Applied force and the force between boxes?

AI Thread Summary
The problem involves three boxes with masses of 5.0 kg, 3.0 kg, and 2.0 kg, where a 50 N force is applied to the 5.0 kg box. The acceleration of the system is calculated to be 5 m/s², leading to contact forces of 25 N for the 5 kg box, 15 N for the 3 kg box, and 10 N for the 2 kg box. The confusion arises regarding the contact force between the 3.0 kg and 2.0 kg boxes, with calculations suggesting it could be 10 N. Ultimately, the correct contact force between the 3.0 kg and 2.0 kg boxes is confirmed to be 10 N. Understanding the distribution of forces and acceleration is key to solving the problem correctly.
Briann
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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.

Fg = m g
Fg = 5 kg * 10 m / s^2
Fg = 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!
 
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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
 
Ah, so the acceleration of the system is 5 m/s^2.

Fmass1 = m * a = 5 kg * 5 m/s^2 = 25 N
Fmass2 = m * a = 3 kg * 5 m/s^2 = 15 N
Fmass3 = 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!
 
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