# Derivation of an acceleration

1. Apr 2, 2013

### keltik

Two boxes are side by side on a floor with friction, from the left a Force acts on them. To be more concise about the picture it is this one on this website:

http://hyperphysics.phy-astr.gsu.edu/hbase/f2m2.html#c1

I dont understand how the acceleration is derived. Especially i dont get which vectors are weighed out against each other. And also from their pictures it is not clear where

I have already tried to "reverse engineer" the given expression for a (on that website):

Step-1: $a = \frac{F- \mu*(m_{1}+m_{2})*g}{(m_{1}+m_{2})}$

Step-2: $a*(m_{1}+m_{2}) = F- \mu*(m_{1}+m_{2})*g$

My question here is wether the left side stands for "$F_{net}$" (or "$F_{result}$") ? That is could i write instead of Step-2, this one:

Step-3: $F_{net} = F- \mu*(m_{1}+m_{2})*g$

And the next question is could i replace F with $F_{push}$?

Step-4: $F_{net} = F_{push}- \mu*(m_{1}+m_{2})*g$

and the mu-Stuff with

Step-5: $F_{net} = F_{push}- f_{m1}-f_{m2}$

If so, why didnt they write it on their website, because i think that it is more intuitive than just spitting out the formula for a? Or is my derivation totally wrong?

2. Apr 2, 2013

### P.Bo

You absolutely can (and must) do that. Any acceleration on the system is going to be a result of the net force, not individual forces. The force is what causes the acceleration not the other way around. Besides you couldn't get multiple accelerations that sum up in a case like this since they're all accelerating at the same rate.

Looks good to me, the F here is implied as the Force that is being applied to the object. It really is the catalyst which makes everything else work. No initial force, then friction won't manifest.

Other than "they didn't want to derive it" I think the reason they went this way, might be simply because the "push" force is what makes everything work and it is the only force that is actively being applied to the system.

However your reverse engineering of what they did looks spot on.