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 4in. wide, 7in. deep, and 2ft long holds up 43240lb. What load would a beam 8in. wide, 6in. deep, and 10ft. long, of the same material, support? Round your answer to the nearest integer if necessary.

Now that the value of the constant has been gotten, lets calculate for [tex]L[/tex] again when [tex]w=8in[/tex], [tex]d=6in[/tex], and [tex]l=10ft=120in[/tex]

[tex]L=\frac{(5294.693878)(8)(6)^2}{120}[/tex]

[tex]L\sim 12707lb[/tex]

msm555 msm555 · Answer 2 · 2024-02-22T04:50:24+01:00

Answer:

120707 pounds

Step-by-step explanation:

Let's express the relationship described in the problem using the joint and inverse variation.

The relationship is given as:

[tex] L \propto \dfrac{w \cdot d^2}{l} [/tex]

To incorporate the constant of proportionality, we write it as:

[tex] L = k \cdot \dfrac{w \cdot d^2}{l} [/tex]

Given that a wooden beam 4 inches wide, 7 inches deep, and 2 feet = 2×12 = 24 in long holds up 43240 pounds, we can substitute these values to find [tex]k[/tex]:

[tex] 43240 = k \cdot \dfrac{4 \cdot 7^2}{24} [/tex]

Now, solve for [tex]k[/tex]:

[tex] k = \dfrac{43240 \cdot 24}{4 \cdot 7^2} [/tex]

[tex] k = \dfrac{259440}{49}[/tex]

Now that we have [tex]k[/tex], we can use it to find the load [tex]L[/tex] for the new beam with the dimensions 8 inches wide, 6 inches deep, and 10 feet = 10×12 in/ft = 120 in long:

[tex] L = k \cdot \dfrac{8 \cdot 6^2}{120} [/tex]

Substitute the value of [tex]k[/tex] into the equation:

[tex] L = \dfrac{ 259440}{49} \cdot \dfrac{8 \cdot 6^2}{120} [/tex]

[tex] L = \dfrac{ 259440}{49} \cdot \dfrac{288}{120} [/tex]

[tex] L = 12707.26531 [/tex]

[tex] L = 120707 \textsf{( in nearest integer)}[/tex]

Therefore, the load that the new beam would support is approximately 120707 pounds.