Question

As the result of a wind blowing normal to the plane of the rectangular sign, a uniform pressure of $175\mathrm{N}/{\mathrm{m}}^2$ is exerted in the direction shown in the figure. Determine the moment of the resulting force at about point $0$. Express your result as a vector using the coordinates shown.

Answer 1

Step 1: Given information

Uniform pressure, $\mathrm{p}=175\mathrm{N}/{\mathrm{m}}^2$

Width is $600\mathrm{mm}$.

Height is $750\mathrm{mm}$.

Step 2: To find

We have to determine the result as a vector using the coordinates shown.

Step 3: Calculate

The net force will be:

$\mathrm{F}=\mathrm{pa}$

F is the force.

p is the pressure.

a is the acceleration.

By substituting the values, we get

$$\begin{gathered} =175\times (600\times 750)\times {10}^{-6} \\ =78.75\mathrm{N} \end{gathered}$$

By using the figure,

The direction of force be $78.75(-\hat{\mathrm{i}})$

Now,

$\hat{\mathrm{r}}=100+\frac{750}{2}$

$1.375\mathrm{m}\left(\hat{\mathrm{j}}\right)$

$\overrightarrow{\mathrm{\tau }}=\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{F}}$

By substituting the values,

$$\begin{gathered} =1.375\left(\hat{\mathrm{j}}\right)\times 78.75(-\hat{\mathrm{i}}) \\ =108.28\hat{\mathrm{k}} \end{gathered}$$

Therefore, the result as a vector will be $108.28\hat{\mathrm{k}}$.