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Step-by-Step Detailed Solution
Step 1: Identify the forces acting on the block
The block of mass $m$ is placed on a smooth wedge inclined at an angle $\theta$. The forces acting on the block in an inertial frame are:
• The force of gravity ($mg$) acting vertically downwards.
• The normal reaction ($N$) from the wedge acting perpendicular to the wedge surface.
Step 2: Consider the horizontal acceleration and pseudo force
Since the wedge-block system is accelerated horizontally to prevent the block from slipping, in the non-inertial reference frame of the wedge, a pseudo force acts on the block horizontally in the direction opposite to the wedge’s acceleration. Let this pseudo force be $ma$, where $a$ is the horizontal acceleration of the wedge.
Step 3: Resolve forces along the plane of the wedge
In the wedge’s reference frame, the block is at rest on the inclined surface, meaning its net acceleration along the plane is zero. The resolved components along the plane and perpendicular to the plane are considered with the pseudo force acting along the horizontal.
Step 4: Balance forces perpendicular to the inclined surface
The normal force $N$ lies perpendicular to the wedge surface. We can decompose the weight $mg$ into components parallel and perpendicular to the inclination, and also place the pseudo force $ma$ horizontally. When the block is on the verge of not slipping, the net force along the plane is zero, and the net force perpendicular to the plane balances as follows:
By geometric considerations or by resolving the combined effect of $mg$ and $ma$ perpendicular to the wedge, we find:
$$
N = \frac{mg}{\cos \theta}.
$$
Step 5: Conclude the normal reaction
Thus, the force exerted by the wedge on the block - the normal reaction $N$ - is:
$$
N = \frac{mg}{\cos \theta}.
$$
