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Thread: four displacement vectors problem.

  1. #1
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    four displacement vectors problem.

    heres another physics displacement problem i'm having trouble with. I also included a drawing diagram to make it easier to see what is going on.

    The magnitudes of the four displacement vectors shown in the drawing are A = 11.0 m, B = 11.0 m, C = 12.0 m, and D = 32.0 m.
    what is the magnitude and directional angle for the resultant that occurs when these vectors are added together.

    magnitude is:
    and the directional angle is (in degrees)
    (from the +x axis)

    http://img339.imageshack.us/img339/2716/p142xd0.gif

    thanks for any help you can offer here.
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    Quote Originally Posted by rcmango View Post
    heres another physics displacement problem i'm having trouble with. I also included a drawing diagram to make it easier to see what is going on.

    The magnitudes of the four displacement vectors shown in the drawing are A = 11.0 m, B = 11.0 m, C = 12.0 m, and D = 32.0 m.
    what is the magnitude and directional angle for the resultant that occurs when these vectors are added together.

    magnitude is:
    and the directional angle is (in degrees)
    (from the +x axis)

    http://img339.imageshack.us/img339/2716/p142xd0.gif

    thanks for any help you can offer here.
    You have the given coordinate axes, so break the vectors down into their components:
    $\displaystyle A_x = -11.0 \cdot cos(20^o)$ and $\displaystyle A_y = 11.0 \cdot sin(20^o)$

    $\displaystyle B_x = 11.0 \cdot cos(90^o)$ and $\displaystyle B_y = 11.0 \cdot sin(90^o)$

    etc.

    Then add all the x components and y components together.
    $\displaystyle R_x = A_x + B_x + C_x + D_x = 0.40276$
    and
    $\displaystyle R_y = A_y + B_y + C_y + D_y = -16.6341$

    The magnitude of the resultant vector is
    $\displaystyle R = \sqrt{R_x^2 + R_y^2} = 16.639$

    The reference angle is
    $\displaystyle \theta = tan^{-1} \left ( \left | \frac{R_y}{R_x} \right | \right ) = 88.613^o$
    and is in the fourth quadrant (as $\displaystyle R_x$ is positive and $\displaystyle R_y$ is negative.)

    So your vector is 16.6 m at 88.6 degrees below the +x axis.

    -Dan
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