Answer:
The load varies jointly as the width w, the square of the depth d, and inversely as the length l, this can be represented below as
[tex]\begin{gathered} L\propto\frac{wd^2}{l} \\ when\text{ the proportionality sign changes to equal to, a constant k is introduced} \\ L=\frac{kwd^2}{l} \end{gathered}[/tex]From the question, the given values are
[tex]w=8in,d=2in,l=16ft,L=7335lb[/tex]By substituting the values, we will have
[tex]\begin{gathered} L=\frac{kwd^{2}}{l} \\ 7335=\frac{k\times8\times2^2}{16} \\ 7335=2k \\ \frac{2k}{2}=\frac{7335}{2} \\ k=3667.5 \end{gathered}[/tex]Substitute the value of k to get the equation connecting the w,d,l and L
[tex]\begin{gathered} L=\frac{kwd^{2}}{l} \\ L=\frac{3667.5wd^2}{l} \end{gathered}[/tex]To get the value of the load, we will substitute the value of
[tex]w=9in,d=6in,l=19ft[/tex][tex]\begin{gathered} L=\frac{3,667.5wd^{2}}{l} \\ L=\frac{3.667.5\times9\times6^2}{19} \\ L=62541lb \end{gathered}[/tex]Hence,
The final answer is =62541 lb
