2004 AIPMT Laws of MotionPseudo Force and Accelerated Frames Medium Question 58 of 72 in Laws of Motion

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 $mg$
C $\dfrac{mg}{\cos\theta}$
D $mg\cos\theta$

The answer is C.

Solution

The wedge is smooth, so the only force it exerts on the block is the normal reaction N, perpendicular to the inclined surface.

Forces on the block: weight mg vertically downward and normal reaction N at angle $\theta$ to the vertical.
The block does not slip, so it moves with the wedge with a purely horizontal acceleration a. Its vertical acceleration is zero.
Vertical balance: $N\cos\theta = mg$
$\Rightarrow N = \dfrac{mg}{\cos\theta}$
(Horizontal: $N\sin\theta = ma$, which gives the required acceleration $a = g\tan\theta$.)
So the force exerted by the wedge on the block is $\dfrac{mg}{\cos\theta}$.

Concept Normal reaction on block on accelerated wedge

Laws of Motion in past papers

72 questions from this chapter have appeared across 27 exam years.