Question

A cylinder and wedge of same masses with a vertical face, touching each other, move along two smooth inclined planes forming the some angle $\alpha$ and $\beta$ respectively with the horizontal. Determine the force of normal $N$ (in newton) exerted by the wedge on the cylinder, neglecting the friction between them. If $m=\frac{1}{\sqrt{(}3)}kg$, $\alpha ={60}^0,\beta ={30}^0$ and $g=10m/s^2$

Answer 1

Since wedge is at equilibrium

$\therefore N=-mg\cos \alpha =$ force exerted by wedge on cylinder)

$\therefore N=\frac{1}{\sqrt{3}}\times 10\times \frac{1}{2}=\frac{5}{\sqrt{3}}N$

$\therefore$ Normal force $=\frac{5}{\sqrt{3}}N$