Question

A large 10,000 L aquarium is supported by four wood posts (Douglas fir) at the corners. Each post has a square 4.0 cm * 4.0 cm cross-section and is 80 cm tall. By how much is each post compressed by the weight of the aquarium?

Short answer

Each post is compressed by$1.23mm$.

Step-by-step solution

Given information

The area of cross-section is $A=0.04m\times 0.04m=1.6\times {10}^{-3}{m}^2$

The height of the aquarium is $L=80cm=0.80m$

Young's modulus of the wood is$Y=1\times {10}^{10}N/m^2$

Weight of the aquarium

$1L={10}^{-3}{m}^3$

The volume of the aquarium is

$$\begin{gathered} V=10000\times {10}^{-3} \\ =10m^3 \end{gathered}$$

The weight of the aquarium is

$$\begin{gathered} F_G=mg \\ ={\rho }_{water}Vg \\ =1000\times 10\times 9.8 \\ =9.8\times {10}^{4}N \end{gathered}$$

Weight supported by each wood post is

$$\begin{gathered} =\frac{1}{4}\left(9.8\times {10}^4\right) \\ =2.45\times {10}^{4}N \end{gathered}$$

Compression in each post

Young's modulus is given by $Y=\frac{{\displaystyle \raisebox{1ex}{$F$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}}{{\displaystyle \raisebox{1ex}{$∆L$}\!\left/ \!\raisebox{-1ex}{$L$}\right.}}$.

Compression in the post is

$$\begin{gathered} ∆L=\frac{F}{A}\frac{L}{Y} \\ =\frac{2.45\times {10}^4}{1.6\times {10}^{-3}}\frac{0.80}{1\times {10}^{10}} \\ =1.23\times {10}^{-3}m \\ =1.23mm \end{gathered}$$

Therefore, each post is compressed by $1.23mm$by the weight of the aquarium.