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 (mass m) stationary with respect to the wedge (mass M), we analyze the forces acting on each block.

1. Common Acceleration:
   Since the block m is stationary with respect to the wedge M, both blocks move together with the same horizontal acceleration, let's call it a.

2. Free Body Diagram (FBD) for Block m:

   • Vertical forces:
     • Weight: mg acting downwards.
     • Friction force: f acting upwards, provided by the wedge, opposing the tendency of m to slide down.
   • Horizontal forces:
     • Normal force: N acting horizontally to the right, exerted by the wedge M on block m.

   For the block m to be stationary vertically, the net vertical force must be zero:
   f = mg (Equation 1)

   For the block m to accelerate horizontally with a, the net horizontal force must be ma:
   N = ma (Equation 2)

3. Condition for Minimum Force F:
   To keep the block m from sliding down, the static friction f must be sufficient to balance its weight mg. The maximum possible static friction is f_max = μN, where μ is the coefficient of friction between the wedge and the block.
   For the minimum force F, the block m is on the verge of sliding downwards, meaning the friction force f is at its maximum possible value:
   f = μN (Equation 3)

4. Solving for Acceleration a:
   Substitute Equation 1 into Equation 3:
   mg = μN

   Now, substitute Equation 2 (N = ma) into this equation:
   mg = μ(ma)
   Cancel m from both sides (assuming m ≠ 0):
   g = μa
   Therefore, the required acceleration a is:
   a = g/μ

5. Free Body Diagram (FBD) for the Combined System (M + m):
   Consider the wedge M and block m as a single system of total mass (M + m).

   • Horizontal forces:
     • Applied force: F acting horizontally to the right.
   • Vertical forces:
     • Total weight: (M + m)g acting downwards.
     • Normal force from the ground: N_ground acting upwards. (Assuming the ground is frictionless, there are no horizontal forces from the ground).

   Applying Newton's second law in the horizontal direction for the combined system:
   F = (M + m)a (Equation 4)

6. Calculating Minimum Force F:
   Substitute the value of a from step 4 into Equation 4:
   F = (M + m)(g/μ)

   Thus, the minimum horizontal force F required is:
   F = (M + m)g / μ

The block m stays stationary relative to the wedge M only if they both accelerate together. The normal force N exerted by the wedge on the block provides the necessary horizontal acceleration a for block m (N = ma). To prevent block m from falling, the upward static friction f must balance its weight mg (f = mg). For the minimum applied force F, the friction must be at its maximum possible value, f = μN. Combining these, we find a = g/μ. Then, applying Newton's second law to the entire system (M+m) gives F = (M+m)a, leading to the result F = (M+m)g/μ.