Chapter 10, Problem 43P

To determine

To Calculate: The net force on the roof.

Answer to Problem 43P

  $189.6\text{N}$

Explanation of Solution

Given info:

The velocity of the flowing wind, $\mathrm{v}=35\text{m/s}$

The area of the roof, $\mathrm{A}=240{\text{m}}^{\text{2}}$

Formula used:

Bernoulli’s equation:

  ${\left({\mathrm{P}}_1+\frac{1}{2}\mathrm{\rho }{\mathrm{v}}_1{}^2+\mathrm{\rho }\mathrm{g}{\mathrm{h}}_1\right)}_{\text{bottom}}={\left({\mathrm{P}}_2+\frac{1}{2}\mathrm{\rho }{\mathrm{v}}_2{}^2+\mathrm{\rho }\mathrm{g}{\mathrm{h}}_2\right)}_{\text{Top}}$

Here, ${\mathrm{P}}_1$ is the pressure at the bottom of roof.

  ${\mathrm{v}}_1$ is the velocity of the wind at the bottom of roof.

  ${\mathrm{P}}_2$ is the pressure at the top of roof.

  ${\mathrm{v}}_2$ is the velocity of the wind at the top of roof.

  ${\mathrm{h}}_1$ is the height at the bottom.

  ${\mathrm{h}}_2$ is the height at the top.

Calculation:

Write the expression for the Bernoulli’s theorem.

  ${\left({\mathrm{P}}_1+\frac{1}{2}\mathrm{\rho }{\mathrm{v}}_1{}^2+\mathrm{\rho }\mathrm{g}{\mathrm{h}}_1\right)}_{\text{bottom}}={\left({\mathrm{P}}_2+\frac{1}{2}\mathrm{\rho }{\mathrm{v}}_2{}^2+\mathrm{\rho }\mathrm{g}{\mathrm{h}}_2\right)}_{\text{Top}}$

Substitute the values in the above expression.

  $\begin{array}{l}{\mathrm{P}}_1+0+0={\mathrm{P}}_2+\frac{1}{2}\mathrm{\rho }{\mathrm{v}}_2{}^2+0\\ {\mathrm{P}}_1-{\mathrm{P}}_2=\frac{1}{2}\mathrm{\rho }{\mathrm{v}}_2{}^2\end{array}$

Here, ${\mathrm{P}}_2$ is the atmospheric pressure.

Thus,

  $\begin{array}{c}{\mathrm{P}}_1-{\mathrm{P}}_{\text{atm}}=\frac{1}{2}\mathrm{\rho }{\mathrm{v}}_2{}^2\\ =\frac{1}{2}\times 1.29\times {\left(35\right)}^2\\ =790.12{\text{N/m}}^{\text{2}}\end{array}$

Write the expression showing relation between the pressure and force.

  $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$

This net force will be:

  $\begin{array}{l}\left({\mathrm{P}}_1-{\mathrm{P}}_{\text{atm}}\right)=\frac{\mathrm{F}}{\mathrm{A}}\\ \mathrm{F}=\left({\mathrm{P}}_1-{\mathrm{P}}_{\text{atm}}\right)\mathrm{A}\end{array}$

Substitute the values in the above expression.

  $\begin{array}{c}\mathrm{F}=790.12\times 240\\ =189628\text{N}\\ \text{=189}\text{.6 kN}\end{array}$

Conclusion:

The net force is $189.6\text{N}$ .