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Find the net force with the given position vector

  1. Sep 13, 2010 #1
    1.The position vector of a 3000N helicopter is given by:
    r=(.02 m/(s^3))(t^3)i + (2.2m/s)tj - (0.06 m/(s^2))(t^2)k
    Find the net force on the helicopter at t = 5s.



    2. I really have no clue where to begin.



    3. I started doing this:
    r= (.02m/(s^3))(5s^3)i + (2.2m/s)(5s)j - (.06m/(s^2)*(5s)^2k
    r= 2.5mi + 11m + 1.5m
    then I drew a 3d graph with each axis labeled and two right triangles but have no clue
    where to go from here.
     
  2. jcsd
  3. Sep 14, 2010 #2

    rl.bhat

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    The given problem can be written as

    r = a*t^3*i + b*t*j -c*t^2*k

    Find d(r/dt and d^2(r)/dt^2. That gives you the acceleration. Find the magnitude of a.

    Weight of helicopter is given. Find its mass. Then the force on the helicopter = m*a
     
  4. Sep 14, 2010 #3
    Thank you so mu rl.bhat.

    After the 2nd derivative I get a= (.12m/s^3)(t) - .12m/s^2
    = (.6m/s^2)- .12m/s^2= -.6 m/s^2
    F=m*a m=F/a m=3000n/9.81m/s^2=305.8kg

    m*a=F 305.8 kg * -.6m/s^2=-183.48N Is that thought process correct?
     
  5. Sep 14, 2010 #4

    rl.bhat

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    After the 2nd derivative I get a= (.12m/s^3)(t) - .12m/s^2

    It should be

    a= (.12m/s^3)(t)*i - .12m/s^2*k

    Find the resultant a.
     
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