Finding shift in centre of gravuty

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To find the minimum lateral shift in the center of gravity for a standing man carrying a 14 kg pail of water, the combined mass of the man and the pail must be considered. The man weighs 80 kg, and to balance the load on each foot, the center of mass needs to be calculated by treating the man and the pail as separate point masses. The calculation shows that the center of mass is located 31 cm from the man's original center of gravity. To achieve an equal load on each foot, the man must shift approximately 5 cm to the left. Understanding the center of mass concept is crucial for solving this problem effectively.
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Homework Statement



Find the minimum lateral shift in position of the centre of gravity of a standing man whose mass is 80 kg, when is carrying a 14 kg pail of water. Assume that he wishes to have an equal load on each foot. How is this shift in his centre of gravity accomplished?

Homework Equations



F=0

http://img339.imageshack.us/img339/1712/pailgg5.png

The Attempt at a Solution



I don't know where to start! With the pail, he has to have 58 kg on each foot. There's the weight of the bucket and the weight of the man, but since he's shifting, should I divide him in half? Like 40 kg on each foot?
 
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Start by finding the center of gravity of "man + pail". (It will be some distance to the left of the man's center of gravity.)
 
How do I do that? :redface:
 
fizzyy said:
How do I do that?
Treat the pail and the man as two separate point masses and find the center of mass of both. (All you care about is the horizontal position of the center of mass of both compared to the original position of the man's center of mass.)

If you don't know how to calculate the center of mass, read this: http://hyperphysics.phy-astr.gsu.edu/HBASE/cm.html"
 
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Aye, okay.

x = 14(0) + 80 (36) / 14 + 80
x = 31 cm

Thanks Doc Al.
 
Good. So how far must he shift to the side?
 
I believe.. 5 cm to his left? Hopefully I'm understanding this now.
 
Sounds good to me.
 
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