Question

A mass m is placed on a wedge (triangular block) of mass M. The wedge moves on a smooth horizontal surface. What should be the force F applied on the wedge to the right, so that m remains stationary with respect to wedge? (Ignore any friction)

• A. $g \tanθ$
• B. $mg\cosθ$
• C. $(M+m)g\tanθ$
• D. $(M+m)g\cosecθ$

Hint:

The acceleration of the wedge must be $g\tanθ$ for the block to not slide.

Answer 1

The Correct Option is C

To solve this problem, we need to ensure that the mass \(m\) remains stationary with respect to the wedge when a force \(F\) is applied. This means that the acceleration of the wedge and the effective acceleration of the block \(m\) due to gravity along the incline must be the same.

Explanation and Derivation:

1. The gravitational force on the block \(m\) has a component along the incline, which is \(mg \sin \theta\), where \(\theta\) is the angle of the incline.
2. To keep the block stationary with respect to the wedge, the horizontal component of the gravitational force must be balanced by the horizontal force \(F\) applied on the wedge. This means we need to equate the horizontal acceleration caused by \(F\) to the horizontal component of the gravitational acceleration of \(m\).
3. The acceleration of the wedge \(a\) due to the force \(F\) is given by:
4. \(a = \frac{F}{M+m}\)
5. For mass \(m\) to remain stationary, we equate:
6. \(g \sin \theta = \frac{F}{M+m} \cos \theta\)
7. Simplifying for \(F\), we get:
8. \(F = (M+m)g \tan \theta\)

Conclusion:

The correct force \(F\) required to keep the mass \(m\) stationary on the wedge is \((M+m)g \tan \theta\).

Thus, the correct answer is \((M+m)g\tan\theta\).