Question

The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and inversely as the length, I. A wooden beam8 in. wide, 2 in. deep, and 16 ft long holds up 7335 lb. What load would a beam 9 in. wide, 6 in. deep and 19 ft long of the same material support? (Round off youranswer to the nearest pound.)

Answer 1

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

`\begin{gathered} L\propto(wd^2)/(l) \\ when\text{ the proportionality sign changes to equal to, a constant k is introduced} \\ L=(kwd^2)/(l) \end{gathered}`

From the question, the given values are

`w=8in,d=2in,l=16ft,L=7335lb`

By substituting the values, we will have

`\begin{gathered} L=(kwd^(2))/(l) \\ 7335=(k*8*2^2)/(16) \\ 7335=2k \\ (2k)/(2)=(7335)/(2) \\ k=3667.5 \end{gathered}`

Substitute the value of k to get the equation connecting the w,d,l and L

`\begin{gathered} L=(kwd^(2))/(l) \\ L=(3667.5wd^2)/(l) \end{gathered}`

To get the value of the load, we will substitute the value of

`w=9in,d=6in,l=19ft`
`\begin{gathered} L=(3,667.5wd^(2))/(l) \\ L=(3.667.5*9*6^2)/(19) \\ L=62541lb \end{gathered}`

Hence,

The final answer is =62541 lb