# Two blocks, a pulley, and an inclined plane.

1. Oct 28, 2011

### tharock220

1. The problem statement, all variables and given/known data
An 8kg block rests on an inclined plane where theta=37 degrees. The coefficient of kinetic friction on the plane is .23. The 8kg block is connect by a massless, frictionless pully to a hanging 16kg block. The blocks are released from rest. What is the acceleration.

2. Relevant equations

I think F=ma

3. The attempt at a solution

So I used F=ma for each block.

8kg block.

T-sin(37)*8*9.8-cos(37)*8*9.8*.23=8*a eq1

16kg block.

T-9.8*16=-16*a eq2

so eq1-eq2 =

9.8*16-sin(37)*8*9.8-cos(37)*8*9.8*.23=24a.

so 95.2=24a

leaving a=3.97.

When we checked the answer it said a = 1.3. What are we doing wrong???

2. Oct 28, 2011

### grzz

I also got 3.97.

3. Oct 28, 2011

### Ibix

Agreed. I spent a few minutes checking if there was a transcription error ($\theta=90-37$, transposed masses, both of the above) and I can't get 1.3. The best I can do is that a coefficient of friction of 1.23 gets you an acceleration of 1.36ms^-2, and you'd have to coat the ramp in glue to get that.

Unless someone answers with something that we're all doing wrong, I'd just double check that you haven't mis-read something in the question and hand it in.

Last edited: Oct 28, 2011
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