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Question essentially in title. Why do we use the dot symbol when writing $W = \int \mathbf F\cdot \mathrm d\mathbf s$? I understand that $\mathbf F$ and $\mathrm d\mathbf s$ are vectors and that probably has something to do with it, but when I read this I don't see "the integral of $\mathbf F$ with respect to $\mathrm d\mathbf s$", I see "find $\mathbf F\cdot \mathrm d\mathbf s$, and THEN take the integral of that". It's kind of tripping me up, and I'm just wondering why the dot is there?

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Consider that $\vec{ds} = \hat{r}ds$ where $\hat{r}$ is a unit vector pointing along the path of the object, and $ds$ is just the regular differential line element. Now also note that at every point $p$ along the path you can separate the component of the force that's along the path from the component perpendicular to the path: $$\vec{F}(p) = F_\parallel(p)\cdot\hat{r} + F_\perp(p)\cdot\hat{r}_{\perp}$$

Where $\hat{r}_{\perp}$ is just a unit vector along the axis perpendicular to the path. So the job of $\vec{ds}$ is simply to pick the component that's along the path at every point. In other words at a point $p$ again: $$\vec{F}(p)\cdot\vec{ds} = \vec{F}(p) \cdot \hat{r}(p)ds = F_{\parallel}(p) ds $$

Note that at every point $F_{\parallel}(p)$ may be either positive, zero or negative. I had the magnitude of the parallel component $|\vec{F_{\parallel}}(p)|$ before this edit as the result, which is incorrect! The force may also be directed opposite to the path and this should contribute negatively to the overall work.

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    $\begingroup$ OHHHHHHHHHH this makes so much sense thank you $\endgroup$ Commented Feb 9, 2023 at 19:49
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    $\begingroup$ @JamshidBatswani if it's indeed helpful, please confirm the answer by clicking (or pressing) on the check mark below the vote buttons. $\endgroup$ Commented Feb 11, 2023 at 5:39
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    $\begingroup$ A perfect answer $\endgroup$
    – khaxan
    Commented Feb 11, 2023 at 7:11
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$\mathbf{F}\cdot d\mathbf{s}$ is the dot product of the vectors $\mathbf{F}$ and $d\mathbf{s}$. The result is a scalar, in this case the work $dW$. Its geometric definition is $$\mathbf{F}\cdot d\mathbf{s} = ||\mathbf{F}||\ ||d\mathbf{s}||\ \cos\theta$$ where $||\mathbf{F}||$ and $||d\mathbf{s}||$ are the magnitudes of the vectors $\mathbf{F}$ and $d\mathbf{s}$, and $\theta$ is the angle between these two vectors.

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