0
$\begingroup$

Consider a moving particle m under the influence of gravity. What's the work done by gravity in moving m from A to B?

$$ \int_A^B \vec{F}\cdot \vec{dS}= \int_{rA}^{rB} \vec{F}\,dr $$

I din't understand this change of variables. Why the work can be calculated using dr instead of dS without considering the angle between dS and dr?

Thank you!

$\endgroup$
  • $\begingroup$ What are $S$ and $r$? $\endgroup$ – user7530 Apr 21 '17 at 16:08
  • 2
    $\begingroup$ This is likely a central force problem where the gravitational force is in the r direction $\endgroup$ – Paul Apr 21 '17 at 16:11
0
$\begingroup$

Imagine that the path connecting $A$ and $B$ is a curve $c$. At any point ${\bf r}$ the direction tangent to the curve $c$ can be written as the linear combination of the radial unitary vector $\hat{\bf r}$, and a perpendicular vector $\hat{\bf r}_\perp$ (in three dimensions, this is actually two vectors).

$$ d{\bf S} = dr\hat{\bf r} + dq \hat{\bf r}_\perp $$

Now, the gravitational force is radial so ${\bf F} = F{\bf \hat{r}}$, this means that

$$ {\bf F}\cdot d{\bf S} = (F{\bf \hat{r}})\cdot (dr\hat{\bf r} + dq \hat{\bf r}_\perp) = Fdr $$

In other words

$$ \int_A^B {\bf F}\cdot d{\bf S} = \int_{r_A}^{r^B}F dr $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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