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Does Normal Force change with new Force diagram?

  1. Nov 26, 2014 #1
    All degrees are measured from the positive x-axis.

    FG = Force of gravity
    FN = Normal force
    FT = Force of tension


    A skier is being pulled up a 15° incline by a rope. I'm solving for FT. The only forces acting on the skier are FG, FN, and FT.

    Attempt at a solution:

    Initially, my free-body diagram shows the FG at 270°, FN at 105°, and FT at 15°.

    Let's say I redraw my free-body diagram such that FG points at 255° FN at 90°, and FT at 0° (or 360°).

    FyG =-mgcosθ. What I want to know is, does my FyN = +mgcosθ, or just +mg?
  2. jcsd
  3. Nov 26, 2014 #2


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    If the angle is 90 degrees, what's cosine of 90 degrees?

    Make sure you don't just blindly apply equations, you must know where those equations come from. If you rotate your diagram, changing the "theta", your equations must necessarily take this into account.
  4. Nov 26, 2014 #3
    Thanks for the reply, Matterwave.

    I understand what you are saying. It's hard for me to look at a straight line (like in the case of FG at 270°) and apply trigonometric values.

    I can see that if we correlate sinθ with the y-axis, and cosθ with the x-axis, that FG at 270° = mgsinθ where sin(270°) = -1, thus we typically refer to FG as -mg, associating the negative sign with the direction of gravitational acceleration.

    Thus, the answer to my question is FyN = +mgcosθ.
  5. Nov 28, 2014 #4


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    Think to yourself... What is the component of FG that acts in the FT direction? and what is the component of FG that acts at 90 degrees to FT. Draw these it on your diagram and you will see they form a right angle triangle. Only then apply the trig.

    Personally I don't think redrawing the diagram rotated 15 degrees makes it easier to understand.
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