Force Acceleration of Two Identical Blocks

[Condensate]

[Solved] Force Accelerating Two Identical Touching Blocks

Homework Statement

The two blocks are accelerating at a constant rate due to the force of the hand, and air resistance is ignored.

I'll go beyond the question above and try to find why F_B and F_A are less than F_hand mathematically, since that would greatly help my conceptual understanding.

My mathematical assumptions (not given in the question, just there so that I can try and figure out the above mathematically):
μ = 0.2
Acceleration= 3
F_hand= 200 N
m=0.05 kg
Normal Force= 0.5 N

Homework Equations

(only for X forces):
Force(A)_net = F_hand + F_friction + F_B
Force(B)_net = F_A + F_friction
F_friction= μF_N
F = ma

3a. Conceptual chart attempt at a solution (Boxes are just to show the scaling and boxes for A are not equivalent to those for B)

3b. Mathematical attempt at a solution
Finding F_friction for Block A using F_friction= μF:: Can be repeated for Block B

• F_friction for Block A= μF_N
• F_friction for Block A= 0.2 x 0.5 (Subsitituting in the values)
• F_friction for Block A= 0.1

Finding Force of Block B on Block A

• F = ma = Force_net
• Force(A)_net = F_hand + F_friction + F_B
• Force(A)_net = F_hand + F_friction + F_B= ma (Substituting net force of Block A for f)
• Force(A)_net = 200 + (-0.1) + F_B= 0.15 (Substituting in the values)
• 199.9 + F_B= 0.15
• F_B= -199.75 => F_A=199.75

Finding F_friction on Block B

• Force(B)_net = F_A + F_friction= ma
• 199.75 + F_friction= 0.15 (Substituting in the values)
• F_friction= -199.6

Finding F_friction when treating both boxes as one large mass (100 g) using F_friction= μF:

• F_friction= μF_N
• F_friction= 0.2 x 1 (Substituting in the values)
• F_friction= 0.2

Finding F_friction when treating both boxes as one large mass (100 g) using Force = ma = Net Force:

• Force_net = F_hand + F_friction= ma
• 200 + F_friction = 0.3 (Substituting in the values)
• F_friction = -199.7

4. What it comes down to:
[STRIKE]-Was I wrong in assuming that F= ma = Force_net of an object?
-How come the force of A exerted by Block A on Object B is less than Force of the Hand on Object A (and is this even true?) Is there a mathematical way to figure this out? My attempt gave different frictions for both blocks.
-Do any of the assumptions appear to mess up the math?
-Do the force diagrams accurately portray acceleration and the horizontal forces on both of the blocks?[/STRIKE]

So, any amount of help from you guys would be super appreciated!

Edit: When finding the actual acceleration from a single 100 g block and using that in the 50 g block equations, everything fit perfectly.

 

[Mindscrape]

It just tells you that you or whoever wrote this problem over-constrained it. Method 1 and 3 are identical. Method 2 should give you the right force of friction given. The trouble is that the acceleration would be wrong in method 2, and the force of friction is wrong in methods 1 and 3. It's like me saying shoot this cannonball 40m forward, at an acceleration of 40m/s^2 for one second; something has to give.

 

[Condensate]

Thanks for the reply. Yeah, I think it was a fault of mine. :)

There's going to be no way that plugging an arbitrary number in for 'a' will give consistent frictions or even F_A depending on which is being solved for.

So, the only question I have left is whether it is possible to:
Determine the acceleration of the objects by way of finding the friction on the two blocks as one solid mass, and then plugging that into Force_net = F_hand + F_friction= ma and finding 'a.' The 'a' can then be used in either of the two separate block equations to solve for the force of A or B on the other block.
Does that seem alright?

Edit: Okay, it's all good now.