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Balance Problem

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A plank is supported from two points. How far can a gril that weights 48 kg walk on the plank before the plank starts to rotate?. Middle of plank is its gravity point.

    Picture: I have added a picture that might tell a bit more(not so good at English :/)

    https://www.physicsforums.com/attachment.php?attachmentid=24296&stc=1&d=1268318353


    m1= mass of the plank = 80kg
    m2= mass of gril = 48kg
    L plank = 5m
    Support point A= 2.5m from center of plank mass/gravity
    Support point B= 0.3m from center of plank mass/gravity


    2. Relevant equations
    Summ F=0
    Summ M=0
    G=mg
    M=Fr


    3. The attempt at a solution

    I have tryed a few equitions but have not had any luck.

    Na= support point A
    Nb= support point B
    Gp= Plank gravity center
    Gg= Girl Gravity
    Na+Nb-Gp-Gg=0


    Thx for Help
     

    Attached Files:

  2. jcsd
  3. Mar 11, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi procol! Welcome to PF! :smile:

    (btw, she's a"girl" … but i like her hair! :biggrin:)

    Hint: when the plank starts to turn, Na will be zero.

    So take moments about B. :wink:
     
  4. Mar 11, 2010 #3
    Re: Welcome to PF!

    Ok thx alot I will try now :D
     
  5. Mar 11, 2010 #4
    I counted the force which point B applys on the plank if the gril is not standing on it and i got to 700N.
    I used that in this equition: Nb x Nr = Gg x r -------> rg = Nb x Nr / Gg

    (700N x 0.3m) / (9.81m/s^2 x 48kg) = 0.44597m....

    Is this Right :D?
     
  6. Mar 11, 2010 #5

    tiny-tim

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    Hi procol! :smile:
    Yes, except for the 700N … I don't understand where you got that from. :confused:
     
  7. Mar 11, 2010 #6
    The 700N I got from counting what support forces the two points A and B have to support the plank without it rotateing or moving when the gril is not standing on the plank.

    F=0

    Na + Nb - Gp = 0

    M=0

    -Na x ra + Nb x rb = 0

    Na x ra = Nb x rb

    Nax 2.5m = Nb x 0.3m

    2.5Na - 0.3Nb = 0

    Making equition par of both equitions with force and force momment:

    2.5Na - 0.3Nb = 0
    Na + Nb - Gp = 0

    2.5Na - 0.3Nb = 0
    Na + Nb - Gp = 0 x 0.3

    2.5Na - 0.3Nb = 0
    0.3Na + 0.3Nb - 0.3Gp = 0

    2.8Na - 0.3Gp = 0

    2.8Na = 0.3Gp

    Na = (3/28)Gp

    G = 80kg x 9.81m/s^2 = 784.8N

    Na = 784.8N x (3/28) = 84.085N

    Gp - Na = Nb

    784.8N - 84.085N = 700,714N = 700N there I got it.
     
  8. Mar 11, 2010 #7

    tiny-tim

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    No, the forces when the girl isn't there are irrelevant.

    Just use the forces on your diagram. :smile:

    (and remember, when the plank starts to turn, Na will be zero)
     
  9. Mar 11, 2010 #8
    ok :D ty
     
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