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Answer :
A similar beam, 2/3 foot wide, 1/4 foot high, and 20 feet long, can support approximately 3.25 tons when the original beam supports 10 tons.
First, let's set up the relationship based on the given information.
Let:
- W be the maximum weight that the beam can support (in tons).
- w be the width of the beam (in feet).
- h be the height of the beam (in feet).
- l be the length of the beam (in feet).
We are told that the maximum weight varies jointly as the width and the square of the height and inversely as the length. So, we can write this relationship as an equation:
[tex]\[ W = k \times \frac{wh^2}{l} \][/tex]
where k is a constant of variation.
Given that a beam [tex]\( \frac{1}{3} \)[/tex] foot wide, [tex]\( \frac{1}{2} \)[/tex] foot high, and 13 feet long can support 10 tons, we can use this information to find k:
[tex]\[ 10 = k \times \frac{\frac{1}{3} \times \left(\frac{1}{2}\right)^2}{13} \][/tex]
Solving for k:
[tex]\[ k = 10 \times \frac{13}{\left(\frac{1}{3} \times \left(\frac{1}{2}\right)^2\right)} \][/tex]
[tex]\[ k = 10 \times \frac{13}{\left(\frac{1}{3} \times \frac{1}{4}\right)} \][/tex]
[tex]\[ k = 10 \times \frac{13}{\frac{1}{12}} \][/tex]
[tex]\[ k = 10 \times 13 \times 12 \][/tex]
k = 1560
Now, we can use this value of k to find the maximum weight that a similar beam can support when the beam is [tex]\( \frac{2}{3} \)[/tex] foot wide, [tex]\( \frac{1}{4} \)[/tex] foot high, and 20 feet long:
[tex]\[ W = 1560 \times \frac{\frac{2}{3} \times \left(\frac{1}{4}\right)^2}{20} \]\[ W = 1560 \times \frac{\frac{2}{3} \times \frac{1}{16}}{20} \]\[ W = 1560 \times \frac{2}{3} \times \frac{1}{16} \times \frac{1}{20} \]\[ W = 1560 \times \frac{1}{480} \]\[ W = 3.25 \][/tex]
So, the maximum weight that a similar beam can support is approximately 3.25 tons.
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