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2 man on a boat

  1. Aug 25, 2004 #1
    2 man, mass m1 and m2 stands on a boat mass M at L distance apart on static water. they slowly swop their places. find the distance the boat move due to this.

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    i know that the centre of mass remains unchange, and at start,
    Xcm=m1x1+m2x2+Mxb/m1+m2+M

    then after changing place, i let y be the distance moved, Xcm becomes
    Xcm'=m1(x2+y)+m2(x1+y)+M(xb+y)/m1+m2+M

    do i simply divide the 2 equations? then i'll end up with lotsa varibles....

    then i am pretty much stuck here.. pls offer ur advice, hints, help ... thx in advance :)
     
  2. jcsd
  3. Aug 25, 2004 #2

    HallsofIvy

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    First, state clearly what it is you are trying to determine What do you mean by "distance the boat moves"? What coordinate system are you using?

    The center of mass of the entire system (boat and men) remains unchanged but the center of mass of the boat alone WILL move- that's the motion you want to find. I would suggest taking some fixed point (that will not move as the boat moves) in front of the bow of the boat as x= 0 and measure everything from that.

    I do notice that there is no L in your equations. It seems to me that the distance between the two men is of importance.

    The center of mass of the system, initially, is
    Xcm= (m1x1+m2x2+Mxb)/(m1+m2+M) as you say, but, taking x1 to be the position of the man in the bow, the position of the other man, x2= x1+ L so this is the same as Xcm= (m1x1+m2(x1+L)+Mxb)/(m1+m2+M)

    After the men exchange places we have
    Xcm= (m2x1+m1(x1+L)+Mxb2)/(m1+m2+M) - the only difference is that m1 and m2 have swapped places and xb2 is the new position of the center of mass of the boat alone. No need to divide anything- since Xcm is the same in both equations, set the right sides equal:

    (m1x1+m2(x1+L)+Mxb)/(m1+m2+M) = (m2x1+m1(x1+L)+Mxb2)/(m1+m2+M) and solve for xb- xb2.

    There are not "lotsa variables"- all except xb and xb2 are given constants. xb- xb2 is the distance the boat moves.
     
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