Applied force and the force between boxes?

[Briann]

Homework Statement

Three boxes rest side-by-side on a smooth, horizontal floor. Their masses are 5.0 kg, 3.0 kg, and 2.0 kg, with the 3.0-kg mass in the center. A force of 50 N pushes on the 5.0-kg mass, which pushes against the other two masses. What is the contact force between the 3.0-kg mass and the 2.0 kg mass?

a) 25 N
b) 10 N
c) 40 N
d) 0 N

Homework Equations

Maybe F = m a?

The Attempt at a Solution

[ 5 kg ] [ 3 kg ] [ 2 kg ]

I would think that gravity would diminish the applied force of 50 N.

F_g = m g
F_g = 5 kg * 10 m / s^2
F_g = 50 N

So the 50 N force would be subtracted from the applied force of 50 N, leaving 0 N, right?

I'm confused. Can someone help me?

Thank you!

 

[dacruick]

50 Newtons here is pushing 10 kg. so you know the acceleration of the system. Since each box will have an acceleration, and a different mass, you have what you need to calculate the applied force for each individual box

 

[Briann]

Ah, so the acceleration of the system is 5 m/s^2.

F_mass1 = m * a = 5 kg * 5 m/s^2 = 25 N
F_mass2 = m * a = 3 kg * 5 m/s^2 = 15 N
F_mass3 = m * a = 2 kg * 5 m/s^2 = 10 N

Because the contact forces between mass2 (the 3-kg mass) and mass3 (the 2-kg mass) are 15 N and 10 N respectively, does that mean that I should subtract 15 and 10 and get 5?

Or is 10 N already correct?

Thanks again!