Newtons Second Law For A System Of Particles

AI Thread Summary
The problem involves a 5.4 kg dog on a 16 kg flatboat, initially 6.1 m from shore, walking 2.1 m towards the shore. The center of mass of the system must remain constant, leading to the equation that balances the initial and final positions of the dog and the boat. The calculations show that as the dog moves, the center of mass shifts, requiring adjustments in the boat's position. The final position of the dog from the shore is determined to be approximately 4.71 m, but initial calculations were incorrect due to misinterpretation of the center of mass movement. The solution emphasizes the importance of considering both the dog’s and the boat’s movements in the system.
richard karn
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Homework Statement



In figure (a), a 5.4 kg dog stands on a 16 kg flatboat at distance D = 6.1 m from the shore. It walks 2.1 m along the boat toward shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore.


Homework Equations



Center of mass = (x1m1 + x2m2)/(m1+m2)


The Attempt at a Solution



Since the center of mass can't change i set the two equal.

x1 (position of dog) = 6.1

x2 (center of canoe) = x

(6.1*5.4 + 16x2)/21.4 =

(5.4*(6.1-2.1) + 16(x2+Δcom))/21.4

which gives the change in the center of mass of the boat is .70875 so the dog only moves 2.1-.70875 and starting 6.1 meters away he is 4.70875 but that's wrong
 
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When the dog moves towards the shore, the center of gravity of the boat moves away from the boat.Along with that the dog also moves away from the the boat. If x is the distance moves, the equation becomes

[(6.1- 2.1+x)*5.4 + 16*(x2 + x)])/21.4 = the initial center of mass.

Now solve for x.
 
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