What is the solution to the cow tipping problem?

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SUMMARY

The cow tipping problem involves calculating the conditions under which a cow can be tipped over by applying force at the optimal angle. The final equation derived is (F) = (mg (x/2)) / (a+b), where (F) is the force applied, (mg) is the weight of the cow, (x) is the width of the cow, and (a) and (b) are the segments of the diagonal of the rectangle surrounding the cow. The necessity of "a" in the equation is to maintain dimensional consistency when calculating the lever arm distance. Understanding the geometry and physics behind the problem is crucial for accurate calculations.

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  • Understanding of basic physics concepts, specifically lever mechanics.
  • Familiarity with trigonometric functions, particularly cosine.
  • Knowledge of center of mass and its implications in physics.
  • Ability to interpret and manipulate algebraic equations.
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  • Study the principles of lever mechanics in physics.
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Physics students, engineers, and anyone interested in the mechanics of force application and lever systems will benefit from this discussion.

asadpasat
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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)
 
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OK, I'll bite. What's the "cow tipping problem"? A diagram and statement of the problem would be nice.
 
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
 
lightgrav said:
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
Got it. Thanks!
 
asadpasat said:
(F)(a+b)= mg a cosθ [ I don't understand why is "a" necessary]
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.
 
Doc Al said:
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.
It seems like a long way from solving for (x/2) by cos, and then substituting cos, just to get (x/2).
 
asadpasat said:
It seems like a long way from solving for (x/2) by cos, and then substituting cos.
I would have went directly to x/2, since you were given that up front.
 
Doc Al said:
I would have went directly to x/2, since you were given that up front.
ok, great. Thanks
 

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