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Dynamics in two dimensions problem

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-10-15_12-28-8.png
    2. Relevant equations
    xf = x0 + v0cos(θ)t + (1/2)axt2
    yf = y0 + v0sinθt - (1/2)gt2
    F = ma
    3. The attempt at a solution

    I first tried to find the acceleration in the x direction so I could solve for the force
    xf = x0 + v0cos(θ)t + (1/2)axt2

    In order to find the time I used the equation for the y coordinate
    yf = y0 + v0sinθt - (1/2)gt2

    The initial and final y positions will both be zero so
    0 = 0 + v0sin(θ) - (1/2)gt2
    t = 2v0sin(θ)/ g = .702 sec

    Then I plugged time back into the x equation

    2.9 = v0cos(θ)(.702) + (1/2)a(.702)2
    a = -2.23

    Since I know the acceleration I can find the force by Newtons second law

    F = ma

    F = .015(-2.23) = -.033 N

    It appears that I am a sign off from the correct answer
     

    Attached Files:

  2. jcsd
  3. Oct 15, 2016 #2
    0 = 0 + v0sin(θ) - (1/2)gt2 t = 2v0sin(θ)/ g = .702 sec

    Don't you need to take the square root here?
     
  4. Oct 15, 2016 #3

    PeroK

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    What do you mean by "a sign off"?
     
  5. Oct 15, 2016 #4
    The correct answer is 0.034 N so it looks like I'm getting the negative of the right answer.
     
  6. Oct 15, 2016 #5

    PeroK

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    And what does the question ask for?
     
  7. Oct 15, 2016 #6
    Oh I see. Since it's asking for the magnitude it's just asking for the value and not the sign, right?
     
  8. Oct 15, 2016 #7

    PeroK

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    Well, I would say it's asking for the magnitude! Which is always a positive number. The negative sign is actually a direction: opposite the direction of motion, in this case.

    "Value" suggests magnitude and direction to me.
     
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