Question

A cubical tank of side $3.0m$is evacuated of air. That is, the number of molecules of air inside the tank is insignificant compared to the number in the same volume outside. The atmospheric pressure is ${10}^{5}Pa$. What is the crushing force exerted by the atmosphere on the tank?

Answer 1

Given:

$$\begin{gathered} Sideof\tan k\left(a\right)=3.0m \\ Atmosphericpressure\left(P\right)={10}^{5}Pa \end{gathered}$$

Step 1: Calculating the Area of 1 face of the cube

$$\begin{gathered} Areaofcube=Sid{e}^2 \\ =a^2={(3.0m)}^2=9m^2 \end{gathered}$$

Step 2: Calculating force on 1 face of the cube

$$\begin{gathered} Force=Pressure\times Area \\ {F}_{1face}={10}^{5}Pa\times 9m^2 \\ =9\times {10}^{5}N \end{gathered}$$

Step 3: Calculating the total crushing force

$$\begin{gathered} TotalForce\left(F\right)=Forceononeface\times No.offaces \\ \Rightarrow F=F_{1face}\times 6 \\ =9\times {10}^5\times 6 \\ =5.4\times {10}^{6}N \end{gathered}$$

Hence, the total crushing force on the cubical tank is $5.4\times {10}^{6}N$.