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Increased acceleration in a pully system

  1. Aug 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A window washer pulls themself upward using a bucket/single pulley system. Their weight with the bucket is 65 Kg.
    a) How hard must she pull doward to raise herself slowly at a constant speed?
    b) If they increase the force by 15% what will her accelertion be?

    2. Relevant equations
    2Ft - mg = ma

    a = (Mb - Mp)/(Mb+Mp) .... I think
    3. The attempt at a solution
    a) I've got no problem...

    to get constant speed accel = 0 so....

    2Ft - mg = 0
    2Ft = mg
    Ft = mg/2 -> (65)(9.8)/2 = 320 N

    b) This is where I think I have the wrong equation to attempt this as
    (320)(0.15)= 48, so ..... [(368-320)/(368+320)](9.8)= 0.68 m/s^2

    answer in the book is 1.5 m/s^2 so need something else but don't know what to use.
     
  2. jcsd
  3. Aug 8, 2012 #2

    ehild

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    The equation in red is irrelevant here. The person exerts F force, and the tension in the chord equals to Ft=F. The person in the bucket is pulled by 2Ft upward, and gravity pulls downward. So the equation you wrote 2Ft - mg = ma is correct.

    The other equation refers to the case of a pulley with two masses.

    ehild
     
  4. Aug 8, 2012 #3
    Your new upward force is 2(320+48)
     
  5. Aug 9, 2012 #4
    So am I literally taking (9.8 m/s^2)(0.15)= 1.47 m/s^2 to get the acceleration?
     
  6. Aug 9, 2012 #5

    ehild

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    Yes.

    ehild
     
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