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## 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?

[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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