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Walking on a boat

  1. Jul 21, 2014 #1
    hi all, here's the problem
    "A dog sits on the left end of a boat of length L that is initially adjacent to a dock to its right. The dog then runs toward the dock, but stops at the end of the boat. If the boat is H times heavier than the dog, how close does the dog get to the dock?

    Ignore any drag force from the water. "

    So, on the internet I found the answer that it's L/(1+H) but it didn't provide any explanation

    could someone help me understand how to approach problems like this. Basically I'm really stuck on problems like " walking on a block on a friction less surface " . So I know it's got something to do with the fact that the Center of Mass doesn't change position, only the block ( boat in that case ) and the object moving( dog). Could someone help me out with this ? thank you so much if you do :)
  2. jcsd
  3. Jul 21, 2014 #2


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    If the dog moves 1 unit distance to the right, the boat beneath his feet must move 1/H unit distance to the left to keep the center of gravity at the same place. The net movement of the dog relative to the boat is 1+1/H. The net movement of the boat relative to the water is 1/H.

    Express this ratio as boat movement (relative to dock) divided by dog movement (relative to boat)

    [itex]\frac {\frac {1} {H}}{1 + \frac {1}{H}} [/itex] = [itex]\frac {\frac {1} {H} }{\frac{H}{H} + \frac {1}{H}}[/itex] = [itex]\frac {\frac {1} {H} }{\frac{H + 1}{H}} [/itex] = [itex]\frac{1}{H+1}[/itex]

    If the dog moves distance L from the left end of the boat to the right, how far does the right end of the boat move away from the dock?
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