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Distance After Release- Problem on Forces

  1. Oct 30, 2009 #1
    1. The problem statement, all variables and given/known data

    In figure A, the coefficient of friction between block A and the tabletop is 0.3, Mass of block A= 30 kg, Mass of block B=20 kg. How far will block B drop in the first 3 s after the system is released from rest?

    here is the figure :D

    http://img137.imageshack.us/img137/5771/imagenz.png [Broken]

    2. Relevant equations

    ....

    3. The attempt at a solution

    *we use gravity= 10 (not 9.8 )

    w (weight)
    Nf (normal force)
    ΣF (semation of forces)
    f (friction)
    T (tension)

    BLOCK A

    ΣFy= 0

    -w + NF = 0
    Therefore w = NF

    W= mg
    = 30 (10)
    = 300 N

    Therefore NF = 300 N

    f= μNF
    = 0.3 (300 N)
    f= 90 N

    ΣFx= ma

    -f + T = ma
    -90 + T = ma
    -90 + T= 30 (a)
    T= 90 + 30a

    BLOCK B

    W=mg
    W=20 (10)
    W= 200 N

    ΣFy= ma

    T-w=ma
    T- 200 N = 20a
    T= 200+20a

    T=T

    90+30a = 200 + 20a
    90- 200 = 20a – 30a
    -110 = -10a
    ------- -------
    -10 -10
    a= 11 m/s^2

    y= Vo + Voyt + at^2/2
    y= 0 + 0 + 11 (3) 2 /2
    y = 49.5 m

    I got 49.5 but my professor said it was wrong :(can someone tell me where I went wrong? and what IS the correct answer?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 30, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You made a sign error in setting up your Fy equation. You need to be consistent in your use of "a" for acceleration. If mass A moves to the right, mass B moves down. Thus if the acceleration of mass A is "a", then the acceleration of mass B is "-a".

    Note how your acceleration is greater than gravity! That can't be right. :wink:

    Just fix that equation for block B above and you'll be fine.
     
  4. Oct 30, 2009 #3
    Oh o_O so the only mistake I made was not making the acceleration of block B negative? :D so if I changed it to negative, then it'll make all the difference ^^? I asked through yahoo answers too and one replied that the answer was 9.7 m :D is it most likely correct? I couldn't understand his explanation that well that's why I'm still confused -__-
     
  5. Oct 30, 2009 #4

    Doc Al

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    Staff: Mentor

    Yes. Try it and see. Correct the equation as I indicated; find the correct acceleration; then use it to find the distance traveled.
     
  6. Oct 30, 2009 #5
    So I tried doing it like this...

    T-w=m(-a)
    T- 200 N = 20(-a)
    T= 200+20(-a)

    Is this correct?
     
  7. Oct 30, 2009 #6

    Doc Al

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    Staff: Mentor

    So far, so good.
     
  8. Oct 30, 2009 #7
    So this is the result right? :D

    T=T

    90+30a = 200 + 20(-a)
    90- 200 = 20(-a) – 30a

    so from what I can understand... 20(-a) becomes -20a right ? :D or is it wrong?

    so...
    90- 200 = -20a – 30a
    -110 = -50a
    ---------------
    -50 -50

    a= 2.2 m/s^2 ? :D
     
  9. Oct 30, 2009 #8

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  10. Oct 30, 2009 #9
    thank you thank you thank you so much!! I got the correct answer!!! xDD thank you for being so patient :)
     
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