A block of mass m is placed on a smooth wedge of inclination . Th… - Vidyadip

MCQ

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\cos \theta $
B
$mg\sin \theta $
C
$mg$
✓
$mg/\cos \theta $

✓

Answer

Correct option: D.

$mg/\cos \theta $

d
(d) When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right.

For the condition of equilibrium

$mg\,\sin \theta = ma\cos \theta $ $⇒$ $a = \frac{{g\sin \theta }}{{\cos \theta }}$

$\therefore $ Force exerted by the wedge on the block

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

$R = mg\cos \theta + m\left( {\frac{{g\sin \theta }}{{\cos \theta }}} \right)\sin \theta $

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

$R = \frac{{mg}}{{\cos \theta }}$

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