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

[affans]

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)

 

[quantumdude]

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).

 

[affans]

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)

 

[quantumdude]

That's what I got too.