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2 people in a canoe, switch places, canoe moves, find mass of one person

  1. Oct 24, 2008 #1
    Ricardo, mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 20 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moved 45 cm relative to a submerged log during the exchange and calculates Carmelita's mass, which she has not told him. What is it?

    I don't really get those sort of problems, so I tried cutting and pasting into a problem similar to this one. This is what I tried to do:

    xcom = xcom

    80(0) + 20(1.5) + 3m = 80(3-x) + 20(1.5-x) + m(3-x)

    After plugging 0.45 for x, I ended up with m = 500 kg.

    What's the right way to do this problem?
    Last edited: Oct 24, 2008
  2. jcsd
  3. Oct 24, 2008 #2

    Doc Al

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    Staff: Mentor

    You seem to understand that the center of mass of the system does not change. What I'd do is find the change in the center of mass (mΔx) due to each object's movement--the three contributions must add to zero. Hint: Measure the movements with respect to the submerged log.
  4. Oct 24, 2008 #3
    So I tried to find the change in position of com. It didn't work to well. This is what I tried:

    xcom = (80(0) + 20(1.5) + 3m) / (100 + 3m)

    xcom = (80(0.45) + 20(1.5 + 0.45) + 3.45m) / (100 + 3m)

    The I subtracted the first from the second and set it equal to 0.45.

    I guess that was a bad attempt. What hints can you give me to get on the right track?
  5. Oct 24, 2008 #4

    Doc Al

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    Just what I told you: Find mΔx for each object, measuring Δx with respect to the submerged log. Also, since Ricardo is heavier, if he moves to the right, which way does the canoe move?
  6. Oct 24, 2008 #5
    I had the same setup equation as before, except this time I changed the positions.

    I assume that at Ricando is originally at x = 0, and the log is also there. When Ricardo and Carmelita switched places, the canoe moves left 0.45 m. So this time I made the equation:

    (80(0.45) + 20(1.5-0.45) + 0.45m) / (100 + m) = (225 - 0.45m) / (100 + m)

    The original equation is: xcom = (80(0) + 20(1.5) + 3m) / (100 + m)

    I subtracted the original from the switched

    (225 - 0.45m - 30 - 3m) / 100 = 0.45
    195 - 3.45m = 45 + 0.45m
    m = 38.46 kg

    Is this what you mean by mΔx?
  7. Oct 24, 2008 #6

    Doc Al

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    Here's what I mean by mΔx:

    The canoe (m = 20) moves 45cm left with respect to the log (Δx = -0.45), so mΔx = - 20*0.45.

    Ricardo moves 3 m to the right with respect to the canoe, so how far does he move with respect to the log? Use that logic to figure out mΔx for him.

    Then do similar thinking for Carmelita.

    Note: The change in center of mass of the system (canoe and two people) will equal ∑(mΔx)/(Total mass). But since we know that the change is zero, we can just say ∑(mΔx) = 0. (I hope that makes sense.)
  8. Oct 24, 2008 #7
    I assumed that Δx would be the original - 0.45. I can't think of it in any other way. Then I guess my first set up is correct since I figured what you mean by com doesn't change.

    So would Ricardo's Δx be 3.0 - 0.45?

    So is the equation

    80(0) + 20(1.5) + 3m = 80(2.55) + 21 - 0.45m

    m = 56.52 kg

  9. Oct 24, 2008 #8

    Doc Al

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    Looks good.

    Here's how I would have done it:

    canoe: mΔx = -(20)*(0.45)
    man: mΔx = (80)*(3-0.45)
    woman: mΔx = (m)*(-3.45)

    Add it up and set equal to zero. (Equivalent to what you did.)
  10. Oct 24, 2008 #9
    Thank you so much! I understand this type of problem a lot better now.
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