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How far has the boat moved and in which direction?

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    Jane and her little brother John are fishing, sitting at opposite ends of a boat (Jane on the left, John on the right). Jane is having incredible luck! John, not so much. Jane offers to switch places in the boat with John. Jane has mass of 50 kg, john has a mass 30kg, and the boat is 2.0m long with a mass of 20 kg. When Jane and John have switched places, how far has the boat moved and in which direction?

    2. Relevant equations
    1/M (m1x1+ m2x2)
    3. The attempt at a solution
    1/20 (50(1)+30(1))=4.0m
     
  2. jcsd
  3. Oct 15, 2016 #2

    gneill

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    Did you have a question?

    Can you explain your use of your relevant equation? What does it represent?
    How did you assign values to x1 and x2? How does the length of the boat come into play?
     
  4. Oct 15, 2016 #3
    the question is how far has the boat moved and in which direction?
    the x1 and x2 are the centre of masses of the object
     
  5. Oct 15, 2016 #4

    gneill

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    That's the problem statement. What is your question? Are you happy with your solution?
    Why are they both "1" ? What is the coordinate system? Where are these "1's" being measured from?
     
  6. Oct 15, 2016 #5
    i'm wondering if the solution i did was correct. im unsure what the centre of mass would be, i assumed they were both one which is half the boat
     
  7. Oct 15, 2016 #6

    haruspex

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    The boat is only 2m long. If it moves 4m, the mass centre of the system has certainly moved.
    Your x1 and x2 are presumably displacements from somewhere - where? should they both be positive?
    Your equation seems to be more hopeful than based on any principles. Try to describe the logic behind it.
     
  8. Oct 15, 2016 #7
    I used that equation because that is what my prof gave us in lecture. Not a lot of explanation in how to use it though. So I see where you think I'm hopeful with it. No one direction should be negative, since the people move in opposite detection.
     
  9. Oct 15, 2016 #8

    haruspex

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    An equation is meaningless without a context defining what all the variables represent.
    I can offer you a context for the Prof's equation. M= m1+m2; m1 and m2 are masses at (signed) offsets x1 and x2 respectively from some reference point along the line of their centres. The result is the offset of the common mass centre from that same reference.
    In the present problem you have three masses. Can you see how to extend the equation to handle that? Choose a reference point. Can you now work out what numbers to plug in?
     
  10. Oct 16, 2016 #9
    i solved the question on my own, no further explanation needed
     
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