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Homework Help: Center of mass question

  1. Nov 11, 2006 #1
    this question has really stumped me.

    23 people on a boat rest on water without friction. Each person has an average mass of 70 kg, and the boat itself weighs 10^4 kg. The entire party walks the entire 8 m distance of the boat from bow to stern. How far (in meters) does the boat move?

    i know i have to use this equation somwhere:

    x_cm = ( m1x1 + m2x2 ) / ( m1 + m2 )

    so far i think
    m1 = 23 * 70
    x1 = 0
    m2 = ?
    x2 = ?

    can anyone help me?

    Last edited: Nov 12, 2006
  2. jcsd
  3. Nov 12, 2006 #2
    You have data to calculate the center of mass before the people move. Since, there are no external forces acting on the c.m., will he change?


    [tex]x_{cm}_{i} = x_{cm}_{f}[/tex]​
  4. Nov 12, 2006 #3
    is it like this?

    in an isolated system, momentum is conserved. so, m1v1=m2v2, were m1 is the mass of boat, v1 is it's velocity, m2 is the mass of all the people, v2 is the velocity of them,

    d2/t=m1/m2 x d1/t
    d2=m1/m2 x d1

    since every1 covers d1, assume that the center of mass is over 1point that covers the 8meters....

    does this make sense?
  5. Nov 12, 2006 #4
    no i dont think im makes sense cuz we're not dealing with velocity here
  6. Nov 12, 2006 #5
    so i got this so far but i dunno which is which to plug in

    [tex]m_{1}x_{1}_{i} + m_{2}x_{2}_{i} = m_{1}x_{1}_{f} + m{2}x_{2}_{f}[/tex]​

    am i on the right track?
    Last edited: Nov 12, 2006
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