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Newtons Law Problem involving kinematic equation

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A duck has a mass of 3.0 kg. As the duck paddles, a force of 0.06 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.13 N in a direction of 60° south of east. When these forces begin to act, the velocity of the duck is 0.12 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.0 s while the forces are acting.


    2. Relevant equations

    ƩF = m*a
    D = VoT + (1/2)aT^2

    3. The attempt at a solution

    Fx = 0.06 + cos(60)*.13 = .125 m/s
    Fy = sin(60)*.13 = .113 m/s

    sqrt(.125^2 + .113^2) = .1685N

    .1685/3.0 = 0.0561 m/s^2

    D = (0.12)(2) + (1/2)(.0561)(2)^2 = .3522
    This is where I am confused, as I think I thought I was doing everything correct up until this point. When I enter this into the answer sheet it says it is wrong. I mistakenly used .13 for Fy instead of .113 and it gave me .326 m which is the correct answer.

    Also I know to find the direction you use arctan(Fy/Fx) but I can't find it...

    Any help is greatly appreciated.

    Thank you.
     
  2. jcsd
  3. Oct 4, 2011 #2

    gneill

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    Staff: Mentor

    Careful with the units! Those are forces, not velocities.
    That's better! That's the magnitude of the net force.
    Okay, that's the magnitude of the acceleration. But what are you going to do with that?
    Uh oh :frown:

    The acceleration vector and initial velocity vector do not lie in the same direction. So you can't just use their magnitudes to find the displacement.

    Instead, keep the velocity and acceleration broken into components and perform the displacement calculation separately on each; Find Δx and Δy, then combine the results into a single displacement. These components will also give you the direction of the displacement.
     
  4. Oct 4, 2011 #3
    So do you mean

    x = (0.12)(2)+(1/2)(.125)(2)^2 = .49
    y = (0.12)(2)+(1/2)(.113)(2)^2 = .466

    x+y = .49 + .466 = .956

    then do arctan(.466/.49)
     
  5. Oct 4, 2011 #4

    gneill

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    Staff: Mentor

    Almost. The initial velocity will also have separate components (and they're not the same). What are they?
    Also, be sure that you have accelerations to plug into the equation (not forces).
    Also, beware of the signs on your components! Draw a diagram: the current is applying its force in a direction 60° SOUTH of east.
     
  6. Oct 4, 2011 #5
    The initial velocity for x would be 0.12 and for y=0.

    x acceleration would be .125/3.0 because of F/M = A
    y acceleration would be .113/3.0 " "

    x = .041 m/s^2
    y = .037 m/s^2

    x = (0.12)(2)+(1/2)(.041)(2)^2 = .322
    y = (0)(2)+(1/2)(.037)(2)^2 = .074

    then would you do sqrt[.322^2 + .074^2] = .330
    or
    .322 + .074 = .396
     
    Last edited: Oct 4, 2011
  7. Oct 4, 2011 #6

    gneill

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    Staff: Mentor

    Did you check the signs for the components? What does an angle "South of East" imply?
    Other than the component sign issue, you might want to keep a few more decimal places for intermediate results. I think that truncation and rounding errors are creeping into your results.
    Definitely the square root of the sum of the squares. The components of a Cartesian vector always "add in quadrature" that way.
     
  8. Oct 4, 2011 #7
    Thanks for being so patient with me and all your help. I finally got it. Keeping separate the velocitys and acceleration components really helped.
     
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