Question

A block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block will be ($g$ is acceleration due to gravity)

• A. $mg \cos \theta;$
• B. $mg \sin\theta;$
• C. $mg$
• D. $\frac {mg} {\cos\, \theta}$

Answer 1

Answer: (D)

The wedge is given an acceleration to the left.
$\therefore $ The block has a pseudo acceleration to the right, pressing against the wedge because of which the block is not moving.

$\therefore mg\,\sin\,\theta=ma \,\cos\,\theta$
or $a=\frac{g\,\sin\,\theta}{\cos\,\theta}$
Total reaction of the wedge on the block is
$N=mg \,\cos\,\theta+ma \,\sin\,\theta$.
or $N=mg\,\cos\, \theta+\frac{mg\,\sin\,\theta \cdot \sin\,\theta}{\cos\,\theta}$.
$N=\frac{mg(\cos^2 \,\theta+\sin^2\,\theta)}{\cos\,\theta}=\frac{mg}{\cos\,\theta}$