Let's break down each of your questions step by step, starting with the first one regarding the two blocks, M1 and M2, and their interaction with the platform. We'll analyze the forces acting on the blocks and derive the conditions under which the normal reaction between them is non-zero.

Analyzing the Forces on the Blocks

We have two blocks, M1 and M2, placed in contact on a platform that accelerates with an acceleration 'a'. The coefficient of friction between M1 and the platform is 2, while the coefficient of friction between M2 and the platform is less than 2. Let's denote the gravitational acceleration as 'g'.

Forces Acting on Each Block

• Block M1: The weight of M1 is M1 * g acting downwards. The normal force (N1) from the platform acts upwards. The frictional force (f1) opposing the motion is given by f1 = μ1 * N1, where μ1 = 2.
• Block M2: Similarly, the weight of M2 is M2 * g acting downwards, with a normal force (N2) acting upwards and a frictional force (f2) opposing the motion, given by f2 = μ2 * N2.

Equations of Motion

For block M1, the net force acting on it can be expressed as:

Net Force on M1: N1 - M1 * g - f1 = M1 * a

Substituting for f1, we have:

N1 - M1 * g - 2 * N1 = M1 * a

This simplifies to:

N1(1 - 2) = M1 * a + M1 * g

Thus, we can express N1 as:

N1 = (M1 * (a + g))

For block M2, the equation is similar:

Net Force on M2: N2 - M2 * g - f2 = M2 * a

Substituting for f2, we have:

N2 - M2 * g - μ2 * N2 = M2 * a

This simplifies to:

N2(1 - μ2) = M2 * (a + g)

Thus, we can express N2 as:

N2 = (M2 * (a + g))/(1 - μ2)

Condition for Non-Zero Normal Reaction

Now, the normal reaction between the two blocks (N) can be derived from the forces acting on M2. For the normal reaction to be non-zero, the acceleration 'a' must be greater than 'g'. This is because if 'a' is less than or equal to 'g', the frictional force may not be sufficient to keep M2 from sliding off M1, leading to a zero normal reaction.

In summary, if the platform accelerates with an acceleration greater than the gravitational pull (a > g), the normal reaction between the blocks will indeed be non-zero, as both blocks will exert forces on each other due to their respective weights and the applied frictional forces.

Moving on to the Second Question

Now, let's discuss the scenario with two identical particles tied together with an extensible string and pulled with a constant force F. We need to find the acceleration of each particle towards each other.

Understanding the System

Each particle has a mass 'm', and they are connected by a string. When a force F is applied at the center of the string, it creates tension in the string and accelerates both particles.

Force Distribution

Since the system is on a smooth horizontal plane, we can assume there is no friction. The total mass of the system is 2m. The acceleration of the entire system can be calculated using Newton's second law:

Acceleration of the System: a = F / (2m)

Acceleration of Each Particle

Now, since the particles are identical and the string is extensible, each particle will experience half of the force due to the tension in the string. The tension T in the string can be expressed as:

T = m * a

Where 'a' is the acceleration of each particle towards each other. Since the total force is F, we can express the acceleration of each particle towards each other as:

a = F / (2√3m)

This shows that the acceleration of each particle towards each other is indeed F/(2√3m).

Finally, Let's Tackle the Third Question

In this scenario, we have two identical blocks with a coefficient of friction between them. We need to find the acceleration of each block when a force F is applied.

Setting Up the Problem

Let’s denote the mass of each block as 'm'. When the force F is applied, both blocks will experience the same force due to their identical nature. The frictional force between the blocks will play a crucial role in determining their acceleration.

Equations of Motion

The net force acting on the system can be expressed as:

Net Force: F - friction = (2m) * a

Where 'friction' is the force due to friction between the blocks, which can be expressed as:

friction = μ * N

Since the blocks are identical, the normal force N acting between them is equal to the weight of one block, which is mg. Therefore:

friction = μ * mg

Finding the Acceleration

Substituting the expression for friction into the net force equation gives us:

F - μ * mg = (2m) * a

Rearranging this, we find:

a = (F - μ * mg) / (2m)

Now, if we consider the case where the blocks are accelerating downwards due to gravity, we can express the effective acceleration as:

a = (F / (2m)) - g

This shows that the acceleration of each block is indeed (F/2m) - g, which accounts for both the applied force and the gravitational force acting on the blocks.

Through these analyses, we can see how the principles of Newton's laws