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Finding direction of net force and direction of travel

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    A 0.35 kg particle moves in an xy plane according to x(t) = - 11 + 1 t - 6 t3 and y(t) = 19 + 3 t - 8 t2, with x and y in meters and t in seconds. At t = 0.7 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel?

    2. Relevant equations
    Fx=m(ax)
    Fy=m(ay)

    3. The attempt at a solution
    I got the first part right by taking the second derivative of each plane and applying it to the equations.
    (ax)= dvx/dt = -36t m/s^2
    (ay) = dvy/dt = -16 m/s^2

    and by applying the relevant equations
    Fx= -8.82
    Fy= -5.6

    By using Pythagoras theorem I found that the net force to be
    Fnet= 10.448
    Which is right.

    After drawing a picture
    in part b my attempt to solve was
    tan^-1(theta)= Fy/Fx

    Which is tan^-1(theta) = -5.6/-8.82
    theta = 32.41
    which I got wrong

    ___________________________________________

    Part C
    My attempt was applying in the velocity functions of both planes which are

    Vx= 1-18t^2
    Vy= 3-16t

    and finding that

    Vx= -7.82
    Vy= -8.2

    After that
    tan^-1(theta)= -8.2/-7.82
    theta= 46.358

    which I also got wrong.


    Thanks in advance.
     
  2. jcsd
  3. Sep 27, 2015 #2

    Student100

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    For part B, what is the angle you found in reference to?
     
  4. Sep 27, 2015 #3
    Is this an online homework platform of some sort? If so, they may be sticklers for things like significant figures and so forth. To actually work through the problem, I'd have to take just a little time here.
     
  5. Sep 27, 2015 #4
    For part B, remember that the given interval is from -180 degrees to 180 degrees. Using the inverse trig functions will give you an angle, but they will always be between 0 and 90. It is up to you to determine from which axis the angle should be measured from and make the adjustment. Which quadrant does the total force lie in?
     
  6. Sep 27, 2015 #5

    Student100

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    Significant figures don't matter right now. For raw computations with no analogue in experiment like this they don't ever matter except to teach people how to use them.
     
  7. Sep 27, 2015 #6
    Based on the drawing it should be in the 3rd quadrant.
    I tried the negative angle and it was wrong.
     
  8. Sep 27, 2015 #7
    In reference to the positive x axis I suppose.

    I am just confused now if I am supposed to use the negative angle or not because the net force is under the +x axis
     
  9. Sep 27, 2015 #8

    Student100

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    What negative angle did you try?

    The angle you found? No it isn't. It is in the third quadrant, like you stated, however.
     
  10. Sep 27, 2015 #9
    I think it was a wild try, makes no sense but I just added the negative sign.

    What do you propose I should try
     
  11. Sep 27, 2015 #10

    Student100

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    First you need to realize what the 32 whatever degree angle is in reference too. Then you need to draw an angle from the positive x axis to the vector that results in an angle between [180, -180].

    Draw a picture.
     
  12. Sep 27, 2015 #11
    Snapshot.jpg

    I am sorry for the drawing's quality, it's my first time using this site.


    It makes sense that the angle relative to the +x axis is negative and it's less than 90.

    but the angle I get is positive. What possible negative angle I could get ?

    I tried using sin(theta) = Fy/F

    sin^-1(theta) = -5.6/10.44
    theta = -32.41

    since sin and cosine give angles from (-180,180] shouldn't this be right ?
     
  13. Sep 27, 2015 #12

    Student100

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    No worries on drawing quality, but your picture is wrong. If you want to add $$F_y\hat{y}$$ and $$F_x\hat{x}$$ to produce $$\vec{F}$$ how would add them together graphically?
     
  14. Sep 27, 2015 #13
    Snapshot.jpg

    I realized my mistake, this one should be right ? So our angle is greater than 90 and is negative.
     
  15. Sep 27, 2015 #14

    Student100

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    Correct, write a formula for that angle now, what is the 32 in reference to?
     
  16. Sep 27, 2015 #15
    It is in reference to the -x axis.

    Theta = 180-32= 148 degrees in negative
     
    Last edited: Sep 27, 2015
  17. Sep 27, 2015 #16

    Student100

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    That works, I would just say -180+32 = -148

    Now retry part C!
     
  18. Sep 27, 2015 #17
    Snapshot.jpg

    These are the velocity vectors and their sum

    Shouldn't the angle be the same ? Or is it the same problem with the drawing ?
     
  19. Sep 27, 2015 #18

    Student100

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    Same error with the drawling, redraw the resultant after adding the two component vectors.
     
  20. Sep 27, 2015 #19
    Nice, thank you very much!

    We had a whole chapter about vectors and we learned that we were supposed to do it the way I did it, why is it wrong for the velocity vector to be this way ?
     
  21. Sep 27, 2015 #20

    Student100

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    Really, you guys learned to do it the wrong way then. Most likely you missed something or got confused when they were teaching it.

    A vectors component can be represented by a vector itself by multiplying it with a base vector (what you were drawing.) Then to add them graphically you can place the head of one at the tail of the other, which produces 1/2 of a parallelogram. The resultant is from the beginning and terminal ends of the parallelogram.

    So the way you're drawing it is the wrong way for any vector! What you're actually drawing is the displacement vector between two points.
     
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