Problem Statement:
Three blocks A, B, and C of equal mass m are placed one over the other on a smooth horizontal ground. The coefficient of friction between any two blocks A, B, and C is 1/2. We need to find the maximum value of the mass of block D so that the blocks A, B, and C move without slipping over each other.

Assumptions:
1. The blocks A, B, C, and D are in vertical alignment.
2. The blocks A, B, and C are in contact with each other and the ground.
3. The coefficient of friction is the same for all pairs of blocks.

Solution:

1. Analysis of Block A:
Since the three blocks A, B, and C have equal masses and are in contact with each other, the forces acting on block A are as follows:
- Weight (mg) acting vertically downward.
- Normal force (N) exerted by block B.
- Frictional force (f1) exerted by block B.

The equation of motion for block A can be written as:
mg - f1 = ma ...(1)

2. Analysis of Block B:
The forces acting on block B are as follows:
- Weight (mg) acting vertically downward.
- Normal force (N) exerted by block C.
- Frictional forces (f1 and f2) exerted by blocks A and C, respectively.

The equation of motion for block B can be written as:
mg - f1 - f2 = ma ...(2)

3. Analysis of Block C:
The forces acting on block C are as follows:
- Weight (mg) acting vertically downward.
- Frictional force (f2) exerted by block B.

The equation of motion for block C can be written as:
mg - f2 = ma ...(3)

4. Conditions for No Slipping:
For the blocks A, B, and C to move without slipping over each other, the frictional forces between the blocks should be at their maximum values. This occurs when the blocks are on the verge of slipping.

The maximum frictional force (f_max) can be given by:
f_max = μN

5. Solving the Equations:
Using the equation of motion (3) for block C and the condition for no slipping, we can write:
f2 = μN ...(4)

Substituting equation (4) into equation (2), we get:
mg - f1 - μN = ma ...(5)

Substituting equation (5) into equation (1), we get:
mg - (mg - ma - μN) = ma
ma + μN = mg ...(6)

Adding equations (5) and (6), we get:
2ma + 2μN = 2mg

Since the masses of blocks A, B, and C are equal, we can write:
2m(a + μN) = 2mg

Simplifying the equation, we get:
a + μN = g

Substituting the value of μN from equation (4), we get:
a + f2 = g

6. Maximum Mass of Block D:
To find the maximum mass of block D