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Net force of a vector

  1. Oct 27, 2007 #1
    Consider 3 force vectors F1, F2, and F3. The vector F1 has magnitude F1 = 55N and direction θ = 41°; the vector F2 has magnitude F2 = 20N and direction θ = - 140°; and the vector F3 has magnitude F3 = 17N and direction θ = 140°. All the direction angles θ are measured from the positive x axis: counter-clockwise for θ > 0 and clockwise forr θ < 0.

    A. What is the magnitude F or the net force vector F = F1+F2+F3? Answer in units of N.

    B. What is the direction of the net force vector F? State your answer as an angle θ between -180° and +180°. Answer in units of °.

    Horizontal: F*sin(θ) = x
    Vertical F*cos(θ) = y

    I found the x resultants to be 10.93 + 12.85 + 36.08 = 59.86
    y resultants: 13.0227 + 41.50 + 15.32 = 69.8427
    Then I added them and got 129.70745, but that is wrong.

    Please help me solve this question, I have no idea how to do it.
    Last edited: Oct 27, 2007
  2. jcsd
  3. Oct 27, 2007 #2
    You did everything correctly up until adding them up. You can't just add them, you have to use the distance formula. You can use the origin as your "other" point for this.
    Last edited: Oct 27, 2007
  4. Oct 27, 2007 #3
    you're right to resolve the forces into components, but you're going wrong once you've done that stage. You can't just add x components to y components, that's the point. What you want is the length of the "resultant" of these two summed components..

    think pythagoras...
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