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Homework Help: Finding magnitude of net force

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data
    A 2.60 kilogram mass is moving in a plane, with its x and y coordinates given by x = 5.40t2 - 1.30 and y = 2.75t3 + 1.80, where x and y are in meters and t is in seconds. Calculate the magnitude of the net force acting on this mass at t = 2.10 seconds.


    2. Relevant equations
    Our teacher has been making us use the position update formula and the momentum principle:
    [tex]\Delta[/tex]r=Vavg[tex]\Delta[/tex]t
    [tex]\Delta[/tex]P=Fnet[tex]\Delta[/tex]t



    3. The attempt at a solution

    I can figure out the velocity at a given point in both the y and x directions by finding the derivative:
    dx/dt=8.4t
    dy/dt=8.25t2

    But I don't know how to apply this. I really don't know where to start.

    Edit: sorry, dx/dt should be 10.8t
     
    Last edited: Jan 30, 2010
  2. jcsd
  3. Jan 30, 2010 #2
    Well the second derivative of the position functions or the first derivative of the velocity functions should give you a(t). Then the second law F=ma. So you have then you should have the forces in the x and y direction at the time given. Add those two vectors together... pythag. theorem.
     
  4. Jan 30, 2010 #3
    Thanks for the reply. I figured out that way of solving this problem after I posted it. However, our teacher won't let us use f=ma yet, as we haven't gotten to it in our class yet.

    I no longer NEED help on this problem since I have the correct answer, but does anyone know how to do it with only the position update formula and momentum principle?
     
  5. Jan 31, 2010 #4

    gabbagabbahey

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    Homework Helper
    Gold Member

    Once you know the components of velocity, just use them to calculate the momentum at two different times, say [itex]t=t_0[/itex] and [itex]t=t_0+\Delta t[/itex], and substitute it into the momentum principle formula you gave. The smaller you make [itex]\Delta t[/itex], the more accurate your result will be; and in the limit that [itex]\Delta t\to 0[/itex], your result becomes exact (taking this limit is exactly the same thing as taking the derivative [itex]\frac{d\textbf{P}}{dt}[/itex], so this should be no surprise to you).
     
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