How Far From Shore Is the Dog After Walking on the Boat?

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A dog weighing 10.8 lb walks 8.5 ft toward shore on a flatboat weighing 46.4 lb, initially 21.4 ft from shore. The center of mass of the dog and boat system remains stationary due to conservation of momentum, leading to the boat moving under the dog. Calculations show the boat moves 1.6 ft in the water, while the dog moves 8.5 ft on the boat. Consequently, the dog is 14.5 ft from shore after walking. Understanding the center of mass is crucial for solving this problem effectively.
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Homework Statement


A dog weighing 10.8 lb, is standing on a flatboat so that he is 21.4 ft from shore. He walks 8.5 ft on the boat toward shore and then halts. The boat weighs 46.4 lb, and one can assume there is no friction between it and the water. How far is he from shore at the end of this time? (Hint: The center of mass of the boat + dog does not move. Why?)


Homework Equations





The Attempt at a Solution


Initially I considered that the boat will move exactly as much as the dog does, but I'm no longer sure it will. I am now of the opinion that maybe the dog will move a proportional distance of

8.50 ft x (his weight/total weight)

But I am not sure and would like a second opinion.

Thanks in advance.
 
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Hi JWDavid! :smile:
JWDavid said:
A dog weighing 10.8 lb, is standing on a flatboat so that he is 21.4 ft from shore. He walks 8.5 ft on the boat toward shore and then halts. The boat weighs 46.4 lb, and one can assume there is no friction between it and the water. How far is he from shore at the end of this time? (Hint: The center of mass of the boat + dog does not move. Why?)

Initially I considered that the boat will move exactly as much as the dog does, but I'm no longer sure it will. I am now of the opinion that maybe the dog will move a proportional distance of

8.50 ft x (his weight/total weight)

Nooo … if that were right, then the lighter the dog was, the less he would move …

but if he were a flea, the boat would stay still, and the proportion would be 1 (roughly) :wink:.

Hint: in problems like this, it's usually best to give letters to things … call the dog D, and the centre of the boat B, and the old and new centres of mass P and Q, and then calculate where P and Q are. :smile:
 
So I think I've gotten it: Here's what I did please confirm.

cm1 = (10.8*0 + 46.4*(L/2))/57.2 = .4L
cm2 = (10.8*8.5 + 46.4*(L/2))/57.2 = .4L + 1.6

cm2 - cm1 = 1.6 which ..? means the boat moves under the dog 1.6 feet

so the final answer is 21.4 - 8.5 + 1.6 = 14.5 feet - is this correct?
 
JWDavid said:
cm2 - cm1 = 1.6 which ..? means the boat moves under the dog 1.6 feet

so the final answer is 21.4 - 8.5 + 1.6 = 14.5 feet - is this correct?

Yup! :biggrin:

(except, technically i'd say that the boat moves 1.6 feet in the water, but 8.5 feet under the dog, and the dog moves 8.5 feet on the boat :wink:)

(and don't forget you're expected to answer "The center of mass of the boat + dog does not move. Why?", which you haven't yet shown)
 
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