Topic: Newton's laws, force and inertia, momentum, impulse· Laws of Motion

A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is

A block of mass m is placed on a smooth inclined wedge ABC of inclination $\theta$ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and $\theta$ for the block to remain stationary on the wedge is

block on accelerating wedge ABC

1. $a=\dfrac{g}{\mathrm{cosec}\,\theta}$

2. $a=g\cos\theta$

3. $a=\dfrac{g}{\sin\theta}$

4. $a=g\tan\theta$

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Answer: 4. $a=g\tan\theta$

For the block to stay at rest on the smooth wedge accelerating right with $a$, in the wedge frame use a pseudo force $ma$ on the block.

Forces along the incline: $mg\sin\theta$ down the slope and $ma\cos\theta$ up the slope. Balance: $ma\cos\theta=mg\sin\theta$.

$$a=g\tan\theta$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A block of mass m is placed on a smooth inclined wedge ABC of inclination θ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and θ for the block to remain stationary on the wedge is

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