What is the solution to the cow tipping problem?

  • Thread starter Thread starter asadpasat
  • Start date Start date
AI Thread Summary
The discussion focuses on the physics behind cow tipping, specifically the calculations involved in determining the forces at play. Participants analyze the geometry of a cow's body, using a rectangle and diagonal to find the center of mass and apply the lever equation. There is confusion about the necessity of certain variables, particularly "a," in the equations, which are clarified as essential for maintaining dimensional consistency. The conversation also touches on the efficiency of deriving certain values, with suggestions to simplify the approach. Ultimately, the discussion emphasizes the importance of understanding the underlying physics to solve the cow tipping problem effectively.
asadpasat
Messages
41
Reaction score
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)
 
Physics news on Phys.org
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
 
Back
Top