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The safe load, L, of a wooden beam of width w, height h and length l, supported at both ends, varies directly as the product of the width and the square of the height and inversely as the length. A wooden beam 5 inches wide, 7 inches high and 144 inches long can hold a load of 8740 pounds. What load would a beam 6 inches wide, 9 inches high, and 216 inches long of the same material, support? Round your answer to the nearest integer if necessary.

Sagot :

Since the load L varies directly with the product of width and square of the height h, and inveresly as the length l, so

[tex]\begin{gathered} L=k(\frac{wh^2}{l}) \\ OR \\ \frac{L_1}{L_2}=\frac{w_1}{w_2}\times\frac{h^2_1}{h^2_2}\times\frac{l_2}{l_1} \end{gathered}[/tex]

We will use the second rule

Since L is 8740 pounds when w is 5 in., h is 7 in. and l is 144 in.

[tex]\begin{gathered} L_1=8740 \\ w_1=5 \\ h_1=7 \\ l_1=144 \end{gathered}[/tex]

We need to find L when w is 6 in., h is 9 in. and l is 216 in.

[tex]\begin{gathered} L_2=? \\ w_2=6 \\ h_2=9 \\ l_2=216 \end{gathered}[/tex]

Let us substitute them in the second rule

[tex]\begin{gathered} \frac{8740}{L_2}=\frac{5}{6}\times\frac{7^2}{9^2}\times\frac{216}{144} \\ \frac{8740}{L_2}=\frac{5}{6}\times\frac{49}{81}\times\frac{216}{144} \\ \frac{8740}{L_2}=\frac{245}{324} \end{gathered}[/tex]

By using cross multiplication

[tex]\begin{gathered} 245\times L_2=8740\times324 \\ 245L_2=2831760 \end{gathered}[/tex]

Divide both sides by 245

[tex]\begin{gathered} \frac{245L_2}{245}=\frac{2831760}{245} \\ L_2=11558.20408 \end{gathered}[/tex]

Round it to the nearest integer

[tex]L_2=11558\text{ pounds}[/tex]

The load is 11558 pounds

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