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Cow tipping problem

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  1. Sep 25, 2015 #1
    So I saw the cow tipping problem and I am having trouble figuring out how they got to the final equation.
    Imagine making a rectangle around a cows body. Making a diagonal across the rectangle and center of mass being in center of the diagonal. One half of the diagonal is "a", and second is "b". Angle between the ground and the diagonal is theta. Drawing a Fg from center of mass divides the bottom line of rectangular in half (x/2)
    From lever equation: (Fe)(de)=(Fl)(dl)
    Transforming it: (F)(a+b)= (Fl)(dl)
    (F)(a+b)= mg a cosθ [ I don't understand why is "a" necessary]
    cosθ= (x/2)/a
    (F)(a+b)= mg a ((x/2)/a)
    (F)(a+b)= mg (x/2)
    (F)= (mg (x/2)) / (a+b)
     
  2. jcsd
  3. Sep 25, 2015 #2

    Doc Al

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    Staff: Mentor

    OK, I'll bite. What's the "cow tipping problem"? A diagram and statement of the problem would be nice.
     
  4. Sep 25, 2015 #3

    lightgrav

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    The cow-tipper is pushing with strength Fe , applied to the cow at the top corner, pushing in the optimal direction (perp. to the diagonal).
    in line 3, the "lever-arm" for the gravity Force (that is, distance from pivot perp. to the weight vector) is x/2
     
  5. Sep 25, 2015 #4

    Doc Al

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    Got it. Thanks!
     
  6. Sep 25, 2015 #5

    Doc Al

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    You need the perpendicular distance between mg and the pivot, which is "a cosθ". (Or x/2.) Without the "a" the equation would be dimensionally inconsistent.
     
  7. Sep 25, 2015 #6
    It seems like a long way from solving for (x/2) by cos, and then substituting cos, just to get (x/2).
     
  8. Sep 25, 2015 #7

    Doc Al

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    I would have went directly to x/2, since you were given that up front.
     
  9. Sep 25, 2015 #8
    ok, great. Thanks
     
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