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Static equilibrium question

  1. Jul 27, 2005 #1
    A uniform plank of length 6.00m and mass 30.0 kg rests horizontally across two horizontal bars of a scaffold. the bars are 4.50 m apart, and 1.50m of the plank hangs over one side of the scaffold. How far can a painter of mass 70.0kg walk on the overhanging part of the plank before it tips?

    This is really confusing me. I have the forces identified: mg going down, normal force up, and the weight of the man going down. Im guessing the equation to solve this would look something like n-mg-70kg(d)=0...I really dont know how to incorporate the lengths into it.
     
  2. jcsd
  3. Jul 27, 2005 #2

    Doc Al

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    You need to consider torques about the pivot point. As the painter walks further, at some point the other end of the plank begins to rise off of its support. At that point the torque due to the painter's weight exactly balances the torque due to the plank's weight. That's the equation you want.
     
  4. Jul 27, 2005 #3
    I dont know how to set up the equation to solve it...and again, where do the lengths (the distance b/t the bars, 1.50m hanging off...) fit in?
     
  5. Jul 27, 2005 #4

    Doc Al

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    You need only worry about forces that create a torque about the pivot point. There are only two: the weight of the painter and the weight of the plank. Since the plank is in equilibrium, the torques must be equal and opposite.

    To find the torque due to the weight of the painter: Multiply his weight by the distance he is from the pivot point. Since that distance is the unknown you asked to find, just call it "d".

    To find the torque due to the weight of the plank: First realize that the weight of the plank acts at the center of the plank (since it's uniform). Find the distance from that center to the pivot point. Multiply the weight of the plank by that distance to get the torque.

    Hint: Since weight = mg, don't actually calculate the weight. When you set up your equation, the "g" will cancel from both sides.
     
  6. Jul 28, 2005 #5
    Okay I'm getting closer...do I need to worry about any upward forces, such as the normal force? So far I tried (weight of painter)*d-(weight of plank)(2.25m)=0 but it didnt work, also tried (weight of plank)(3m) (since 3 is half of 6, the length of the plank), but still didnt work. I think I'm leaving something out.
     
  7. Jul 28, 2005 #6

    Doc Al

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    Does the normal force exert a torque? (No.)
    Where did you get the 2.25m from? What's the distance of the middle of the plank to the pivot point? (Draw yourself a picture showing the plank, the supports, and the center of mass of the plank.)
     
  8. Jul 30, 2005 #7
    Solution : If “a” is the pivot point, then at the moment of equilibrium the “reaction” force in “b” is zero. Then , Sum of Forces in Y = Na + Nb – 70g – 30g = 0
    But, Nb = 0, => Na = 100g
    Sum of torques in “b” = 0
    70g * d + Na * 0 - 30g * (4.5/2) + Nb * 4.5 = 0
    => 70g * d + Na * 0 - 30g * (4.5/2) + 0 * 4.5 = 0
    => 70g * d - 30g * (4.5/2) = 0
    => d = [ 30g * (4.5/2) ] / 70g
     
  9. Jul 30, 2005 #8

    Doc Al

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    This correct. The normal force, at the moment when the plank just starts to tip, is totally at point a (the pivot point). But since we are taking torques about point a, the normal force there will exert no torque since it is zero distance from the pivot.


    This is incorrect. You've made the same error as rlmurra2. See my last post for a hint.
     
  10. Jul 30, 2005 #9
    Sorry, a typo error! (can happen..) However, we need to differentiate form and substance
    You detected the typo error (good!) but didn’t “see” the whole exposition (?) … think about ..
    Donald Knuth pay $2.56 to the first person who finds and reports anything that remains typographically, or politically incorrect in his books….
    In this case I will not pay since this typo error is too basic. …

    Just change [ Sum of torques in “b” = 0 ]
    with : [ Sum of torques in “a” = 0 ] ,which is the right equation and too obvious [*]

    All the other exposition is perfect.

    70g * d + Na * 0 - 30g * (4.5/2) + Nb * 4.5 = 0
    => 70g * d + Na * 0 - 30g * (4.5/2) + 0 * 4.5 = 0
    => 70g * d - 30g * (4.5/2) = 0
    => d = [ 30g * (4.5/2) ] / 70g


    [*] This is like chess… Hope now you could ‘see’ better the whole game...

    Quote :"Imagination is more important than knowledge" , Albert Einstein
     
  11. Jul 30, 2005 #10

    Doc Al

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    Still wrong for the same reason. :wink:
     
  12. Jul 31, 2005 #11
    “same reason” ( a typo error ?). Maybe you need more time to understand…
    (Doesn’t means knowledge, intelligence…. some times lack of concentration…)...
    So take your time to process with "more quality"
    Math, Physics… are “Sciences” (with perfect rules and demonstrations) not just “subjective statements”.
    The solution indicated is “the right solution”. Now, if you don’t understand there is nothing that I could do for you. (Maybe study more……)
    note : This is a basic problem explained in any "introductory physic course" (Mechanic Static ) go to any basic book to check your concepts…..
    Same problem is well-explained and documented in hundred of places…
     
  13. Jul 31, 2005 #12

    Doc Al

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    You are making the same mistake that the original poster made. (No, I'm not talking about a typo!) See my comment in post #6.
    :zzz:

    Thanks for the lecture, but all you are doing is further embarrassing yourself. Your solution is still "the wrong solution".
     
  14. Jul 31, 2005 #13

    mukundpa

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    For rlmurra2

    Please dont forget that the weight of the plank is acting at its mid point which is at 3m from each end and thus at 1.5 m from the pivot point. So the solution is simpelly
    (weight of painter)*d-(weight of plank)(1.5m)=0 [1.5m of plank is hanging]

    Sorry Mr.Doc Al:smile:
     
  15. Jul 31, 2005 #14

    Doc Al

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    That is correct, of course. But I wanted rlmurra2 to draw a diagram and discover that for himself. :wink:
     
  16. Jul 31, 2005 #15
    ......................./\.............................../\......................
    ....[70]............[Na].......[30]...............[Nb]....................
    .....\/.............................\/..........................................
    .....<---d------..A.................................B.....................
    =============================================
    .......................^................................^......................
    <------1.5------><-------- 4.5 ---------><------1.5----->

    <------------------------ 7.5 --------------------------->

    => d = [ 30g * (4.5/2) ] / 70g

    My solution was for the above problem ( I made a mistake when I read the statement of the problem,
    thinking that scaffold was used as a typical configuration symetrical....)
    Now I realize that the right section of my draw don't exist (the 1.5 in the right in B
    ... (and also the numbers match)
    So the diagram is :

    ......................./\.............................../\
    ....[70]............[Na]...[30]...................[Nb]
    .....\/.........................\/..........................
    .....<---d------..A.................................B
    =================================
    .......................^................................^
    <------1.5------><-------- 4.5 ---------->

    <--------------------6.0 ---------------->

    Hence :
    70g * d + Na * 0 - 30g * ([6/2]-1.5) + Nb * 4.5 = 0

    => 70g * d + Na * 0 - 30g * ([6/2]-1.5) + 0 * 4.5 = 0
    => 70g * d - 30g * ([6/2]-1.5) = 0

    => d = [ 30g * ([6/2]-1.5) ] / 70g

    General solution (in this particular case) :
    d max = [(L/2)-C]*(m1/m2)

    C : distance of point "A"

    ( "C" mts of the plank hangs over one side of the scaffold. )
     
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