Question

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 will be ($g$ is acceleration due to gravity)

• A. $mg \cos \theta;$
• B. $mg \sin\theta;$
• C. mg
• D. $\frac {mg} {cos\, \theta}$

Answer 1

The Correct Option is D

To solve this problem, we need to determine the force exerted by the wedge on the block when the entire system is accelerated horizontally. The important condition here is that the block does not slip on the wedge, meaning the acceleration must be such that the block remains in relative equilibrium with the wedge.

Firstly, let's analyze the forces acting on the block:

• The gravitational force acting vertically downward: $mg$
• The normal force exerted by the wedge, acting perpendicular to the surface of the wedge: $N$
• A horizontal force due to the acceleration of the wedge. Let's denote this horizontal acceleration as $a$.

Since the block does not slip, the component of gravitational force parallel to the wedge surface must be balanced by the pseudo force due to horizontal acceleration. The components of the force are:

• Normal force $N \cos \theta = mg \cos \theta$ (as the wedge supports the block vertically)
• The force required to prevent slipping, $ma = mg \tan \theta$

From these relationships, we know:

$$ ma = mg \tan \theta $$

Simplifying gives:

$$ a = g \tan \theta $$

The resultant force exerted by the wedge, which combines both normal and friction forces (acting perpendicular and parallel to the incline, respectively), is:

$$ N = \sqrt{(mg)^2 + (ma \cos \theta)^2} $$

Substituting $a = g \tan \theta$:

We calculate the net force exerted on the block as:

$$ N = \frac{mg}{\cos \theta} $$

This calculation accounts for all forces acting on the block while maintaining equilibrium. Hence, the correct answer is the fourth option:

$\frac {mg} {\cos \theta}$

The value of the force exerted by the wedge on the block ensures stability without slipping, correctly incorporating both the normal and horizontal components of acceleration.