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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.)

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 class=

Sagot :

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

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