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Math Help - eww...vectors...

  1. #1
    Junior Member CONFUSED_ONE's Avatar
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    Confused_One lives in the United States.
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    eww...vectors...

    here's the problem:

    The magnitude and direction of two forces acting on an oject are
    -70 lbs, S56[degree]E and
    -50 lbs, N72[degree]E.
    find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.

    i did this problem like 5 times! i keep getting different answers.
    the answer in the back of the book is 108.21 lbs, S77.4[degree]E.
    could someone explain it to me?

    tthhhaannnk you, even for reading my post. thanks.
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  2. #2
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    Lexington, MA (USA)
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    Hello, CONFUSED_ONE!

    I assume you're familiar with trig forms for vectors.

    The magnitude and direction of two forces acting on an object are:
    70 lbs, S 56 E and 50 lbs, N 72 E.

    Find the magnitude, to the nearest hundredth of a pound,
    and the direction angle, to the nearest tenth of a degree,
    of the resultant force.

    Answers: 108.21 lbs, S 77.4 E
    Code:
                  Q
          |       *
          |     *
          |72*
          | * 18
        O * - - - - - - -
          | * 34
          |56*
          |     *
          |       *
          |         *
          |           *
                      P

    The object is at O.
    Rewrite the bearings (angles) relative to the positive x-axis.

    The forces are: \overrightarrow{OP} = [70\cos(-34^o),\,70\sin(-34^o)] = [70\cos34^o,\,-70\sin34^o]

    . . . . . . . and: \overrightarrow{OQ} = [50\cos18^o,\,50\sin18^o]


    The resultant vector is the sum of these vectors: \overrightarrow{OR} \:= \:\overrightarrow{OP} + \overrightarrow{OQ}

    . . \overrightarrow{OR} \;= \;[50\cos18^o + 70\cos34^o,\,50\sin18^o - 70\sin34^o]

    . . \overrightarrow{OR}\;\approx\;[105.585,\,-23.693]

    The magnitude is: |\overrightarrow{OR}| \;= \;\sqrt{105.585^2 + (-23.693)^2} \;\approx 108.21


    The resultant makes an angle \theta with the positive x-axis where:
    . . \tan\theta \:= \:\frac{23.693}{105.585} \;=\;0.224397405\quad\Rightarrow\quad\theta \,\approx \,12.6^o

    Therefore, the direction is: S 77.4 E

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