physics

An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a positive magnitude of 6.7N; a second force has a magnitude of 4.6 N and points in the negative y direction. (1)"Find magnitude of the third force acting on the object.
(2) Find the direction of the third force acting on the object in terms of pheta=-----------degrees from the + x direction.

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  1. constant velocity means no net force. So add the three forces, and equal zero.

    6.7x -4.6y + z=0
    let z have two components, in the x direction, and the y direction.

    then zx=-6.7x
    and zy=4.6y
    but zx= ZcosTheta and zy=ZsinTheta
    so ZcosTheta=-6.7
    ZsinTheta=4.6
    which makes (dividing the first equation into the second)
    tanTheta=-.687
    Theta=180-34.5 degrees where theta is counterclockwise from the x axis.
    Z sinTheta=4.6, so you can solve for Z

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    bobpursley
  2. 1,2. F1+F2+F3 = M*a = M*0 = 0.
    6.7 - 4.6i + F3 = 0.
    F3 = -6.7 + 4.6i = 8.13 N.[34.5o] N. of W. = 145.5o CCW from +x-axis.

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