1890 lb
Explanation:
Using the variation relationship mentioned in 30:
[tex]\begin{gathered} \text{load, L varies jointly as the width w and square of the depth d and inversely as the length l} \\ L\text{ }\alpha\text{ }\frac{\text{w}d^2}{l} \\ \text{Equating the variation:} \\ L\text{ }=\text{ k}\frac{\text{w}d^2}{l} \end{gathered}[/tex][tex]\begin{gathered} In\text{ question 31:} \\ w\text{ = 2 in},\text{ d = 8 in, l = 14 ft},\text{ L = 2400 lb} \\ we\text{ convert length in ft to inches:} \\ l\text{ = }168in \\ \text{substituting the variables:} \\ 2400\text{ = k}\times\frac{2\times8^2}{168} \end{gathered}[/tex][tex]\begin{gathered} 2400\text{ = }\frac{128k}{168} \\ 2400(168)\text{ = 128k} \\ k\text{ = }\frac{2400(168)\text{ }}{128} \\ k\text{ = 3150} \\ \text{The relationship becomes:} \\ L\text{ =}\frac{3150wd^2}{l} \end{gathered}[/tex][tex]\begin{gathered} \text{when the dimension is 3in by 6in by 15ft:} \\ w\text{ = 3 in, d = 6 in, l = 15 ft, L = ?} \\ l\text{ = 15 ft, }l\text{ in inches = }180\text{ in} \\ we\text{ substitute for the variables in the equation showing the relationship:} \\ L\text{ = }\frac{3150\times3\times6^2}{180} \end{gathered}[/tex][tex]L\text{ = 1890 lb}[/tex]
It would support a load of 1890 lb
