# How to calculate work done when movement is not in the direction of the force? [closed]

I was trying to solve this problem: 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?

## closed as off-topic by ACuriousMind♦Jan 1 '17 at 21:00

• This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• Is math stack exchange more appropriate for this question? – jim Dec 30 '16 at 20:59
• @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. – sammy gerbil Jan 2 '17 at 4:35
• @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. – sammy gerbil Jan 2 '17 at 15:10
• @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. – sammy gerbil Jan 2 '17 at 15:16
• @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. – sammy gerbil Jan 2 '17 at 19:40

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 $F=[-F_1,F_2]$.)

• 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 ??? – Hilbert Dec 30 '16 at 18:32
• 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. – user130529 Dec 30 '16 at 18:40

Here $\vec{F} \cdot d\vec{s} = FR d\theta \cos(\pi-\theta)$. Because when the object is at angle $\theta$ from the vertical, the angle between $\vec{F}$ and $d\vec{s}$ is $(\pi-\theta)$. Refer to the diagram below (ds is exaggerated) : Now this is can be easily integrated using $\cos(\pi-\theta ) = -\cos\theta$. But the limits you have written are wrong, $\frac{\pi} {6}$ should actually be the lower limit and 0 be the upper.

• Probably the downvote is because of the poor presentation of the answer. I was quite tempted to downvote it myself, but I suppose one DV is enough to make the point. – sammy gerbil Dec 31 '16 at 2:42
• Can you tell how to improve the presentation! – Sagar Kaushik Dec 31 '16 at 5:26
• @sammygerbil Do you think the other 2 answers have better presentation? And can you tell me how can I make my answer more presentable? – Sagar Kaushik Dec 31 '16 at 10:38
• You could use graphics software for the diagram. Paint, GeoGebra, among others, are free and easy to use. Yes I think the other answers are better presented, but they have their own shortcomings. Your mixture of type and handwriting/sketching is more difficult to follow. – sammy gerbil Dec 31 '16 at 19:44
• @sammygerbil Yeah I hear that a lot about my handwriting. Thanks for the software names. I will try to make the relevant image but later. – Sagar Kaushik Dec 31 '16 at 19:53

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 In Space

EDIT

OP asks : this means that the work done over AB is equal to the work done over the purple path in the Figure below? 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 : $$\int\limits_{\rm{arc\:AB}} \mathrm{d}\mathbf{s}$$

• 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. – sammy gerbil Dec 31 '16 at 19:47
• @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. – Frobenius Jan 1 '17 at 7:24
• To the user who downvoted my answer without a comment to explain why : I don't care so much about reputation so continue downvoting. – Frobenius Jan 1 '17 at 21:06
• 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! – sammy gerbil Jan 1 '17 at 21:24
• @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. – Frobenius Jan 1 '17 at 21:43