Answer :
To solve this problem, we need to understand how the maximum weight a beam can support varies with its dimensions. The problem states that the maximum weight the beam can support varies:
- Jointly with the width ([tex]\(w\)[/tex])
- Jointly with the square of the height ([tex]\(h^2\)[/tex])
- Inversely with the length ([tex]\(l\)[/tex])
The relationship can be expressed with the formula:
[tex]\[ \text{max\_weight} = k \times \frac{w \times h^2}{l} \][/tex]
where [tex]\(k\)[/tex] is the constant of variation.
Step 1: Find the constant of variation ([tex]\(k\)[/tex]).
For the known beam:
- Width ([tex]\(w_1\)[/tex]) = [tex]\(\frac{1}{3}\)[/tex] foot
- Height ([tex]\(h_1\)[/tex]) = [tex]\(\frac{1}{4}\)[/tex] foot
- Length ([tex]\(l_1\)[/tex]) = 12 feet
- Maximum weight supported ([tex]\(\text{max\_weight}_1\)[/tex]) = 13 tons
Plug these values into the formula:
[tex]\[ 13 = k \times \frac{\frac{1}{3} \times \left(\frac{1}{4}\right)^2}{12} \][/tex]
Simplify:
[tex]\[ 13 = k \times \frac{\frac{1}{3} \times \frac{1}{16}}{12} \][/tex]
[tex]\[ 13 = k \times \frac{1}{576} \][/tex]
Solve for [tex]\(k\)[/tex]:
[tex]\[ k = 13 \times 576 = 7488 \][/tex]
Step 2: Use the constant [tex]\(k\)[/tex] to find the maximum weight the similar beam can support.
For the similar beam:
- Width ([tex]\(w_2\)[/tex]) = [tex]\(\frac{1}{3}\)[/tex] foot
- Height ([tex]\(h_2\)[/tex]) = [tex]\(\frac{1}{2}\)[/tex] foot
- Length ([tex]\(l_2\)[/tex]) = 15 feet
Plug these values and [tex]\(k\)[/tex] into the formula:
[tex]\[ \text{max\_weight}_2 = 7488 \times \frac{\frac{1}{3} \times \left(\frac{1}{2}\right)^2}{15} \][/tex]
Simplify:
[tex]\[ \text{max\_weight}_2 = 7488 \times \frac{\frac{1}{3} \times \frac{1}{4}}{15} \][/tex]
[tex]\[ \text{max\_weight}_2 = 7488 \times \frac{1}{180} \][/tex]
Calculate:
[tex]\[ \text{max\_weight}_2 = 41.6 \][/tex]
Therefore, the maximum weight that the similar beam can support is 41.6 tons.
- Jointly with the width ([tex]\(w\)[/tex])
- Jointly with the square of the height ([tex]\(h^2\)[/tex])
- Inversely with the length ([tex]\(l\)[/tex])
The relationship can be expressed with the formula:
[tex]\[ \text{max\_weight} = k \times \frac{w \times h^2}{l} \][/tex]
where [tex]\(k\)[/tex] is the constant of variation.
Step 1: Find the constant of variation ([tex]\(k\)[/tex]).
For the known beam:
- Width ([tex]\(w_1\)[/tex]) = [tex]\(\frac{1}{3}\)[/tex] foot
- Height ([tex]\(h_1\)[/tex]) = [tex]\(\frac{1}{4}\)[/tex] foot
- Length ([tex]\(l_1\)[/tex]) = 12 feet
- Maximum weight supported ([tex]\(\text{max\_weight}_1\)[/tex]) = 13 tons
Plug these values into the formula:
[tex]\[ 13 = k \times \frac{\frac{1}{3} \times \left(\frac{1}{4}\right)^2}{12} \][/tex]
Simplify:
[tex]\[ 13 = k \times \frac{\frac{1}{3} \times \frac{1}{16}}{12} \][/tex]
[tex]\[ 13 = k \times \frac{1}{576} \][/tex]
Solve for [tex]\(k\)[/tex]:
[tex]\[ k = 13 \times 576 = 7488 \][/tex]
Step 2: Use the constant [tex]\(k\)[/tex] to find the maximum weight the similar beam can support.
For the similar beam:
- Width ([tex]\(w_2\)[/tex]) = [tex]\(\frac{1}{3}\)[/tex] foot
- Height ([tex]\(h_2\)[/tex]) = [tex]\(\frac{1}{2}\)[/tex] foot
- Length ([tex]\(l_2\)[/tex]) = 15 feet
Plug these values and [tex]\(k\)[/tex] into the formula:
[tex]\[ \text{max\_weight}_2 = 7488 \times \frac{\frac{1}{3} \times \left(\frac{1}{2}\right)^2}{15} \][/tex]
Simplify:
[tex]\[ \text{max\_weight}_2 = 7488 \times \frac{\frac{1}{3} \times \frac{1}{4}}{15} \][/tex]
[tex]\[ \text{max\_weight}_2 = 7488 \times \frac{1}{180} \][/tex]
Calculate:
[tex]\[ \text{max\_weight}_2 = 41.6 \][/tex]
Therefore, the maximum weight that the similar beam can support is 41.6 tons.
