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Force in Equilibrium

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data

    Untitled.png
    2. Relevant equations


    3. The attempt at a solution
    I'm not sure about which topic it is asked. I think it's something about the centre of mass . Because the statement " The pole and the box do not change form" . The answer is
    (1 + tan Θ)/(1 - tan Θ)
     
  2. jcsd
  3. Jan 20, 2015 #2

    BvU

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    What are (all!) the relevant equations for static equilibrium ?

    Who do you think will have to carry more in the right picture ? B or A ? Why ?

    The "don't change form" remark is to reassure you the center of gravity stays in the same place in the box/pole combo.
     
  4. Jan 21, 2015 #3
    A will carry more, I think it's because he's in higher position.
    And is there any special formula for static equilibrium ? I just draw a triangle and use the formula F=ma.
     
  5. Jan 21, 2015 #4

    BvU

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    Hehe, you never had to carry something heavy down the stairs, I suppose ?

    The conditions for equilibrium I hinted at in post #1 are in the first place ##\Sigma \vec F = 0## So that with your formula a = 0 ##\Rightarrow## v = constant. v = 0 remains v = 0. No translation.

    But that's not enough. You also want no rotation, in other words: ##\Sigma \vec \tau = 0##.
    And now the positions where the forces act come in the expressions.
     
    Last edited: Jan 21, 2015
  6. Jan 21, 2015 #5
    F = ma comes into question when acceleration comes into question. Here, a = 0 and v = 0. All forces balance each other out. FA and FB act vertically upward. In which direction do you think 'W' acts?
     
  7. Jan 21, 2015 #6

    DEvens

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    The forces are vertical. That is a constraint on the answer given by the question statement. In an actual case of carrying a couch down a ramp the people doing the carrying probably would not arrange their hand-holds that way, but the problem requires it.

    The pole is not rotating while it is being carried.
     
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