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Two blocks, a pulley, and an inclined plane.

  • Thread starter tharock220
  • Start date
  • #1

Homework Statement


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.


Homework Equations



I think F=ma

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???
 

Answers and Replies

  • #2
993
13
I also got 3.97.
 
  • #3
Ibix
Science Advisor
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Agreed. I spent a few minutes checking if there was a transcription error ([itex]\theta=90-37[/itex], 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:

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