Where am I going wrong?

  • #1

Main Question or Discussion Point

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?
 
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Answers and Replies

  • #2
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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.
 
  • #3
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.
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.
 
  • #4
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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.
 
  • #5
Redbelly98
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[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.
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.

Thus Work = F.D.tan(theta).
With θ = 0, this is correct.

Where am I going wrong?
In thinking that you can construct a triangle out of the force and displacement vectors.
 
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