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    Two forces, (4i-5j)N and (pi+pj)N, act on a particle P of mass 'm' kg. The resultant of the two forces is 'R'. Given that 'R' acts in a direction which is parallel to the vector (i-2j), show that '2p + q + 3 = 0'

    I completely don't understand what to do or where to begin, can someone show me the steps to this question please

    On an unrelated matter to that question, when I find the 'deceleration' of a particle, do I put the figure as a negative number and when I use it in one of the 'S U V A T' equations do I also put it in as a negative?

    This Question is pure evil! I did it earlier today, I'll try to explain.

    because it is parallel to that, but you dont know the length, we will call the length N,
    N(i-2j)= resultant
    (4i-5j)+(pi+qj) = Ni-2Nj

    Then split to i and j, then similtaneous equation N out.

    Hope that Helps


    in general the gradient is rise/run.

    You know it's parallel to (i-2j) so the gradient of that is -2/1.

    Equating the the forces in terms of i and j gives (4+p)i and (q-5)j.

    gradient is rise/run (j/i) therfore q-5/4+p=-2 (gradient of the parallel vector)
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