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CTR Mass - dog walking on boat

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    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?)


    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Nov 18, 2008 #2

    tiny-tim

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    Hi JWDavid! :smile:
    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:
     
  4. Nov 18, 2008 #3
    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?
     
  5. Nov 18, 2008 #4

    tiny-tim

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    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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