Question

the safe load, $l$, of a wooden beam supported at both ends varies jointly as the width, $w$, and the square of the depth, $d$, and inversely as the length, $l$. a wooden beam 8 in. wide, 8 in. deep, and 5 ft long holds up 39515 lb. what load would a beam 9 in. wide, 5 in. deep, and 15 ft long, of the same material, support? round your answer to the nearest integer if necessary.

Explanation:

Step1: Write the joint - variation formula

The joint - variation formula is $L = k\frac{wd^{2}}{L}$, where $L$ is the safe load, $w$ is the width, $d$ is the depth, $L$ is the length, and $k$ is the constant of variation. First, convert the length of the first beam to inches. Since $1$ ft=$12$ inches, a $5$ - ft long beam is $5\times12 = 60$ inches long.

Step2: Find the value of the constant $k$

Substitute $w = 8$ inches, $d = 8$ inches, $L = 60$ inches, and $L=39515$ lb into the formula $L = k\frac{wd^{2}}{L}$:
\[39515=k\frac{8\times8^{2}}{60}\]
\[39515 = k\frac{8\times64}{60}\]
\[39515=k\frac{512}{60}\]
\[k=\frac{39515\times60}{512}\]
\[k=\frac{2370900}{512}\]
\[k = 4630.6640625\]

Step3: Find the safe load of the second beam

Convert the length of the second beam to inches. A $15$ - ft long beam is $15\times12 = 180$ inches long. Substitute $w = 9$ inches, $d = 5$ inches, $L = 180$ inches, and $k = 4630.6640625$ into the formula $L = k\frac{wd^{2}}{L}$:
\[L=4630.6640625\times\frac{9\times5^{2}}{180}\]
\[L=4630.6640625\times\frac{9\times25}{180}\]
\[L=4630.6640625\times\frac{225}{180}\]
\[L=4630.6640625\times1.25\]
\[L = 5788.330078125\approx5788\]

Answer:

$5788$