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Homework Help: CENTER OF MASS: finding final position of skater 3 given final positions of others

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Three ice skaters are on a perfectly smooth frictionless lake. They are together at rest at the middle of the lake when they push off on each other. When skater #1 (m1 = 80 kg) is 6.00 meters EAST of the starting position, and skater #2 (m2 = 60 kg) is 6.00 meters NORTH of the starting position, how far away is skater #3 (m3 = 40) kg from the starting position?


    My first intuition was to solve it this way:

    Xcom = ((80kg)(6m)+(60kg)(6m) + (40kg)(X3))/180 kg

    If I set Xcom to zero, X3 = 3m. Is it that simple?
    However, I think I need to break it apart into components, but I'm still unclear about how to do this.

    Please help!
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    Doc Al

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    Staff: Mentor

    Re: CENTER OF MASS: finding final position of skater 3 given final positions of other

    Yes, treat the x (East) and y (North) components separately. What are the x and y components of the position of masses #1 and #2?
     
  4. Nov 6, 2011 #3
    Re: CENTER OF MASS: finding final position of skater 3 given final positions of other

    Xcom = ((80kg)(+6m) + [STRIKE](60kg)(0m)[/STRIKE] + (40kg)(X3))/180 kg

    Ycom = ([STRIKE](80kg)(0m)[/STRIKE] + (60kg)(+6m) + (40kg)(X3))/180 kg

    Where do I go from here?
     
    Last edited: Nov 6, 2011
  5. Nov 6, 2011 #4

    Doc Al

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    Staff: Mentor

    Re: CENTER OF MASS: finding final position of skater 3 given final positions of other

    Good.
    That's exactly what you're supposed to do.
    Show how you got that.
     
  6. Nov 6, 2011 #5
    Re: CENTER OF MASS: finding final position of skater 3 given final positions of other

    Doc,

    I didn't find X3=3m by breaking it up into components. I found X3 by setting Xcom = 0 = ((80kg)(6m)+(60kg)(6m)+(40kg)(X3))/180kg in my original format and solving for it algebraically.

    But if I break it into components and set Xcom and Ycom to zero:

    Xcom=0= ((80kg)(6m) + 0 + (40kg)(X3))/180
    I get..Xx3= -12 m

    Ycom=0=((0 + (60kg)(6m) + (40kg)(X3))/180
    I get..Xy3 = -9 m

    I'm missing the point.
     
  7. Nov 7, 2011 #6

    Doc Al

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    Staff: Mentor

    Re: CENTER OF MASS: finding final position of skater 3 given final positions of other

    Your original approach is incorrect, since it mixes up the x and y coordinates.

    Good! You've just found the position coordinates of mass #3:

    (X3, Y3) = (-12, -9)

    Use those coordinates to find its distance from the starting point.
     
  8. Nov 7, 2011 #7
    Re: CENTER OF MASS: finding final position of skater 3 given final positions of other

    Thank you very much Doc!
     
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