Forces acting on a block which is lying on another block

[Like Tony Stark]

Homework Statement

In the picture, the coefficient of static and dynamic friction for all contact surfaces are $0.4$ and $0.14$, respectively. The mass of $A$ is $73 kg$ and the mass of $B$ is $22 kg$. Determine the maximum force so that B doesn't slide. Then, suppose that the force applied to the system is greater than the force calculated previously, how much time would it take for the blocks to move $10 m$?

Relevant Equations

Newton's equations

Well, I'm having trouble with the free body diagrams. For $A$ we have
$y)$ weight, normal force, contact force with $B$, $F . sin(36.8°)$. And the acceleration is $0$ because we want to calculate the maximun force before moving.

$\Sigma \vec F = m . a_y$
$\vec N_A + \vec F . sin(36.8°) - \vec W_A -\vec Fc= 0$

$x)$ $F . cos(36.8°)$, friction force with the ground and friction force with $B$ and the acceleration is 0.

$F. cos(36.8°) - Fr - Fr_B=0$

$Fr$ and $Fr_B$ have the same direction, don't they? Because the friction with $B$ must be pushing $B$ towards, and then the reaction force is in the opposite direction.

For $B$ we have:
$y)$ weight and contact force (normal force)

$\vec N_B - \vec W_B =0$

$x)$ friction force with $A$
But is this force pointing to the right? And then, is the string applying a force on $B$?And then when I want to answer the second question, do they move equally? I mean, if $A$ traveled $10 m$ so did $B$

 

[haruspex]

normal force, contact force with B

The contact force with B could include a frictional component in general. Better to say normal force from ground, normal force from B.

the acceleration is 0 because we want to calculate the maximun force before moving.

No, it's the maximum force before B slides on A. A will be moving by then.

 

[Like Tony Stark]

The contact force with B could include a frictional component in general. Better to say normal force from ground, normal force from B.

No, it's the maximum force before B slides on A. A will be moving by then.

So the free body diagram for A is
$x) F.cos(36,8°)-Fr-Fr_{AB}=m.a_x$
$y) F.sin(36,8°)+N-P-F_{AB}=m.a_y$

And for B is
$x) Fr_{AB}=m.a_x$
$y) F_{AB}-P=m.a_y$

Is this correct?

And for the $y)$ axis of both bodies the acceleration is equal to 0, isn't it? Because they don't move upwards. Is this right?

 

[haruspex]

Is this correct?

Yes, except what is P in these equations? Not the same in both, I assume.

And for the y) axis of both bodies the acceleration is equal to 0

Yes.

 

[Like Tony Stark]

Yes, except what is P in these equations? Not the same in both, I assume.

Yes.

Sorry, $P$ is weight
Thanks!

 

[haruspex]

Sorry, $P$ is weight
Thanks!

Ok, but different weights for the two blocks, right?

 

[Like Tony Stark]

Ok, but different weights for the two blocks, right?

Yess

 

[haruspex]

Yess

Ok, so can you solve the equations?