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 ($g$ is acceleration due to gravity) will be:

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

Answer 1

The correct option is C $\frac{mg}{\cos \theta }$


FBD of $m$ w.r.t. incline plane.

$ma\cos \theta =mg\sin \theta ....\left(i\right)$
$N=mg\cos \theta +ma\sin \theta ....\left(ii\right)$

$\therefore N=mg\cos \theta +m\sin \theta \times g\times \frac{\sin \theta }{\cos \theta }$
$\Rightarrow N=\frac{mg}{\cos \theta }$