A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first block. The surfaces of the blocks are rough. A constant force of magnitude F is applied to the first block.

Assume that the upper block does not slip on the lower block, and find the magnitude of the acceleration of each block in terms of m and F.

F = 4m*a.

a = F/4m.

3M

To find the magnitude of acceleration for each block, we need to consider the forces acting on each block.

For the lower block with mass 3m:
1. The force applied (F) causes an acceleration (a1) on the lower block.
2. The weight of the lower block (3mg) acts vertically downward.
3. The normal force (N1) from the upper block acts vertically upward.
4. The friction force (friction1) acts horizontally in the opposite direction to the applied force.

For the upper block with mass m:
1. The force of friction (friction2) between the upper and lower block acts horizontally.
2. The force applied (F) causes an acceleration (a2) on the upper block.
3. The weight of the upper block (mg) acts vertically downward.
4. The normal force (N2) from the lower block acts vertically upward.

Now let's define the equations of motion for each block:

For the lower block:
ΣF1 = ma1

1. F - friction1 = 3ma1 (1) [Sum of horizontal forces]
2. N1 - 3mg = 0 (2) [Sum of vertical forces]

For the upper block:
ΣF2 = ma2

1. friction2 - F = ma2 (3) [Sum of horizontal forces]
2. N2 - mg = 0 (4) [Sum of vertical forces]

To solve these equations, we need to find the friction forces. The friction force is given by:

friction1 = μ1N1 (5) [Friction force for the lower block]
friction2 = μ2N2 (6) [Friction force for the upper block]

where μ1 and μ2 are the coefficients of friction for the lower and upper block interfaces, respectively.

From equations (2) and (4), we have:
N1 = 3mg (7)
N2 = mg (8)

Substituting equations (7) and (8) into equations (5) and (6), we get:
friction1 = μ1(3mg) (9)
friction2 = μ2(mg) (10)

Substituting equations (9) and (10) into equations (1) and (3), we obtain:
F - μ1(3mg) = 3ma1 (11)
μ2(mg) - F = ma2 (12)

Now, we can solve equations (11) and (12) simultaneously to find the acceleration of each block in terms of m and F.

Let's manipulate equation (11) and equation (12) to solve for a1 and a2:
1. From equation (11): F - μ1(3mg) = 3ma1
Rearranging, we get: a1 = (F - μ1(3mg))/(3m) (13)

2. From equation (12): μ2(mg) - F = ma2
Rearranging, we get: a2 = (μ2(mg) - F)/m (14)

Thus, the magnitude of acceleration for the lower block (a1) is given by:
a1 = (F - μ1(3mg))/(3m)

And the magnitude of acceleration for the upper block (a2) is given by:
a2 = (μ2(mg) - F)/m

To find the magnitude of the acceleration of each block, we need to consider the forces acting on each block. Let's analyze the forces separately for each block.

For the first block:
1. The applied force F is the external force acting on the first block.
2. Since there is no friction on the horizontal surface, there are no other forces acting on the first block.

For the second block:
1. The weight of the second block is mg, where g is the acceleration due to gravity.
2. There is a normal force acting on the second block due to the first block.
3. There is a frictional force between the two blocks.

Let's break down the forces and apply Newton's second law (F = ma) to each block.

For the first block:
The only force acting on the first block is the applied force F. Therefore, we have:
F = (3m)a₁

For the second block:
The net force acting on the second block is the difference between the force due to the first block and the frictional force:
(3m)a₁ - μN = ma₂
Here, μ is the coefficient of friction between the two blocks, and N is the normal force.

Since the normal force N is equal to the weight of the second block, we have N = mg.

Now, let's find the expression for the frictional force. The frictional force depends on the coefficient of friction and the normal force. The coefficient of friction between the two blocks determines the roughness of their interaction. Let's denote it as μ₂.

The frictional force between the two blocks is given by:
Frictional force = μ₂N = μ₂mg

Substituting this into the equation for the net force on the second block, we have:
(3m)a₁ - μ₂mg = ma₂

We can now solve these two equations simultaneously to find the magnitudes of the acceleration for each block.