A block of mass $m$ is placed on a smooth inclined wedge ABC of inclination $ heta$ as shown in the figure. T

Question

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

Accepted Answer

1. **Frame of Reference:** Analyze the problem from the non-inertial frame of reference of the wedge, which is accelerating with acceleration $a$ towards the right.
2. **Pseudo Force:** In this non-inertial frame, a pseudo force of magnitude $ma$ acts on the block of mass $m$. Since the wedge accelerates to the right, the pseudo force acts to the **left**.
3. **Forces on the Block:**
   - Weight ($mg$) acting vertically downwards.
   - Normal Reaction ($N$) acting perpendicular to the incline.
   - Pseudo Force ($ma$) acting horizontally towards the left.
4. **Equilibrium Condition:** For the block to remain stationary relative to the wedge, the net force along the incline must be zero.
   - Component of weight down the incline: $mg \sin 	heta$.
   - Component of pseudo force up the incline: $ma \cos 	heta$.
5. **Calculation:** Equating the components along the incline:
   $$ ma \cos 	heta = mg \sin 	heta $$
   $$ a = g \frac{\sin 	heta}{\cos 	heta} $$
   $$ a = g 	an 	heta $$
   (Reference: NCERT Class 11, Physics Part I, Chapter 5, Laws of Motion).

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