New's Law of Motions Problem helppp

1. Oct 28, 2009

mc8569

New's Law of Motions Problem!! helppp

1. The problem statement, all variables and given/known data
A horizontal force F is applied to a small block of mass m1 to make it slide along the top of a larger block of mass m2 and length l. The coefficient of friction between the blocks is μ. The larger block slides without friction along a horizontal surface. The blocks start from rest with the small block at one end of the larger block as shown

http://i37.tinypic.com/2emh3bn.jpg
^will probably have to zoom in on it, click on the image

question:
Find the acceleration of each block a1 and a2 relative to the horizontal surface.

2. Relevant equations
F = ma

3. The attempt at a solution
I know the acceleration of a1, that was easy.
For a2, I do:
fk = m2a2
a2 = um1g/m2

My concern is, in solving for a2 do we use m2 for the system's mass or do we use (m1 + m2) for the system's mass? I still am having trouble understanding that. I guess if you answer this, my concerns will be addressed:

Given a situation where there is an object on top of an object and you are finding the acceleration for the bottom object, but the object on top is not moving with the same acceleration as the object on the bottom, that is it is slipping above it, do you account the mass in F = ma as (m1 + m2) because m1 is still on top of m2 no matter what, or do you disregard the upper mass(m1) since it is not moving with m2 as a system.

Last edited: Oct 28, 2009
2. Oct 28, 2009

PhanthomJay

Re: New's Law of Motions Problem!! helppp

yes
You must draw free body diagrams. Isolate m2 and indicate the forces acting on it. m1 does not come into play here. What did you do when you calculated the acceleration of m1?

3. Oct 28, 2009

mc8569

Re: New's Law of Motions Problem!! helppp

Okay here are my answers:

a1
F = ma
F - fk = m1a1
a = (F - fk) / m1

a2
fk = m2a2
a2 = fk/m2
a2 = (um1g)/m2

I didn't show work for a1 because I know that is right & for a2, simply understanding the concept for it, I would know if I was right.

My problem with a2 is, would it be the above or
a2 - (um1g)/(m1+m2)

Pertaining to my question, when the two blocks are not moving as one - they don't have the same acceleration therefore are not part of a system - do you still account for both masses when solving for ma in F = ma?

4. Oct 29, 2009

PhanthomJay

Re: New's Law of Motions Problem!! helppp

yes
no
No. That's what I mean about using free body diagrams. When you did part a, you isolated the top block 1 and identified the horizontal forces acting, then used newton 2 in the horizontal direction to solve for the acceleration. You also used just the mass of m1, which is correct.
Now when you isolate block 2 in a free body diagram, again identify the forces acting in the horizonatal direction, and apply Newton 2 to that mass m2. There is no m1 in that equation. Note that there is a normal force of m1(g) acting down on block 2 from block 1, and a gravity force of m2(g) acting down on block 2, and a normal force of (m1 + m2)g acting up on block 2 from the floor, but these are forces in the y direction.