A block of mass $m$ is placed on a smooth wedge of inclination $ heta$. The whole system is accelerated horiz

Question

A block of mass $m$ is placed on a smooth wedge of inclination $	heta$. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block ($g$ is acceleration due to gravity) will be:

Accepted Answer

1. **Identify the Force:** The force exerted by the wedge on the block is the Normal Reaction force ($N$), which acts perpendicular to the surface of the wedge.
2. **Analyze Motion (Inertial Frame):** The block moves horizontally with the same acceleration $a$ as the wedge. It has no vertical acceleration.
3. **Resolve Forces:**
   - Vertical forces: The vertical component of the Normal force ($N \cos	heta$) acts upwards, and gravity ($mg$) acts downwards.
   - Since vertical acceleration is zero, these forces must balance: 
     $$ N \cos	heta = mg $$ [Source 125]
4. **Solve for N:**
   $$ N = \frac{mg}{\cos	heta} $$

*Alternative (Non-Inertial Frame):* Using the concept of pseudo force ($ma$) acting opposite to the acceleration [Source 35]. The components perpendicular to the incline balance out:
$$ N = mg \cos	heta + ma \sin	heta $$
To prevent slipping, the components along the incline balance: $ma \cos	heta = mg \sin	heta \implies a = g 	an	heta$.
Substituting $a$: $N = mg \cos	heta + m(g 	an	heta)\sin	heta = mg(\cos	heta + \sin^2	heta/\cos	heta) = mg/\cos	heta$.

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