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What distance does the boat move

  1. Mar 28, 2004 #1
    The mass of a boat is M=80kg, the mass of a boy is m=36kg. The boy moves from the stern to the bows of the boat. What distance does the boat move, if its length is 2.8m? At such low speeds the water resistance may be neglected. Hint: Center of mass remains at rest.

    Please help me because I am stumped :confused:
  2. jcsd
  3. Mar 28, 2004 #2

    Doc Al

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    This question is best suited for the Homework help section.

    Why not try your own hint? Calculate the center of mass (with respect to the boat) of the boat&boy in these two cases:
    (1) Boy at stern
    (2) Boy at bow

    What must have happened?
  4. Mar 28, 2004 #3
    Hasn't there been a question just like this at least twice in the past week or so?

  5. Mar 28, 2004 #4

    Doc Al

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    I thought so too, cookiemonster. See if you can find it.
  6. Mar 28, 2004 #5
    Well, I can't find them, so here we go again.

    Do you have an equation to find the center of mass of two objects, misstoi21? What is it? What would it be for when the boy is at the stern and what would it be for when the boy is at the bow? Don't these two have to be equal?

    Also, please don't make a new thread in order to post a reply. Please use the "Post Reply" button at the bottom of the page.

  7. Mar 28, 2004 #6


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    The formula for centre of mass is something like this

    [tex]d = \frac{m_1d_1 + m_2d_2}{m_1 + m_2}[/tex]

    d, d1, and d1 are all relative to a given point. I think you have to assume that the centre of mass for the boat is in half-way between the front and back of the boat.

    On second thought it could just be an inertia thing like this

    [tex]m_1 \Delta d_1 + m_2 \Delta d_2 = 0[/tex]

    I'll work on this a bit more.
    Last edited: Mar 28, 2004
  8. Mar 28, 2004 #7


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    Ok instead of trying to figure out some formula then get a value then plug it into another formula, just make 1 giant formula.

    d does not change

    [tex]d = d'[/tex]

    So then switch that around once

    [tex]d - d' = 0[/tex]

    Sub it in once

    [tex]\frac{m_1d_1 + m_2d_2}{m_1 +m_2} - \frac{m_1d_1' + m_2d_2'}{m_1 + m_2} = 0[/tex]

    [tex]\frac{m_1d_1 - m_1d_1' + m_2d_2 - m_2d_2'}{m_1 + m_2} = 0[/tex]

    [tex]\frac{m_1 \Delta d_1 + m_2 \Delta d_2}{m_1 + m_2} = 0[/tex]

    Since the denominator will never affect whether it is zero or not, it can just be cut out of the equation.

    [tex]m_1 \Delta d_1 + m_2 \Delta d_2 = 0[/tex]

    That's the same equation I had above which shows that inertia is conserved. How awesome is that? :smile:
  9. Mar 28, 2004 #8
    Actually, the location of the center of mass of the boat is irrelevant. Its change will be the same regardless of where it is.

  10. Mar 29, 2004 #9
    can anybody send or give me the address of a good site where I can find some good A level notes?about any topic.

    Thank you
  11. Mar 29, 2004 #10


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