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Net force exerted on the two-particle system

  1. Jan 20, 2009 #1
    net force exerted on the system quick question- halllp lol.

    1. The problem statement, all variables and given/known data
    The vector position of a 3.10 g particle moving in the xy plane varies in time according to the following equation.

    r1 = (3i+3j)t + 2jt^2

    At the same time, the vector position of a 5.15 g particle varies according to the following equation.
    r2= 3i-2it^2 -6jt

    For each equation, t is in s and r is in cm. Solve the following when t = 2.00

    (e) Find the net force exerted on the two-particle system.
    i μN
    j μN


    2. Relevant equations

    f=ma

    3. The attempt at a solution

    Took the 2nd derivative to get the acceleration vectors.

    r1 : 4 j (mass is 3.1g)
    r2 : -4 i (mass is 5.15g)



    I tried this:

    ( (m2)(4 i) ) / (m1 + m2)

    and got -1.93515 in the i-hat direction. It is incorrect : "Your answer is off by a multiple of ten."
     
    Last edited: Jan 20, 2009
  2. jcsd
  3. Jan 20, 2009 #2

    Doc Al

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    Re: net force exerted on the system quick question- halllp lol.

    Looks good.

    Not sure what you were going for here.

    What's the net force on each mass?
     
  4. Jan 20, 2009 #3
    Well the vector r2 is the only mass going in the i-hat direction. And since its a system, I divided by the total mass
     
  5. Jan 20, 2009 #4

    Doc Al

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    Were you trying to find one component of the net force?
    Does that make sense? Check units.
     
  6. Jan 20, 2009 #5
    The answers they want is in μN. Newtons I understand, but what is μ doing in there?
     
  7. Jan 20, 2009 #6

    Doc Al

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    Staff: Mentor

    That just means micro = 10-6. (Realize that you're given measurements in cm and grams.)
     
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