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What is your method of knowing when to use sine and cosine in force problems?

  1. Feb 25, 2012 #1
    I am having the hardest time attaching my brain to some sort of method to know when to use sine and cosine on force problems. What is an easy way of remembering which function to use to find the force in the direction of x and force in the direction of y?
     
  2. jcsd
  3. Feb 25, 2012 #2

    cepheid

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    Staff Emeritus
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    First of all, draw a picture of the right-angled triangle, so that you don't have to just visualize it in your head.

    Second of all, use the mnemonic SOH CAH TOA to remember the definitions of the trigonometric ratios.

    Sine = Opposite/Hypotenuse.

    Cosine = Adjacent/Hypotenuse.

    Tangent = Opposite/Adjacent.

    Third of all, realize that in decomposing force vectors, there are usually only ever TWO possible cases.

    Case 1: The angle that you've been given is measured from the horizontal

    Code (Text):



        |\
        | \
        |  \
        |   \
        |    \  
    Fy |     \  F
        |      \
        |       \
        |        \
        |         \
        |         <------ θ
        |_______\
           Fx

     
    In this situation, Fy is the side of the triangle that is opposite from the angle, and Fx is the side of the triangle that is adjacent to the angle. (The total magnitude, F, of the force, is always the hypotenuse). Therefore, it follows that:

    sinθ = Fy/F (opposite side / hypotenuse)

    cosθ = Fx/F (adjacent side / hypotenuse)

    Fy = Fsinθ
    Fx = Fcosθ​



    Case 2: The angle that you've been given is measured from the vertical

    Code (Text):



        |\
        | \
        |  <------ θ
        |   \
        |    \  
    Fy |     \  F
        |      \
        |       \
        |        \
        |         \
        |          \
        |_______\
           Fx

     
    In this situation, Fy is the side of the triangle that is adjacent to the angle, and Fx is the side of the triangle that is opposite from the angle. (The total magnitude, F, of the force, is always the hypotenuse). Therefore, it follows that:

    sinθ = Fx/F (opposite side / hypotenuse)

    cosθ = Fy/F (adjacent side / hypotenuse)

    Fy = Fcosθ
    Fx = Fsinθ​


    So, you can see that, if the angle is measured from the horizontal, then the cosine is associated with the horizontal component, and the sine is associated with the vertical component.

    if the angle is measured from the vertical, then the cosine is associated with the vertical component, and the sine is associated with the horizontal component.
     
  4. Feb 25, 2012 #3

    tiny-tim

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    Homework Helper

    Hi AnthroMecha! :wink:
    I keep telling people …

    it's always cos!! :smile:

    It's always cos of the angle between the force and the direction …

    whenever it looks like sine, that's because you're using the "wrong" angle …

    maybe θ is marked on the diagram, but if the angle you really want is 90°-θ, then you use cos(90°-θ), which of course is sinθ ! :wink:

    (however, a good check o:), when you're using slopes, is to imagine :rolleyes: "what would happen if the slope was 0°?" … would the component vanish (sin0°) or be a maximum (cos0°) ?)​
     
  5. Feb 25, 2012 #4
    This forum always delivers. Thanks guys these are very useful tools.
     
  6. Feb 25, 2012 #5

    SammyS

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    SohCahToa !
     
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