# Finding net force

1. Mar 9, 2009

### okgo

1. The problem statement, all variables and given/known data

2. Relevant equations

F=ma

3. The attempt at a solution
According to Newton's 3rd law, the box pushes back on the person 36N. But I don't know how you find how much force the smaller crate exerts on the larger one.

2. Mar 9, 2009

### LowlyPion

Re: Forces

If there is a net force pushing on both crates of 36N, then the crates are accelerating aren't they?

If the 20 kg crate is accelerating at that rate, then what must the force be on it to make it accelerate at that rate?

3. Mar 10, 2009

### okgo

Re: Forces

I'm not sure I totally understand how they got 30N as an answer. Our test is coming up soon and I was worried if the multiple choice might be closer to the real answer like C. 25 D.33
Weight is a measurement of force so I used w=mg. 20kg*9.8=196N ??

4. Mar 10, 2009

### LowlyPion

Re: Forces

F = m*a

Sure m*g is weight normal to the surface, but you are dealing with horizontal motion which has no interaction with gravity, unless there is friction dependent on the weight.

Your concern here is horizontal forces..

As to where the 30 comes from :

F2 = m2/(m1+m2)*F = 20/24*36 = 30

5. Mar 10, 2009

### LowlyPion

Re: Forces

Or more formally since you know acceleration for the system is a constant a, then you know that

F2/m2 = F/(m1+m2)

Hence

F2 = m2/(m1+m2)*F

where F is the total applied force.