# Problem Solving: Forces

1. Oct 10, 2005

### aceXstudent

Hi there,

I'm having some trouble with this problem right now. We are learning about Newton's First/Second/Third Laws.

With a friction coefficient of 0.15, what is the force that the crates exert on each other if the crate 1 is 75kg and crate 2 is 110kg with a push of 730N?

Can anyone give me pointers or starters?

2. Oct 10, 2005

### cscott

Treat the system as a whole to get acceleration.

3. Oct 10, 2005

### aceXstudent

Errrr... I was looking for the force that each box exerts on each other, not the acceleration...

But thanks anyways.

4. Oct 10, 2005

### cscott

I didn't give you the complete story. You asked for pointers/starters.

Let me elaborate: Treat the system as a whole to find the acceleration of the entire system (use F = ma.) The system moves as a whole, thus each box will have the acceleration that you found. Continue by drawing free-body diagrams for each box and keep in mind that $F_{12} = -F_{21}$ ($F_{12}$ - read "force acted on box two by box one.")

Last edited: Oct 10, 2005
5. Oct 10, 2005

### aceXstudent

-_-, still confused, sorry... acceleration is 2.48m/s^2-- what do i do with that number?

6. Oct 10, 2005

### cscott

I don't know if you read my edited post but use $\Sigma F = ma$ on each box. You know both m and a.

$$\Sigma F_1 = m_1a$$

You now have the net force on box 1. If you drew your free-body diagrams you would also know that

$$\Sigma F_1 = F + F_{21}$$

Our last two expressions are both give us the net for on box 1 so we can equate them

\begin{align*} F + F_{21} &= m_1a \\ F_{21} &= m_1a - F \\ \end{align*}

Tada! You have the force acting on box 1 by box 2. If you read what I said in my last post you should be able to get $F_12$.

Last edited: Oct 10, 2005
7. Oct 10, 2005

### aceXstudent

is 435 correct then?

Last edited: Oct 11, 2005
8. Oct 11, 2005

### cscott

Looking back, I see your problem includes friction. It shouldn't be too hard for you to include that in with what I've given you. Recheck your calculation of acceleration too.