Question

In the adjoining figure, the coefficient of friction between wedge (of mass M) and block (of mass m) is $\mu$.

Find the minimum horizontal force F required to keep the block stationary with respect to wedge.

Answer 1

$\frac{(M+m)g}{\mu}$

Answer 2

To find the minimum horizontal force F required to keep the block stationary with respect to the wedge, we analyze the forces acting on the system.

1. Acceleration of the combined system: Since the block is stationary with respect to the wedge, both the block (mass m) and the wedge (mass M) move together with the same acceleration, let's call it 'a'. Considering the combined system (M + m), the net horizontal force is F. According to Newton's Second Law: $F = (M + m)a$ Therefore, the acceleration of the system is: $a = \frac{F}{M+m}$

2. Free Body Diagram (FBD) of the block (m): Let's consider the forces acting on the block 'm' independently.

• Horizontal forces: The wedge exerts a normal force (N) on the block, pushing it horizontally. This normal force is responsible for accelerating the block along with the wedge. $N = ma$ Substituting the expression for 'a': $N = m \left(\frac{F}{M+m}\right)$

• Vertical forces: The block experiences a downward gravitational force (mg). To keep the block stationary vertically with respect to the wedge, an upward friction force (f) must act on the block, exerted by the wedge. For vertical equilibrium: $f = mg$

3. Condition for minimum force F: The friction force 'f' acting on the block is static friction. For the block to remain stationary, the static friction required must be less than or equal to the maximum possible static friction, which is given by $\mu N$. $f \le \mu N$ For the minimum force F required to keep the block stationary, the friction force will be at its maximum static value, i.e., $f = \mu N$. Substituting $f = mg$: $mg = \mu N$

4. Solving for F: Now, substitute the expression for N from the horizontal forces equation ($N = m \frac{F}{M+m}$) into the friction condition ($mg = \mu N$): $mg = \mu \left( m \frac{F}{M+m} \right)$ We can cancel 'm' from both sides (assuming m ≠ 0): $g = \mu \frac{F}{M+m}$ Rearranging the equation to solve for F: $F = \frac{g(M+m)}{\mu}$

This is the minimum horizontal force F required to keep the block stationary with respect to the wedge.

Explanation of the solution: The block and wedge move together with acceleration $a = F/(M+m)$. The normal force from the wedge on the block provides this acceleration, so $N = ma = mF/(M+m)$. To prevent the block from sliding down, the upward static friction force must balance its weight, so $f = mg$. For the minimum force F, the static friction reaches its maximum value, $f = \mu N$. Equating $mg = \mu N$ and substituting $N$, we get $mg = \mu (mF/(M+m))$, which simplifies to $F = (M+m)g/\mu$.

Answer: The minimum horizontal force F required is $\frac{(M+m)g}{\mu}$.