1. Sep 30, 2007

### splac6996

1. The problem statement, all variables and given/known data

The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.260.

2. Relevant equations

3. The attempt at a solution

I set up my free body diagram and from there my force equations for all three systems I believe that I need to find an equation for tension and from that find accelaration by doing some substitution but I am getting stuck.

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2. Sep 30, 2007

### Dick

You haven't given us enough information to tell where you are getting stuck. What's the total force on each object in terms of the unknown tension T? Keep all of the forces going in the same direction along the rope. Finally remember all of the accelerations are equal.

3. Sep 30, 2007

### splac6996

That is where I am getting stuck I now that the only reactions on the 1.0kg weight and the 3kg weight is Tension and mg but I am not sure how to use that in my equation for my 2.0kg weight which is

$$\Sigma$$Fx=2T-fk=ma
$$\Sigma$$Fy=N-mg=0

4. Sep 30, 2007

### Dick

The two tensions acting on the 2kg weight are likely not equal. You'll get three unknowns (the two unknown tensions and the acceleration). Luckily you have three objects to do a force balance on so you will get three equations.

5. Sep 30, 2007

### splac6996

SO SHOULD MY EQUATION FOR ACCELERATION LOOK LIKE THIS
(T1-T2-fk)/m=a

6. Sep 30, 2007

### Dick

YES BUT BE SURE YOU GET THE FORCE DIRECTIONS CONSISTENT. Like all in the same direction along the rope.

7. Sep 30, 2007

### splac6996

thanks for walking me through that