How Is Acceleration Derived in a Frictional System with Two Boxes?

AI Thread Summary
The discussion focuses on deriving acceleration in a frictional system involving two boxes subjected to an external force. The user questions the relationship between net force and the derived formula for acceleration, specifically whether the left side of their equation represents net force. It is confirmed that the net force is indeed the driving factor for acceleration, and the applied force can be referred to as the push force. The reasoning behind the original formula's presentation on the referenced website is clarified, indicating that it emphasizes the applied force as the primary factor in the system's dynamics. Overall, the user's reverse engineering of the acceleration formula is deemed accurate and intuitive.
keltik
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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 don't understand how the acceleration is derived. Especially i don't 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?
 
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keltik said:
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

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.

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

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

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.
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?

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.
 
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