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Frictional Coefficient

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    WP_20160129_22_40_07_Pro.jpg

    2. Relevant equations
    a=meu*g

    3. The attempt at a solution
    meu= 20/10
    = 2
    The answer's wrong
     
  2. jcsd
  3. Jan 29, 2016 #2

    Suraj M

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    Myu*g? What is that suppose to be?
    Don't just apply the formula you've learnt,
    Identify the normal force
     
  4. Jan 29, 2016 #3
    The normal force here is the driving force of the vehicle, which is it's mass times the acceleration
     
  5. Jan 29, 2016 #4

    cnh1995

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    You've got the concept right. Check the math again.
     
  6. Jan 29, 2016 #5

    Suraj M

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    Exactly so the formula for frictional force is?
    Do you thing myu*g represent the frictional force now?
     
  7. Jan 29, 2016 #6
    The firctional force is meu*normal force, so the frictional force is meu*M*a
    Can I do this:
    The driving force- frictional force = (M1+M2) a
    friction= [F-(M1+M2)a]
    so meu = [F-(M1+M2)a]/M1a
    meu= [M1a - (M1+M2)a]/M1a
    meu = M2/M1
    But then what?
     
  8. Jan 29, 2016 #7
    Actually, 3 of the answers are correct. Static friction is determined by ##N\mu \geq F ## for no slippage, not ##N\mu = F ##
     
  9. Jan 29, 2016 #8

    cnh1995

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    What is the acceleration of the smaller block?
     
  10. Jan 29, 2016 #9
    Isn't the acceleration 20m/s^2 for both the blocks?
     
  11. Jan 29, 2016 #10

    cnh1995

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    Yes. That is what is needed. Write the equation for forces "on" the smaller block. You'll get the value of friction coeff just to avoid slipping. You'll see Chestermiller's reply makes true sense!
     
  12. Jan 29, 2016 #11
    Have you drawn a free body diagram showing the forces acting on the smaller mass, or do you feel like you have advanced to the point where you no longer need to use free body diagrams?
     
  13. Jan 29, 2016 #12
     
  14. Jan 29, 2016 #13
    WP_20160130_002.jpg
    Got it! Thank you!
     
  15. Jan 29, 2016 #14
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