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Math Help - Vector Problem

  1. #1
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    Vector Problem

    Could someone please verify if this problem is done correctly please. If not where did I go wrong.

    The problem:

    Use the component method to find the resultant(magnitude and direction) of the forces listed below.

    A 200 N at 20 degrees N of East
    B 250 N at 180 deg.
    C 500 N at -150 deg. S of East

    My attempt at the solution:

    Ax = 500 cos (-150) = - 433
    Ay = 500 sin (-150) = -250
    Bx = -250
    By = 0
    Cx = 200 cos 20 = 187.9
    Cy = 200 sin 20 = 68.4


    R = (-433-250+187.9)I + (-250 + 0 + 68.4)j
    = (-495.1, -181.6)
    R = sqrt(-495.1^2 + -181.6^2)
    = 527

    Tan Theta = 318.4 / 370.9
    Theta = 20.1 deg.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Apart from getting A and C the wrong way round ... are you sure that 180 degrees is due west and not due south?
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  3. #3
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    Quote Originally Posted by Matt Westwood View Post
    Apart from getting A and C the wrong way round ... are you sure that 180 degrees is due west and not due south?
    Yes it is supposed to be due west.
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  4. #4
    Super Member Matt Westwood's Avatar
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    Sorry, it's just an incredibly unnecessarily badly worded problem.
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  5. #5
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    Quote Originally Posted by Matt Westwood View Post
    Sorry, it's just an incredibly unnecessarily badly worded problem.
    Maybe I should draw a diagram. I think that will help.
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  6. #6
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    Vector Problem-diagram.jpg
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  7. #7
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    Noone can tell me if this problem looks correct? Using the diagram given use the component method to determine the resultant(magnitude and direction). Forget how it was worded before, now all you have to do is look at the diagram.
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  8. #8
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    R_x = 200\cos(20) + 250\cos(180) + 500\cos(-150)

    R_y = 200\sin(20) + 250\sin(180) + 500\sin(-150)<br />

    |R| = \sqrt{R_x^2 + R_y^2}

    since the resultant components end up in quad III ...

    \theta = \arctan\left(\frac{R_y}{R_x}\right) + 180

    where \theta = angle measured from due east (the + x-axis)
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  9. #9
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    Quote Originally Posted by skeeter View Post
    R_x = 200\cos(20) + 250\cos(180) + 500\cos(-150)

    R_y = 200\sin(20) + 250\sin(180) + 500\sin(-150)<br />

    |R| = \sqrt{R_x^2 + R_y^2}

    since the resultant components end up in quad III ...

    \theta = \arctan\left(\frac{R_y}{R_x}\right) + 180

    where \theta = angle measured from due east (the + x-axis)
    Thank you very much!
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