Friction and Newton's second law

1. May 19, 2012

Joshuarr

1. The problem statement, all variables and given/known data
It's attached.

2. Relevant equations

F = m*a
F_f = -μ_k*F_N // [Force of friction] = μ_k*[Normal force]

3. The attempt at a solution
So, I think I managed to solve it -- at least, I got what the book got, but I don't understand part of what I did to get the answer.

This is how I solved it and got the answer the book got:
I drew a free-body diagram, and realized that F_n = 2Mg
Which means F_f = -μ_k*2Mg, Block 2 is being pulled by an effective force to the right of Mg (Since -Mg + 2Mg = Mg).
So the summation of all the forces in the x direction would be:

***: Mg + F_f = Mg - μ_k*2Mg = (5*M)a

Plugging in for a and other known values and some algebra later, I get μ_k = 0.37

And that's the answer the book gives.

My question is why is it m = 5M in Newton's second law (the *** equation). When I first tried to solve it I used the mass of the 2nd block (m = 2M) and I thought that was correct.. apparently not.

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2. May 19, 2012

AudioFlux

think of it in a linear manner (see attached file). You will probably understand it now.

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3. May 19, 2012

Infinitum

The equation relates the total force on the system of the three blocks to their acceleration. If you take m=2M that would be considering only the force on the block of mass 2M, and its force equation would be different(involving tensions in string, etc). The 5M is for the system of all the three blocks taken together.

4. May 19, 2012

Joshuarr

Thanks, that did help a lot. :)

5. May 19, 2012

azizlwl

Normally i calculate for each mass.

Check where the direction of acceleration
1.(3M)-
3Mg - F1=3Ma

2 (2M)
F1-F3-friction=2Ma

3. (M)
F3-Mg=Ma

(1)+(2) +(3)

3Mg-Mg-friction =6Ma
net force=ma(Newton 2nd Law)

Last edited: May 19, 2012
6. May 19, 2012

Joshuarr

lol, I was stuck on a problem involving tensions (I was using the total mass for m), and then I remember your saying something about that. Thanks! :)

7. May 19, 2012

nishkash

Basically when you judge it from the accelerating frame of reference then there will be a pseudo force acting on each of the spheres which amounts to 5Ma.

From a non accelerating frame draw an FBD for each of the blocks seperately and then solve the equations simultaneously. You will get what you were missing earlier.

8. May 19, 2012

Joshuarr

I think your 3M is a mistake (it's 1M, 2M, 2M), but I understand what you're saying.

Oh. So it always ends up as a linear system of equations with n equations n unknowns (where n is the # of bodies), so I can use matrices! The other unknowns in this case would be the two tensions, what you called F1 and F2.

Thanks for helping me generalize the solution to the problem.

9. May 19, 2012

Joshuarr

I'm not sure what you mean by an "accelerating frame of reference." The frame of reference would be the table (which is stationary), no?

I think by spheres you mean blocks, or perhaps that is some concept that I'm unfamiliar with..

Oh.. In this situation I would use the latter approach, right? Since there is a non-accelerating (stationary in fact) frame of reference.

You seem to imply that this problem has an accelerating frame of reference though, but maybe I'm misunderstanding you.

10. May 20, 2012

nishkash

Actually both the approaches are applicable and by accelerating FOR I mean the blocks itself ( sphere was a mistake).