What Are the Dynamics of Two Objects in Motion with No Friction?

[bubblewrap]

Homework Statement

Object B with mass m is sliding on an object A with mass M. Object A is being pulled by a string with acceleration A. There is no friction. (Picture Uploaded)

1) Find the maximum value of A that allows m to stay on M (sliding)
2) Find the x and y value of the acceleration of object B
3) In this case, what is the value of the force T (the force that is pulling Object A)
4) Find the value of the normal force F that is acting on object A.
5) As an alternate version of this question, suppose that force T was given instead of acceleration A.

Homework Equations

Newton's Law; F=ma

The Attempt at a Solution

1) In the case of object B, there are three forces; the normal force N, mg and inertia mA
setting a new coordinate with X and Y where Y is perpendicular to the hypotenuse and X perpendicular to Y, I showed that acceleration in Y direction must equal 0 and the A in this case is the A for 1). The answer I got was (mgcosθ-N)/msinθ

2) Using the coordinate and the forces from 1), I figured out that X direction of the acceleration equals mAcosθ+mgcosθ. The angle between X and x is θ, so x direction would be -(mAcosθ+mgsinθ)cosθ, and y direction would be -(mAcosθ+mgsinθ)sinθ

3) Because there is only one force given from the outside, which is T, T=∑ma for a in the x direction.
T=-(mAcosθ+mgsinθ)cosθ+MA

4) Normal force equals the force acting on the surface which is object B's force in the y direction
F=(mAcosθ+mgsinθ)sinθ
5) This would be a simple change, just writing the answer from 3) like A= f(T)

Overall I have tried to solve all these and came up with answers. But I'm not sure if they are correct.
If someone could check them I would be grateful.

 

[CWatters]

1) Find the maximum value of A that allows m to stay on M (sliding)

1) In the case of object B, there are three forces; the normal force N, mg and inertia mA
setting a new coordinate with X and Y where Y is perpendicular to the hypotenuse and X perpendicular to Y, I showed that acceleration in Y direction must equal 0 and the A in this case is the A for 1). The answer I got was (mgcosθ-N)/msinθ

You can simplify that further.

Hint: If object B is "only just" in contact with A (eg it's just about to leave the surface) what is the value of N ?

 

[haruspex]

1) In the case of object B, there are three forces; the normal force N, mg and inertia mA

This is rather confusing because A is being used both as the label on an object and as an acceleration. I assume you mean an inertial force ma, where a is the acceleration of A.
Looks like you are using a noninertial frame here. It would help if you were to state that at the start.

 

[CWatters]

Once simplified the answer just looks more obviously correct :-)

 

[bubblewrap]

Once simplified the answer just looks more obviously correct :-)

Then N=0 so A=g/tanθ?

 

[haruspex]

Then N=0 so A=g/tanθ?

Looks right to me.

 

[CWatters]

and me.

 

[bubblewrap]

How about the rest? Are they correct as well?