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Work done by a force by finding component of force along the displacement

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)

Work done by a force by finding component of force along the displacement

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 🙏.

Work done by a force by finding component of force along the displacement

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)

Post Closed as "Not suitable for this site" by Frobenius, Miyase, Jon Custer
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gandalf61
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Why isntisn't work Fd sec theta$Fd \sec \theta$?

Corrected what the asker asked. He meant work not force
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Why isnt force $Fd\sec\theta$?work Fd sec theta

Work done by a force by finding component of force along the displacement

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*Displacement(AC)*sec theta$F\sec\theta$ so work $F*Displacement(AC)*\sec \theta $.

I understand why it is cos theta$\cos \theta$ but why cant it be sec theta$\sec \theta$ also?

I would greatly appreciate a solution to this doubt 🙏.

Why isnt force $Fd\sec\theta$?

Work done by a force by finding component of force along the displacement

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*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 🙏.

Why isnt work Fd sec theta

Work done by a force by finding component of force along the displacement

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 🙏.

edited tags; edited title
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Qmechanic
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