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Where am I going wrong?

  1. Apr 16, 2010 #1
    Hello, I am almost surely doing something wrong here, but I think it would be better if I drew a diagram to illustrate what I mean.

    [PLAIN]http://img690.imageshack.us/img690/5830/workh.png [Broken]

    AB is Force, AC is displacement.

    Now I know that work here = Force . Displacement x Cos(theta), by constructing a perpendicular between AB and AC and using sine-cosine functions.

    When theta = 90, component of force along displacement equals 0 - this, I don't get, as since I use trigonometry to derive W = F . D x Cos (theta) I think I can do that here like this:

    [PLAIN]http://img193.imageshack.us/img193/2720/work2i.png [Broken]

    AC = Force, AB = Displacement.

    The angle formerly indicated as (theta) now has the sign of a right angle.

    Now here, tan(theta) = AB/AC or AC x tan(theta) = AB. Therefore F x tan(theta) = D.

    Thus Work = F.D.tan(theta).

    Where am I going wrong?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 16, 2010 #2
    The angle theta is the angle between the displacement vector and the force vector. What you're doing in the second example is changing the meaning of the angle theta.
  4. Apr 16, 2010 #3
    Lol, I know that. But it's not a crime to rename variables. Call that angle alpha or beta or gamma or what you wish.

    I suspect my error lay in using the tan function.
  5. Apr 16, 2010 #4
    What you have drawn in your second diagram is not a perperdicular between AB and AC.
    You have (incorrectly) made the angle at A into a right angle.

    What actually happens if you drop a perpendicular from AB to AC, you drop it from a useful point on AB, say B.
    This meets AC in a new point D, so BD is perp to AC and the angle at D is a right angle.

    The distance AD is the component of the force in the direction of the displacement.
    the work done is then ADxAC = ABcos([tex]\theta[/tex])xAC as required.

    AD is also known as the projection of AB on AC.
  6. Apr 16, 2010 #5


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    No, the way it would really work is:
    tan(θ) = F / (component of F that is parallel to D)​
    That means θ should be zero in the 2nd figure.

    With θ = 0, this is correct.

    In thinking that you can construct a triangle out of the force and displacement vectors.
    Last edited by a moderator: May 4, 2017
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