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Math Help - [SOLVED] Easy Vectors Question

  1. #1
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    [SOLVED] Easy Vectors Question

    Find the magnitude and direction of the equilibrant (resulting force) of each of the following forces: 16 N and 10 N acting at an angle of 10 degrees to each other.
    Please make it as visual as possible... Thanks! (best answerer gets a thanks ^^)
    Last edited by x y z; April 18th 2009 at 07:47 PM.
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    Quote Originally Posted by x y z View Post
    Find the magnitude and direction of the equilibrant (resulting force) of each of the following forces: 16 N and 10 N acting at an angle of 10 degrees to each other.
    There are different ways to solve this question:

    1. Draw the parallelogram of forces. Use the Cosine rule to calculate the length of the larger diagonal. Keep in mind that the adjacent sides of the parallelogram include an angle of 180 - 10 = 170 :

    F_r = \sqrt{10^2+16^2-2 \cdot 10 \cdot 16 \cdot \cos(170^\circ)}

    2. Use vectors:

    Let \overrightarrow{F_1} = (10,0) and

    \overrightarrow{F_2} = 16(\cos(10^\circ), \sin(10^\circ))
    Then

    \overrightarrow{F_r} = (10,0) + 16(\cos(10^\circ), \sin(10^\circ)) = (10+16 \cos(10^\circ), 16 \sin(10^\circ))

    Now calculate the magnitude of \overrightarrow{F_r} :

    | \overrightarrow{F_r} | = \sqrt{\left(10+16 \cos(10^\circ) \right)^2 + \left(16 \sin(10^\circ) \right)^2}
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    Thanks.
    But how do I then get the direction of the vector?
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    Quote Originally Posted by x y z View Post
    Thanks.
    But how do I then get the direction of the vector?
    If you have the vectors \vec u and \vec v then the included angle is calculated by:

    \cos(\angle(\vec u , \vec v)) = \dfrac{\vec u \cdot \vec v}{|\vec u| \cdot |\vec v|}

    With your vectors you get:
    <br />
\cos(\angle(\overrightarrow{F_r} , \overrightarrow{F_1}) = \dfrac{(10+16 \cos(10^\circ), 16 \sin(10^\circ))<br />
 \cdot (10, 0)}{2\cdot \sqrt{80 \cdot \cos(10^\circ)} \cdot 10}

    After your calculator has cooled down a little bit you'll get:

    \angle(\overrightarrow{F_r} , \overrightarrow{F_1}) \approx 6.157^\circ

    Keep in mind that you now have calculated the angle which is included by \overrightarrow{F_r} \text{ and } \overrightarrow{F_1}
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    That's what I got too, but the answer says that it's 174 degrees to the 10N vector. Why did they subtract from 180 degrees?
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    Quote Originally Posted by x y z View Post
    ...
    Please make it as visual as possible... ...
    Quote Originally Posted by x y z View Post
    That's what I got too, but the answer says that it's 174 degrees to the 10N vector. Why did they subtract from 180 degrees?
    I've attached the sketch of the situation ... and honestly I can't explain where the result in your textbook comes from, sorry!
    Attached Thumbnails Attached Thumbnails [SOLVED] Easy Vectors Question-zweikraefte.png  
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    Wow, thank you so much! <3
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  8. #8
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    Quote Originally Posted by x y z View Post
    That's what I got too, but the answer says that it's 174 degrees to the 10N vector. Why did they subtract from 180 degrees?
    I finally found out where the answer in your textbook comes from:

    The most important word in your question is equilibrant (resulting force).

    If you combine the given forces to a resultant force and you should find a third force such that the complete system is in equilibrium then the result must be a force with the magnitude of \overrightarrow{F_1}+\overrightarrow{F_2} and the opposite direction. That is -\left( \overrightarrow{F_1}+\overrightarrow{F_2} \right).

    I've attached a sketch. You easily find the angle in question.
    Attached Thumbnails Attached Thumbnails [SOLVED] Easy Vectors Question-twoforces_equilib.png  
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