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    Question 6 (b) from http://www.examsolutions.net/a-level...uary/paper.php

    I watched the exam solutions solution, but I don't understand how it worked. Can anyone explain a different way of answering this, or just explain the theory...?
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    (Original post by jshark97)
    Question 6 (b) from http://www.examsolutions.net/a-level...uary/paper.php

    I watched the exam solutions solution, but I don't understand how it worked. Can anyone explain a different way of answering this, or just explain the theory...?
    The Examsolutions method is the way I would do it.

    It would be useful if you tell us the exact part of the video where you first didn't understand something.
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    By adding both of the vectors you obtain 4i-5j+pi+qj=(p+4)i+(q-5)j
    Since this vector is in the direction i-2j you obtain the expression
    (p+4)i+(q-5)j=k(i-2j)
    Where k is just a numerical value. As the force can be of greater magnitude than this vector. By splitting it into i and j components.
    p+4=k
    q-5=-2k
    Sub one into other to get
    q-5=-2(p+4)
    so 2p+q+3=0
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    The theory is that the resultant force is found as the vector sum of all the forces acting upon a particle. If a direction is given for the resultant force then it will act in that direction but may be of a different magnitude hence the constant term k.
 
 
 
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