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In the following image if force the triangle PAN was right angle at P then the component of force in the direction of displacement would be $F\sec\theta$ so work $F*Displacement(AC)*\sec \theta $.
I understand why it is $\cos \theta$ but why cant it be $\sec \theta$ also?
I would greatly appreciate a solution to this doubt 🙏.
(the question got removed cause the site thought im asking a hw question but im not)
$\begingroup$When resolving one force into two other forces, the convenient way to do that is to choose for the two other forces to be perpendicular to each other. If you make the right angle be at P, then the two forces will not be perpendicular, then the leg PN will also have some parallel component w.r.t. the displacement, and thus would also have to be included in the computation. The result will be the same, with much more difficult computation.$\endgroup$
I assume you mean why work is not $Fd\sec\theta$. Only work done by a force perpendicular to displacement is zero. In your case the other component is not perpendicular and contributes to work
