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I have a question and I'm not sure which equation to apply to it:

A fluid is at rest in a gravitational field of strength $\mathbf g = −g \underline k$, where $g$ is a positive constant, and both the unit vector $\underline k$ and the $z -axis$ point vertically upward. The fluid pressure, $p$ , and the fluid density, $\rho$ , are related by $∇p = \rho g$.

Show that in the fluid $$\frac {dp}{dz} = −\rho g$$.

Thanks

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    $\begingroup$ in this case there is only a force on negative axis $ z$ because the accleration of gravity which is negative so $ \nabla p= \frac{dp}{dz}$ $\endgroup$ – Jose Garcia May 13 '17 at 14:13
  • $\begingroup$ is there a particular equation to use and then eliminate terms? $\endgroup$ – juper May 13 '17 at 14:17
  • $\begingroup$ I am assuming you meant $\nabla p=\rho \mathbf g$, since otherwise you have $\text{vector}=LHS=RHS=\text{scalar}$. $\endgroup$ – John Doe May 13 '17 at 14:25
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$$\nabla p=\left(\begin{matrix}\frac{\partial p}{\partial x}\\\frac{\partial p}{\partial y}\\\frac{\partial p}{\partial z}\end{matrix}\right)$$

Then $$\frac{\partial p}{\partial z}=\nabla p\cdot \vec k=\rho \vec g\cdot\vec k=-\rho g \vec k\cdot\vec k=-\rho g$$ There is no eliminating terms, this is quite easy to derive from the properties given.

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  • $\begingroup$ ah yes, that's great. I didn't stop to think what $\nabla p$ actually was. Thanks! $\endgroup$ – juper May 13 '17 at 14:25

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