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How to Calculate Torque of a Leaning Rigid Body

  1. Jun 12, 2012 #1
    So this is a conceptual question, it's not a direct homework question, but it does involve how to do a kind of calculation. I hope this isn't the wrong place to post a nonspecific question like this. In the case of a board/stick/ladder leaning against a wall, about an axis O at the bottom of the object where it is in contact with the ground, why is the distance used to determine torque (which is Force x Distance x sin θ) halfway up the ladder? Ladder is 15m long in my example.

    I tried to use integration to calculate the torque, and this is what I ended up with:

    [tex]
    \tau = mg \times x \times \sin\theta = mgsin\theta \cdot x \\

    d\tau = mg \sin\theta \cdot dx \\

    \int d\tau = \int mgsin\theta dx \\

    \int d\tau = mg \sin\theta \int dx \\ = mg \sin\theta \int_0^{15} dx \\

    = mgsin\theta \cdot x |_0^{15} \\

    = 15 \cdot mg \sin\theta


    [/tex]

    So why is the scalar for distance 1/2 of the length rather than the entire length like my calculations? What am I missing? Thanks for any insight.

    !! The mass also changes, so I think I need a differential mass element... more thinking required :/
     
    Last edited: Jun 13, 2012
  2. jcsd
  3. Jun 13, 2012 #2

    Simon Bridge

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    Because gravity acts at the center of mass.
     
  4. Jun 13, 2012 #3
    Or can I use a Riemann sum to calculate the torque? Assume the ladder is divided up into n pieces, and the mass of each piece is its linear density λ times the change in distance.

    [tex]
    \tau = x \cdot F \cdot \sin\theta \\
    F = m \cdot g \\
    m = \lambda \Delta x \\
    \tau_i = x_i \cdot \lambda \Delta x \cdot g \cdot \sin\theta = \lambda g \sin\theta \cdot x_i \cdot \Delta x \\
    \sum_i^n \lambda g \sin\theta \cdot x_i \cdot \Delta x \\
    \lim_{n\to\infty}\sum_i^n \lambda g \sin\theta \cdot x_i \cdot \Delta x = \int_0^{15} \lambda g \sin\theta x dx\\
    = \lambda g \sin\theta \cdot \int_0^{15} x dx = \lambda g \sin\theta \cdot \left[ \dfrac{1}{2}x^2 \right]_0^{15}
    [/tex]

    I don't see this moving towards a validation of the torque being calculated at the half-way point with mg acting on the center of gravity... can anyone nudge me in the right direction?
     
  5. Jun 13, 2012 #4
    I am aware of that, but shouldn't I end up finding the same torque using integration?
     
  6. Jun 13, 2012 #5

    Simon Bridge

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    Well yes - and the integrated part will amount to finding the center of mass.
    So your question amounts to: "how come I keep blowing the math?"
     
  7. Jun 13, 2012 #6

    haruspex

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    Not sure what the first equation is saying, but the second is wrong. The RHS is the mass of the element dx multiplied by g sin(θ). The distance is x sin(θ):
    [tex]
    d\tau = mgxsin\theta \cdot dx \\
    [/tex]
     
  8. Jun 13, 2012 #7

    Simon Bridge

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    * Actually I think you got it right in the reiman sum - the first integral was poorly set up which is why it didn't work.

    Finish the calculation.
    [tex]
    g\lambda\sin(\theta)\int_0^L x.dx = g\lambda\sin(\theta)\cdot \frac{1}{2}x^2 \bigg |_0^L = \frac{g\lambda L^2}{2} \sin(\theta)
    [/tex]
    ... which is what you want - because λL = m, the mass of the ladder.


    In the first derivation, you misidentified the elements - it should have gone from [itex]\tau = mgx\sin(\theta)[/itex] to [itex]d\tau = gx\sin(\theta)\cdot dm[/itex].
    In the second derivation you lost confidence just before the payoff :)
     
    Last edited: Jun 13, 2012
  9. Jun 13, 2012 #8
    You're right! I'm thrilled!!! I need to get better at recognizing things in the answers. This isn't the first time I've gotten something more or less right but didn't see it because I didn't make the connection. Thanks for seeing it through with me!:biggrin: Generalizing the 15 to L was I think the coup de grace that I was missing.
     
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