Chapter 12.8, Problem 12P

To determine

To calculate: The wind pressure on a plane varies jointly as the surface area and the square of the speed of the wind.

Answer to Problem 12P

The speed of the wind is: $s_2=18.018\text{mi/h}$ .

Explanation of Solution

Given information:

The wind pressure on a plane varies jointly as the surface area and the square of the speed of the wind.

What is the speed of the wind when the pressure on a $2$ ft by $2$ ft square is $30$ lb. With a wind speed of $12mi/h$ , the pressure on a $4$ ft by $1.5$ ft rectangle is $20\text{lb}$ .

Formula used:

For varies directly and inversely:

  $P=KA{s}^2$

Calculation:

Consider that What is the speed of the wind when the pressure on a $2$ ft by $2$ ft square is $30$ lb. With a wind speed of $12mi/h$ , the pressure on a $4$ ft by $1.5$ ft rectangle is $20\text{lb}$ .

Recall that varies directly and inversely:

  $P=KA{s}^2$

The values are:

  $P=20,A=4\times 1.5,s=12\text{cm}$

Therefore, the value of $k$ is,

  $\begin{array}{c}P=KA{s}^2\\ 20=K\times \left(4\times 1.5\right)\times {12}^2\\ 20=K\times 6\times 144\\ \frac{20}{6\times 144}=K\end{array}$

So simplified further the value of $k$ is,

  $\begin{array}{c}\frac{20}{6\times 144}=K\\ \frac{20}{864}=K\\ 0.0231=K\end{array}$

Now to find the value of $s$ ,

  $P_2=K{A}_2{s}_2{}^2$

The values are:

  $P_2=40,A=2\times 2,K=0.0231$

Hence, when:

  $\begin{array}{c}{P}_2=K{A}_2{s}_2{}^2\\ 30=0.0231\times \left(2\times 2\right)\times s^2\\ 30=0.0231\times 4\times s^2\\ \frac{30}{0.0231\times 4}=s^2\end{array}$

So,

  $\begin{array}{l}\frac{30}{0.0231\times 4}=s_2{}^2\\ s_2{}^2=\frac{30}{0.0924}\\ s_2{}^2=324.6753\\ s_2=\sqrt{324.6753}\\ s_2=18.018\end{array}$

Thus, the speed of the wind is: $s_2=18.018\text{mi/h}$ .