Question

A block of mass $m$ is placed on a smooth wedge of inclination $\theta$. The whole system is accelerated horizontally so that block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude:

• A. $mg$
• B. $mg\cos \theta$
• C. $mg/\cos \theta$
• D. $mg\tan \theta$

Answer 1

The correct option is C $mg/\cos \theta$

As there is no relative motion between block and wedge:

$ma\cos \theta =mg\sin \theta$

$⟹a=g\tan \theta$

The normal force by wedge on the block is

$\begin{array}{c}N=mg\cos \theta +ma\sin \theta \\ =mg\cos \theta +m\left(g\tan \theta \right)\sin \theta \\ =mg\cos \theta +mg\frac{{\sin }^2\theta }{\cos \theta }\\ =mg\left(\frac{{\cos }^2\theta +{\sin }^2\theta }{\cos \theta }\right)\\ =\frac{mg}{\cos \theta }\end{array}$