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Acceleration coefficient

  1. Oct 10, 2007 #1
    Find the acceleration experienced by each of the two masses shown in the picture below if the coefficient of friction between the 7 kg mass and the plane is .25 ?
    ------------
    In the picture, is the force of fiction go that way ?
    [​IMG]
    m1 = 7kg u = .25
    m2 = 12kg
    F (per) = F (normal)
    F (per) = 68.6 cos 37 = 54.8 N
    F (parallel) = 68.66 sin 37 = 41.3 N
    F net = (12*9.8) - 1.75 = 115.85 N
    115.85 / 19 = 6.1 m/s2
    So 7 kg is accelerate at 6.1 m/s2 and 12 kg is the same but in the negative direction.
    Is this right ?
     
    Last edited: Oct 10, 2007
  2. jcsd
  3. Oct 10, 2007 #2

    learningphysics

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    can you explain this part:

    "F net = (12*9.8) - 1.75 = 115.85 N"

    how are you getting 1.75... and which mass are you taking Fnet for...
     
  4. Oct 10, 2007 #3
    in this problem, there are only 2 force make the block able to slid. (and it is force of gravity on 12 kg and force friction on 7 kg.
    So F net = F (gravity) - F (fiction)
     
  5. Oct 10, 2007 #4

    learningphysics

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    but how do you get 1.75N?

    Also what about the F (parallel) = 68.66 sin 37 = 41.3 N?
     
  6. Oct 10, 2007 #5
    F (friction) = .25 time 7 = 1.75
     
  7. Oct 10, 2007 #6

    learningphysics

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    But friction = 0.25*Fnormal = 0.25*7*9.8*cos37.

    Although it might be faster to do the whole problem at once... I think it's best to divide into 2 sections...

    do the freebody diagram of the first mass, get the equations there... then the second mass, get the equations there... use T for tension...
     
  8. Oct 11, 2007 #7
    ok, I got this
    a = 3.3 m/s2
     
  9. Oct 11, 2007 #8

    learningphysics

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    looks good to me. that's what I get also.
     
  10. Oct 11, 2007 #9
    Yes ... thanks learningphysics
     
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