What is the Force of Contact Between Two Blocks on a Frictionless Table?

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The problem involves two blocks on a frictionless table, with a horizontal force applied to one block. Given the masses of the blocks (m1 = 2.0 kg and m2 = 1.0 kg) and the applied force (3N), the goal is to find the force of contact between the blocks. Using Newton's laws, the equations derived show that the acceleration of the system is 1 m/s². The force of contact between the blocks is calculated to be 1N. This solution aligns with the principles of action and reaction as stated by Newton's third law.
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Homework Statement



Two blocks are in contact on a frictionless table. A horizontol force is applied to one block as shown in Fig 7. If m1 = 2.0 kg, m2 = 1.0 kg, and |F|applied = 3N, find the force of contact between two blocks.

----> --------
| A | B |

note: the height of B is a little smaller.
A= m1
B = m2.
The arrow is the force applied.



Homework Equations





The Attempt at a Solution


Fnet(1) = Fapp - Fm2
-> Fapp = m1a + Fm2 (eqn 1)

Fnet(2) = Fapp + Fm1
Fapp = m2a - Fm1. (eqn 2)

Fm2 + Fm1 = a(m2-m1)
 
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affans said:

The Attempt at a Solution


Fnet(1) = Fapp - Fm2
-> Fapp = m1a + Fm2 (eqn 1)

Good.

Fnet(2) = Fapp + Fm1

No. Fapp doesn't act on block 2. Only Fm1 acts on block 2.

Also, you don't need 3 variables. Newton's 3rd law says that the action of block 1 on block 2 is equal and opposite to the reaction of block 2 on block 1. So Fm1 and Fm2 have the same magnitude, but opposite directions. So just call their common magnitude F. Then you have 2 equations in 2 unknowns. (F and a).
 
From post 1 EQN1: Fapp = m1(a) + Fm2
EQN2: m2(a) = Fm1
but Fm1 = Fm2 (action rxn) = F
therefore

Fapp = m1(a) + m2(b)
3 = a(2 + 1)
a = 1 m/s^2

therefore back to EQN 1
Fapp = m1(a) + F
3 = (2)(1) + F
F = 1N?

Can anyone confirm this answer?
thanks (its not in the back of the textbook)
 
That's what I got too.
 
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