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 $N=mg\cos \theta +\frac{mg\sin \theta \cdot \sin \theta }{\cos \theta }$.
$N=\frac{mg({\cos }^2\theta +{\sin }^2\theta )}{\cos \theta }=\frac{mg}{\cos \theta }$