An office window has dimensions $\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}$by$\mathbf{2}\mathbf{.}\mathbf{1}\mathbf{}\mathit{m}$. As a result of the passage of a storm, the outside air pressure drops to$\mathbf{0}\mathbf{.}\mathbf{96}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$, but inside the pressure is held at$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{a}\mathit{t}\mathit{m}$. What net force pushes out on the window?

1. Dimensions of the window are $3.4\mathrm{m}\times 2.1\mathrm{m}$.
2. Internal pressure,$p_i=1.0\mathrm{atm}$
3. Outer pressure,$p_o=0.96\mathrm{atm}$

We can find the applied pressure on the window from the inside pressure and the outside pressure Then we can find the area from the given dimensions of the window. Using this pressure and area, we can find the net force on the window.

Formula:

Pressure on the body through applied force F on a given area A,

$p=\frac{F}{A}$ (i)

The net applied pressure on the window can be given as:

$$\begin{gathered} p=p_i-p_o \\ =1.0\mathrm{atm}-0.96\mathrm{atm} \\ =0.04\mathrm{atm} \\ =0.04\times 1.013\times {10}^5\mathrm{Pa}\left(\because 1\mathrm{atm}=1.013\times {10}^5\mathrm{Pa}\right) \\ =4052\mathrm{Pa} \end{gathered}$$

Again, area of the window,

$$\begin{gathered} A=3.4\times 2.1 \\ =7.14{\mathrm{m}}^2 \end{gathered}$$

Hence using equation (i) and the above values of p and A, the force on the window can be given as:

$$\begin{gathered} F=p\times A \\ =4052\mathrm{Pa}\times 7.14{\mathrm{m}}^2 \\ =28931.28\mathrm{N} \\ =2.9\times {10}^4\mathrm{N} \end{gathered}$$

Therefore, the net force on the window is $2.9\times {10}^4\mathrm{N}$.