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I know this is a very simple question, and I have searched it too. How to avoid incorrect symbols in calculation results.I don’t understand how to choose the sign of $ds$.

An object moves from a to b,Note the number $b<a$,$W>0$.

enter image description here Take the direction of $i$ as the positive direction,Can I think of $\vec{ds}$ as $\vec{b}-\vec{a}$?

$\int _{a}^{b} {-F}{ds}=W<0$

I know that the choice of ab determines the positive and negative, maybe I should write $\int _{a}^{b} {-F}{ds}=w>0$

BUT enter image description here I changed the direction of i

Write the formula $\int _{a}^{b} {F}{ds}=W<0$

I can judge the positive or negative of the result by the order of the upper and lower limits of the integral and the sign of F. If it does not match, I will take the opposite sign for ds.I don’t want to judge and correct by the prediction result.

I can only modify the direction of ds through the results. It seems that this does not refer to the direction. Does it make sense to take positive and negative directions for ds?(I know that sometimes the direction is important,The positive and negative here refer to the positive and negative in the same direction)

How do I decide the sign of $ds$?Do I made any errors about the vector of ab and ds?

Eli
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能够可能
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  • The problem you have begins earlier, with the coordinates. Are $a$ and $b$ the coordinates of two points? or are they the physical distances of two points from something else? – pglpm Dec 14 '20 at 16:03
  • To be honestly, i'm not sure.The displacement vector of ab starts from the origin of the coordinate. – 能够可能 Dec 14 '20 at 16:08
  • I would recommend, if possible, that first of all you have a chat with your teacher about your doubts. – pglpm Dec 14 '20 at 16:11
  • The vector of ab starts from the origin of the coordinates，But I want to determine the direction of the vector ds, so a-b.The direction of ds refers to the direction of s, or the direction of a-b.I am not sure. – 能够可能 Dec 14 '20 at 16:13
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    Possible duplicate: https://physics.stackexchange.com/questions/443988/why-does-this-line-integral-give-the-wrong-sign/444191#444191. Duplicate because the answer to this question is the same as the answers there: It's all about the parameterization of the path, and how you don't get to choose the direction of the differential line element independent of the limits of integration. – march Dec 14 '20 at 18:16

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First, you should pay attention to the integrand, which is the dot product of two vectors: $$\vec{F}\cdot\mathrm{d}\vec{r}.$$ You can't ignore that. Then you must pay attention to the coordinate system and state d$\vec{r}$ in coordinates appropriate to the problem. In Cartesian or rectangular coordinates $$\mathrm{d}\vec{r}=\hat{i}\mathrm{d}x + \hat{j}\mathrm{d}y + \hat{k}\mathrm{d}z.$$ Here is where many get tripped up. The differentials, like $\mathrm{d}x$, represent the positive coordinate changes and the limits of the definite integral define the actual path. So basically $\mathrm{d}\vec{r}$ tells us the positive direction change for our coordinate system and the integral limits tell us the actual path.

Applying this to your setup: a constant force of magnitude $F$ acts in the negative-$x$ direction on an object which moves from initial position $\vec{r}=(a,0,0)$ to final position $\vec{r}=(b,0,0)$. $$\vec{F}\cdot\mathrm{d}\vec{r} = (-\hat{i})F\cdot(\hat{i}\mathrm{d}x + \hat{j}\mathrm{d}y + \hat{k}\mathrm{d}z) = -F~\mathrm{d}x$$

Because the only integral is along an $x$-path and the limits are not functions of $y$ nor $z$, we only integrate in $x$. $$W = \int_a^b (-F)~\mathrm{d}x=-\left.Fx\right|^b_a = F\left(a-b\right)$$

So, if $b<a$, the work done by the force acting in the negative-$x$ direction is positive, which is proper. If $b>a$, the work is negative, again, proper.

Bill N
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  • Thank you very much for your reply, this answer is great！ – 能够可能 Dec 14 '20 at 16:33
  • could I ask some question？Why dx takes $+\hat{i}$.if $\int _{a}^{b}$,Is it pointing from a to b, or b to a. – 能够可能 Dec 14 '20 at 16:34
  • As I said in the answer, the specific path is determined by the limits. The dx tells us how a positive change in x behaves. The dx is a general term, and the limits are specific to the path. – Bill N Dec 14 '20 at 16:57