How Does Newton's Third Law Apply in a Frictionless Two-Block System?

[ScullyX51]

Homework Statement

Two blocks are accelerated across a horizontal frictionless surface as shown. The coefficient of static friction between the two blocks is 0.7, and the coefficient of kinetic friction is 0.5, use M= 1.0 kg. When F=1.2 N, frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. In this case, what is the acceleratiob of the two block system? (The picture is of mass 2m on the bottom with mass M on top of it, and F pointing to the right)
2) What is the horizontal component of the force the large block exerts on the small block
3) Suppose that F is increased. What is the maximum acceleration that mass 2M can have, without mass M slipping off?

Homework Equations

F=ma

The Attempt at a Solution

I got the first portion of the problem right. I just solved for f=ma, and plugged in the f=1.2. I figured that the coefficients of friction given were not relevant here because the system is accelerating. For an answer I got a=.4 m/s² to the right.
2) For the second part of the question: the horizontal component the large block exerts on the small block, I put .4 to the left. I thought that according to Newton's third law the force is equal in magnitude and opposite in direction, but it is saying this answer is wrong. It says the correct answer is .4 to the right. Why is it the same?

 

[Doc Al]

I don't understand how you are applying Newton's 3rd law in this case.

Realize that both blocks have the same acceleration, which equals 0.4 m/s^2 to the right. To find the force on the top block (which is exerted by the bottom block), apply Newton's 2nd law.

 

[thomate1]

Your solution for the first part of the problem is correct, the acceleration of the two block system is 0.4 m/s^2 to the right. However, for the second part of the problem, the horizontal component of the force the large block exerts on the small block is also 0.4 N to the right. This is because although the forces are equal in magnitude and opposite in direction, they are acting on different masses (2m and M). Therefore, the acceleration of the small block will be equal to the acceleration of the entire system, and the horizontal component of the force it experiences will also be equal to the force applied to the entire system. It is important to remember that Newton's third law applies to individual objects, not the system as a whole.
For the third part of the problem, the maximum acceleration that mass 2M can have without mass M slipping off can be calculated using the coefficient of static friction between the two blocks. The maximum acceleration is given by μs*g, where μs is the coefficient of static friction and g is the acceleration due to gravity. In this case, the maximum acceleration would be 0.7*9.8 = 6.86 m/s^2. If F is increased beyond this value, the small block will start to slip off the larger block.