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    tirumounika's Avatar
    tirumounika Posts: 2, Reputation: 1
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    #1

    Jul 29, 2010, 06:16 AM
    properties of vector products
    Determine the resultant force acting at point A of a concurrent force system
    having force P = 40N and θ = 30 w.r.t positive y-axis, Q = 80N and θ = 45
    w.r.t positive x-axis, S = 60N and θ = 30 w.r.t negative x – axis? (by
    resolution of forces into components and by polygon rule graphical
    method)?
    tirumounika's Avatar
    tirumounika Posts: 2, Reputation: 1
    New Member
     
    #2

    Jul 29, 2010, 06:18 AM

    Determine the resultant force acting at point A of a concurrent force system
    having force P = 40N and θ = 30 w.r.t positive y-axis, Q = 80N and θ = 45
    w.r.t positive x-axis, S = 60N and θ = 30 w.r.t negative x – axis? (by
    resolution of forces into components and by polygon rule graphical
    method)?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Jul 29, 2010, 10:53 AM

    Resolution of forces into components means you determine the x-component and y-component of each force, add them up to get the resultant total forces in the x and y direction, then convert them to a single force vector. To get you started: for vector P the x component is -40 cos(30) = -20 N, and the y component is -40 sin(0) = -20 sqrt(3). Notice each of these is a negative number, because P is acting to the left and downward. Now do the same for Q and S, and add up the x-components to get Total Fx, then the y-components to get Total Fy. To convert the resulting Fx and Fy to a single vector R, use:
    |R| = sqrt(Fx^2 +Fy^2)
    θ = Arctan(Fy/Fx)

    Note - be careful to get the corect quadrant for θ.

    The graphical method is to simply add the three vectors, placing them tip to tail. The resulting composite force is then the vector that goes from the tail of the first vector to the tip of the last, as in the attached fiigure.

    Post back with what you get for an answer, and we'll check it for you.
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