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Q & A: free-body diagrams

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Most recent answer: 01/12/2014
Q:
(See Image: http://i.imgur.com/aWjJ4Lk.png) There is one object with mass -m- sliding on another with mass -M-. There is friction between objects - with coefficient of (mu) and we don't have friction on the floor. Object -m- starts moving with velocity -V0-. Find out when the the objects end moving from each other. (I mean when one object becomes static from reference frame of another object). ---------------------------------------... I tried to solve the problem, solved it but made one mistake, which I don't understand why is wrong: When -m- starts moving, F(friction) starts acting to reverse direction of velocity vector. F(friction) = * m * g. According to Newton's laws, same F(friction) starts acting on second object direction to the right (reverse direction) = * m * g. acceleration of first object is going to be F(fr)/m = *g; ---------------------------------------... acceleration of second object is going to be: (Here is my mistake) a = F(fr) / (M+m) = * m * g / (M+m). the correct answer is: a = F(fr) / M = * m * g / M. ---------------------------------------... why should I use just M and not M+m - I know that top object is pushing the the object below with mg force, so I though I should consider it's mass for finding out the acceleration of the second object. Please help me find out what's the problem? I've been searching problems like this whole day but couldn't find explanation on Google. If you have link to related problem, I would appreciate if you provided it to me, thank you.
- Dimitri (age 16)
Tbilisi, Georgia
A:

You've made a very common mistake by over-thinking the problem. If you just stick to a simple, systematic free-body-diagram method of applying Newton's laws everything works out fine. What are the horizontal forces on big M? Only Ffr. So you just use F=Ma for that object to get the right a. No tricks. Just follow the basic laws.

You've jumped ahead to try to use your correct sense that m+M has something to do with the answer. It does, but if you're careful to follow the simple rules, you'll see that come out automatically in the end, with v= mv0/(m+M), thanks to Newton's third law. That simply says that the final momentum (M+m)v is the same as the initial momentum mv0 because there are no horizontal forces from the outside on the (M+m) system.

The key point is not to get sidetracked from one self-consistent path to the answer by scrambling its math with pieces from another. 

Mike W.


(published on 01/12/2014)

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