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

F = (M+m)g/μ

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 and individual blocks.

1. Consider the system (M+m):

When the block 'm' is stationary with respect to the wedge 'M', both move together with a common acceleration, say 'a'.
The total mass of the system is $(M+m)$.
The net external horizontal force acting on the system is F.
According to Newton's second law for the combined system: $F = (M+m)a \quad \text{(Equation 1)}$

2. Analyze forces on the small block 'm' (Free Body Diagram of m):

• Horizontal forces: The wedge 'M' pushes the block 'm' to the right with a normal force, let's call it $N_1$. Since 'm' is accelerating horizontally with 'a', the net horizontal force on 'm' must be $ma$. $N_1 = ma \quad \text{(Equation 2)}$
• Vertical forces:
  • Gravitational force: $mg$ (acting downwards).
  • Friction force: $f$ (acting upwards, to prevent 'm' from sliding down due to gravity).
    Since 'm' is stationary vertically relative to 'M', the net vertical force on 'm' must be zero. $f = mg \quad \text{(Equation 3)}$

3. Apply friction condition:

For the block 'm' to remain stationary with respect to 'M', the static friction force 'f' must be less than or equal to the maximum static friction force, which is $\mu N_1$. $f \le \mu N_1$ For the minimum horizontal force F, the friction force must be at its maximum possible value, i.e., $f = \mu N_1$. This is the limiting case where the block is just about to slide down. $f = \mu N_1 \quad \text{(Equation 4)}$

4. Combine and solve for F:

Substitute Equation 3 ($f=mg$) into Equation 4: $mg = \mu N_1$ Now, substitute Equation 2 ($N_1=ma$) into this equation: $mg = \mu (ma)$ Cancel 'm' from both sides (assuming $m \neq 0$): $g = \mu a$ Solve for the acceleration 'a': $a = \frac{g}{\mu}$ Finally, substitute this value of 'a' back into Equation 1 to find the minimum force F: $F = (M+m)a$ $F = (M+m)\frac{g}{\mu}$

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