0
$\begingroup$

I have known, work is defined as the scalar product of applied force component in the line of the displacement and the displacement, or the product of force component applied in the direction of displacement and the magnitude of displacement. That becomes $Fxcosθ$.

Now when I change the definition to: scalar product of displacement component in the line of applied force and the displacement, it becomes: $xcosθ.F$

I have just started, so I can't yet determine the following: is there any case where the 2nd definition will be violated?

F = applied force vector, x = displacement vector, θ = angle between these 2 vectors

Improve this question
$\endgroup$
1
  • $\begingroup$ "scalar product of applied force component in the line of the displacement and the displacement" This is slightly confused: the scalar product operation takes two vectors in any direction ($\vec F$ and $\vec x$ in your case) and delivers a number with units, in your case, $|\vec F||\vec x| \cos \theta$. In the phrase of yours that I quoted at the beginning of this comment, the words "component in the line of the displacement" shouldn't be there. They may be useful, though, in explaining to someone what the scalar product is doing. $\endgroup$
    – Philip Wood
    Commented Dec 6 at 16:23

1 Answer 1

Reset to default
3
$\begingroup$

Multiplication is commutative so $Fx\cos\theta=x\cos\theta F$. Additionally, cosine is a symmetric function so if you measure the angle to be the opposite magnitude (i.e., with a minus sign), the expression is invariant ($\cos(-\theta)=\cos\theta$).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Well, I mean there won't be any change in the formula due to any other changes, because all conditions give the same formula for both definitions? $\endgroup$
    – damnOk
    Commented Dec 6 at 15:41
  • $\begingroup$ I was talking of the step preceding your assumption of the 2nd definition yielding (xcosθ)F, i.e., will it ever give any other result? $\endgroup$
    – damnOk
    Commented Dec 6 at 15:44
  • $\begingroup$ @damnOk, can you emphasize more? Like, for which "step preceding" my result? If you draw out the vectors $\vec{x}$ and $\vec{F}$, no matter how you rotate your paper or something the magnitude of each vector and the angle between don't change. Further, what "conditions" do you have in mind? $\endgroup$
    – QPhysl
    Commented Dec 6 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.