frac{W}{2\sin \alpha } \\ = \frac{\mu W}{\sin \alpha } \end{gathered}$$