I was trying to solve this problem:

enter image description here

but I encountred a problem while I was trying to evaluate the work done by $F$ over $AB$ which is:

$$ W= \int_{A}^{B} \vec{F}\cdot d\vec{s} $$

but how to compute this kind of integral? I tried to switch the variabe $d\vec{s}$ with $ Rd\theta$: $$W= \int_{0}^{\pi/6} FR ~d\theta $$

but there isn't a function of $\theta$ inside the integral, so how to evaluate this integral?

  • $\begingroup$ Is math stack exchange more appropriate for this question? $\endgroup$
    – jim
    Dec 30, 2016 at 20:59
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    $\begingroup$ @ACuriousMind : I think you have exceeded your powers as a Moderator by closing this question. The Help Center states that Moderators are exception handlers who close blatantly off-topic questions. This question is not blatantly off-topic and received no close votes other than your own. Please would you reconsider your decision and reopen this question. $\endgroup$ Jan 2, 2017 at 4:35
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    $\begingroup$ @ACuriousMind : see this Meta post by JamalS. Current policy seems to be :"If the question is regarding a mathematical method applied to a physics problem, then it is appropriate for the physics SE." There is a clear physics background to this question. If it is a duplicate, none has been identified. None of the reviewers who cast "Leave Closed" votes had voted to close. Reviewers are reluctant to change the status of a question, and give the benefit of the doubt to current status. $\endgroup$ Jan 2, 2017 at 15:10
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    $\begingroup$ @ACuriousMind : The question might have been closed by Reviewers had they been given the option. Your binding vote deprived them of that decision. That is my point : that you unilaterally closed a question which is not blatantly off topic. If it is not blatantly off topic, the question should be dealt with by the community in the Review Queues. I think the site policy on this is clear. ... I made the same complaint to DavidZ a few days ago. His response was to reopen the question. But I think he also deleted the comments and I cannot find the question. $\endgroup$ Jan 2, 2017 at 15:16
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    $\begingroup$ @ACuriousMind : While the title is blatantly off topic, the content of the question (how to calculate work done) is not. I suggest that the Reviewers who voted to "Leave Closed" looked no further than the title. One Reviewer agreed with me; when my vote and yours are counted that is 4 v 2 in favour of "Leave Closed," although a majority this cannot be described as "blatant." See How is the dot product a generalization of multiplication? for a more "blatantly" pure math question which was not closed. $\endgroup$ Jan 2, 2017 at 19:40

2 Answers 2


You forgot the unit vector $[-\sin\theta,\cos\theta]$ along the path. In your integral $W= \int_{0}^{\pi/6} F\cdot R ~d\theta $, replace $F \cdot R ~d\theta$ by $R F\cdot [-\sin\theta,\cos\theta] ~d\theta$ and you are done ($F\cdot [-\sin\theta,\cos\theta] = -F_1\sin\theta+F_2\cos\theta$ for $F=[-F_1,F_2]$.)

  • $\begingroup$ and what is $ [ -sin \theta , cos \theta ] $ ? is it a vector ? if it's the case that means that I have to integrate a vector ??? $\endgroup$
    – Hilbert
    Dec 30, 2016 at 18:32
  • $\begingroup$ Yes, it is a vector, and $F.[−\sin \theta,\cos \theta] = -F_1\sin \theta+F_2\cos \theta$. The end result is a scalar, not a vector. $\endgroup$
    – user130529
    Dec 30, 2016 at 18:40

A particle is moving between points $\:\mathrm{A}\:$ and $\:\mathrm{B}\:$ of a curve $\:C\:$ and a constant force $\:\mathbf{F}\:$ is applied continuously on it. The work done by $\:\mathbf{F}\:$ between points $\:\mathrm{A}\:$ and $\:\mathrm{B}\:$ is ...

On Plane

enter image description here

In Space

enter image description here enter image description here


OP asks : this means that the work done over AB is equal to the work done over the purple path in the Figure below? enter image description here Answer : No, the work done over AB is equal to the work done over the green path AB', the projection of the curvilinear orbit (here: circular arc) AB on the direction of the constant force $\:\mathbf{F}\:$. This is valid for constant force vector. If the force is not constant in magnitude and/or direction then you must study about line integrals.

The purple path AB (segment) is the integral : $$ \int\limits_{\rm{arc\:AB}} \mathrm{d}\mathbf{s} $$

  • $\begingroup$ Jan Lalinsky's comment seems to have disappeared, leaving your reference to it meaningless. Best to reproduce comments if you quote them, because they are liable to be removed by the author or a Moderator. $\endgroup$ Dec 31, 2016 at 19:47
  • $\begingroup$ @sammy gerbil : Happy New Year. Thanks for your attention. The deleted Ján Lalinský's comment was this one : The scalar product $\:\overrightarrow{F}\cdot \overrightarrow{ds}\:$ can be expressed as $\:F\cdot ds'$, where $\:ds'\:$ is horizontal component of the displacement $\: \overrightarrow{ds}$. Then integrate over $\:s'$. I think the deletion was done by moderator due to the policy "don't post answers in comments". Anyway, I deleted this reference in my answer. $\endgroup$
    – Frobenius
    Jan 1, 2017 at 7:24
  • $\begingroup$ To the user who downvoted my answer without a comment to explain why : I don't care so much about reputation so continue downvoting. $\endgroup$
    – Frobenius
    Jan 1, 2017 at 21:06
  • $\begingroup$ Perhaps the down-voter did not get a pair of 3D goggles in his/her Christmas Stocking, so is not able to appreciate your 2nd diagram! $\endgroup$ Jan 1, 2017 at 21:24
  • $\begingroup$ @sammy gerbil : I accept downvoting by a user but I wish to have his comment explaining why. I suspect that this downvote is done by a moderator who reasonably has not time to explain. In such a case I understand the absence of an explanation comment. $\endgroup$
    – Frobenius
    Jan 1, 2017 at 21:43

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