Edits to “Statement”, “Answer”
en/4.5.12.md
+3 −2
| @@ -1,7 +1,8 @@ | |||
| ### Statement | |||
| − | $4.5.12.$ [Insert the problem statement] | ||
| + | $4.5.12.$ a. The sum of the forces acting on the volume of liquid highlighted in the figure is zero. Using this fact, determine the height to which the liquid rises along a vertical wall. The contact angle is $\theta$. The surface tension and density of the liquid are $\sigma$ and $\rho$, respectively. | ||
| + | b. To what height will water rise along a vertical wall that it wets completely? | ||
| ### Solution | |||
| The sum of the forces acting on the volume of liquid highlighted in the figure is zero,so let's write force balance to the $x$ axis: | |||
| \begin{equation} | |||
| \sigma L-\sigma L\sin\theta=P_{avg}Lh | |||
| \end{equation} | |||
| where $P_{avg}$ is an average pressure of the liquid.As pressure linearly increasing with depth we get: | |||
| \begin{equation} | |||
| P_{avg}=\frac{1}{2}\rho gh | |||
| \end{equation} | |||
| Plugging eq(2) into eq(1) we get rise height of the liquid: | |||
| \begin{equation} | |||
| h=\sqrt{\frac{2\sigma(1-\sin\theta)}{\rho g}} | |||
| \end{equation} | |||
| In the case of complete wetting, the contact angle is zero,so: | |||
| \begin{equation} | |||
| h=\sqrt{\frac{2\sigma}{\rho g}} | |||
| \end{equation} | |||
| For the water $\sigma=0.073 \frac{N}{m}$,$\rho=1000\frac{kg}{m^3}$ and $g=9.8\frac{m}{s^2}$,so: | |||
| \begin{equation} | |||
| h \approx 3.9mm | |||
| \end{equation} | |||
| @@ -46,4 +47,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | a.$h=\sqrt{\frac{2\sigma(1-\sin\theta)}{\rho g}}$ b.3.9mm | ||
| @@ -1,7 +1,8 @@ | |||
| ### Statement | ### Statement | ||
| $4.5.12.$ [Insert the problem statement] | $4.5.12.$ a. The sum of the forces acting on the volume of liquid highlighted in the figure is zero. Using this fact, determine the height to which the liquid rises along a vertical wall. The contact angle is $\theta$. The surface tension and density of the liquid are $\sigma$ and $\rho$, respectively. | ||
| b. To what height will water rise along a vertical wall that it wets completely? | |||
| ### Solution | ### Solution | ||
| The sum of the forces acting on the volume of liquid highlighted in the figure is zero,so let's write force balance to the $x$ axis: | The sum of the forces acting on the volume of liquid highlighted in the figure is zero,so let's write force balance to the $x$ axis: | ||
| \begin{equation} | \begin{equation} | ||
| \sigma L-\sigma L\sin\theta=P_{avg}Lh | \sigma L-\sigma L\sin\theta=P_{avg}Lh | ||
| \end{equation} | \end{equation} | ||
| where $P_{avg}$ is an average pressure of the liquid.As pressure linearly increasing with depth we get: | where $P_{avg}$ is an average pressure of the liquid.As pressure linearly increasing with depth we get: | ||
| \begin{equation} | \begin{equation} | ||
| P_{avg}=\frac{1}{2}\rho gh | P_{avg}=\frac{1}{2}\rho gh | ||
| \end{equation} | \end{equation} | ||
| Plugging eq(2) into eq(1) we get rise height of the liquid: | Plugging eq(2) into eq(1) we get rise height of the liquid: | ||
| \begin{equation} | \begin{equation} | ||
| h=\sqrt{\frac{2\sigma(1-\sin\theta)}{\rho g}} | h=\sqrt{\frac{2\sigma(1-\sin\theta)}{\rho g}} | ||
| \end{equation} | \end{equation} | ||
| In the case of complete wetting, the contact angle is zero,so: | In the case of complete wetting, the contact angle is zero,so: | ||
| \begin{equation} | \begin{equation} | ||
| h=\sqrt{\frac{2\sigma}{\rho g}} | h=\sqrt{\frac{2\sigma}{\rho g}} | ||
| \end{equation} | \end{equation} | ||
| For the water $\sigma=0.073 \frac{N}{m}$,$\rho=1000\frac{kg}{m^3}$ and $g=9.8\frac{m}{s^2}$,so: | For the water $\sigma=0.073 \frac{N}{m}$,$\rho=1000\frac{kg}{m^3}$ and $g=9.8\frac{m}{s^2}$,so: | ||
| \begin{equation} | \begin{equation} | ||
| h \approx 3.9mm | h \approx 3.9mm | ||
| \end{equation} | \end{equation} | ||
| @@ -46,4 +47,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | a.$h=\sqrt{\frac{2\sigma(1-\sin\theta)}{\rho g}}$ b.3.9mm | ||