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Archimedes Derivation: Law of the lever

  1. Jul 26, 2015 #1
    Hi, I'm watching a talk about Archimedes Law of the lever, and I was wondering: Does anyone know how the formula d=w/2w/b-a) was derived from the lecture drawing at 9.00 min in? The speaker just skips the algebraic derivation.

     
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  3. Jul 26, 2015 #2

    BvU

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    With "d=w/2w/b-a)" you mean d=W/(2w) (b-a) ?

    It's a simple torque balance (see at 8:51).

    On the left you have ##d \times w + {a\over 2}\times pW##, on the right ##{b\over 2}\times qW##,

    and ##\ \ a = {p\over l}\ \ ## and ## b = {q\over l}\ \ ## so that $$
    d \times w = \left ( {b^2 \over 2l} - { a^2 \over 2l} \right ) \times W.
    $$Now use ##\ \ b^2-a^2 = (a-b) (a+b) = (a-b)\; l\ \ ## and there you are !
     
  4. Jul 26, 2015 #3
    Thank you very much! There is one more thing i can't seem to understand: I understand that little w is the weight of the object on the lever, but right next to little w there is a big W pointing down (on the left side). What does the big W mean? And what is it doing on the left side?
     
  5. Jul 26, 2015 #4

    BvU

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    W is the weight of the beam on which the weight small w is resting.
    A fraction ##{a\over l}## of W is on the left side, and a fraction ##{b\over l}## of W is to the right of the balance point
     
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