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    Ive been unable to answer this question -

    Forces P, Q and R act on a particle at O in the plane of the co-ordinate axis Ox, Oy. Force P acts along Ox, Q acts along Oy, and R acts at angle Ѳ with Ox, in the first quadrant. Caculate the magnitude of the resultant force and the angle it makes with Ox when:

    P=3N, Q=4N, R=5N, Ѳ=60 degrees

    I know this is a long question and is one which is quite difficult to answer on a computer, but any time and help is greatly appreciated!


    EDIT: sorry i posted in the wrong forum, ive posted in the maths forum instead.

    Resolve the forces into the two components, the x-axis and the y-axis.

    Resolving horizontally = Force P + RcosѲ = 3 + 5cos60 = 5.5N
    Resolving vertically = Force Q + RsinѲ = 4 + 5sin60 = 8.33N

    Now you've got just two perpendicular forces which allows you to use Pythagoras to work out the resultant.

    Resultant force = {\sqrt{5.5^2 + 8.33^2}} = ~10N

    To find the angle it makes with the horizontal, use Trigonometry.

    TanѲ = opp/adj

    Ѳ = arctan(8.33/5.5) = 56.6
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