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Math Help - i need help with vectors!.......

  1. #1
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    i need help with vectors!.......

    Physics:
    For the vectors A and B in the figure use the method of components to find the magnitude and direction of a.) THe vector sum A + B, b.) the vector difference A - B, c.) The vector difference B - A.

    in the figure
    A is 12 m West
    B is 18 m 37 \circ N of E



    im not sure how to do vector sum but i think it will be from the head to tail...
    but i dunno how to form it... same as vector difference... can u help me with this?

    thank you! in advance
    Last edited by ^_^Engineer_Adam^_^; June 16th 2007 at 03:19 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Physics:
    For the vectors A and B in the figure use the method of components to find the magnitude and direction of a.) THe vector sum A + B, b.) the vector difference A - B, c.) The vector difference B - A.

    in the figure
    A is 12 m West
    B is 18 m 37 \circ N of E



    im not sure how to do vector sum but i think it will be from the head to tail...
    but i dunno how to form it... same as vector difference... can u help me with this?

    thank you! in advance
    Component method.

    A+B
    First choose a coordinate system:
    I'll take N as +y and E as +x.

    Now resolve each vector into components in this coordinate system. ( \theta is the angle between the vector and the +x axis.)

    A_x = Acos(\theta) = (12 \, m)cos(180) = -12 \, m
    A_y = Asin(\theta) = (12 \, m)sin(180) = 0 \, m
    B_x = Bcos(\theta) = (18 \, m)cos(37) = 14.3754 \, m
    B_y = Bsin(\theta) = (18 \, m)sin(37) = 10.8327 \, m

    Now, add the x components together to form the x component of the resultant R_x and add the y components together to form the y component of the resultant R_y.
    R_x = A_x + B_x = -12 \, m + 14.3754 \, m = 2.3754 \, m
    R_y = A_y + B_y = 0 \, m + 10.8327 \, m = 10.8327 \, m

    I recommend drawing a sketch at this point, but note that both components of R are positive. Thus R is a vector in Quadrant I. So to find the length and the angle:
    R = \sqrt{R_x^2 + R_y^2} = 11.0901 \, m
    \theta_R = atn \left | \frac{R_y}{R_x} \right | = 77.6317 ^o
    (Note: Finding \theta_R using the absolute value bars gives the reference angle, from which you can extract the angle if you know what quadrant it is in.)

    So A + B is 11 m at 78 degrees above the + x axis, otherwise known as 78 degrees N of E.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    To do A - B consider the following hint:
    A - B = A + (-B)

    The components of the vector -B are simply the negatives of the components of the vector B.

    -Dan
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  4. #4
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    Hello, ^_^Engineer_Adam^_^!

    For the vectors A and B in the figure, use the method of components
    to find the magnitude and direction of:
    . . a) the vector sum \vec A + \vec B
    . . b) the vector difference \vec A - \vec B
    . . c) the vector difference \vec B - \vec A

    In the figure: A is 12 m West, B is 18 m 37 N of E
    You're expected to be able write the vectors in component form.

    . . . . . \vec A \:=\:\langle \text{-}12,\:0\rangle\qquad\qquad\vec B \;=\;\langle 18\cos37^o,\;18\sin37^o\rangle


    (a)\;\;\vec A + \vec B \;=\;\langle\text{-}12- 18\cos37^o,\;18\sin37^o\rangle

    Then: . |\vec A + \vec B| \;=\;\sqrt{(\text{-}12-18\cos37^o)^2 + (18\sin37^o)^2} \;\approx\;\boxed{28.5}

    And: . \tan\theta \;=\;\frac{18\sin37^o}{\text{-}12-18\cos37^o} \;=\;-0.410710523\quad\Rightarrow\quad\theta \:\approx\:-22.3^o,\:157.7^o

    Since \vec A+ \vec B is in Quadrant 2: . \theta \:=\:\boxed{157.7^o}


    Get the idea?

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    thankssss!!!
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