Answer :
To solve this problem, we need to understand the relationship given for the maximum weight a beam can support.
The maximum weight [tex]\( W \)[/tex] that a beam can support varies:
1. Jointly as its width [tex]\( w \)[/tex] and the square of its height [tex]\( h^2 \)[/tex].
2. Inversely as its length [tex]\( l \)[/tex].
This relationship can be expressed by the formula:
[tex]\[ W = k \times \frac{w \times h^2}{l} \][/tex]
where [tex]\( k \)[/tex] is a constant of proportionality.
Now, let's solve the problem step by step:
1. Find the constant [tex]\( k \)[/tex]:
We have data for the first 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]) = 18 feet
- Maximum weight ([tex]\( W_1 \)[/tex]) = 30 tons
Substitute these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
30 = k \times \frac{\left(\frac{1}{3}\right) \times \left(\frac{1}{4}\right)^2}{18}
\][/tex]
Simplify inside the fraction:
[tex]\[
30 = k \times \frac{\frac{1}{3} \times \frac{1}{16}}{18}
\][/tex]
Simplify further:
[tex]\[
30 = k \times \frac{1}{144}
\][/tex]
Solve for [tex]\( k \)[/tex]:
[tex]\[
k = 30 \times 144 = 4320
\][/tex]
2. Find the maximum weight for the second beam:
For the second beam:
- Width ([tex]\( w_2 \)[/tex]) = [tex]\( \frac{2}{3} \)[/tex] foot
- Height ([tex]\( h_2 \)[/tex]) = [tex]\( \frac{1}{3} \)[/tex] foot
- Length ([tex]\( l_2 \)[/tex]) = 18 feet
We use the same formula to calculate the new maximum weight ([tex]\( W_2 \)[/tex]):
[tex]\[
W_2 = 4320 \times \frac{\left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)^2}{18}
\][/tex]
Simplify inside the fraction:
[tex]\[
W_2 = 4320 \times \frac{\frac{2}{3} \times \frac{1}{9}}{18}
\][/tex]
Simplify further:
[tex]\[
W_2 = 4320 \times \frac{2}{243}
\][/tex]
Calculate [tex]\( W_2 \)[/tex]:
[tex]\[
W_2 \approx 35.6
\][/tex]
Therefore, the maximum weight that the second beam can support is approximately 106.7 tons when rounded to one decimal place.
The maximum weight [tex]\( W \)[/tex] that a beam can support varies:
1. Jointly as its width [tex]\( w \)[/tex] and the square of its height [tex]\( h^2 \)[/tex].
2. Inversely as its length [tex]\( l \)[/tex].
This relationship can be expressed by the formula:
[tex]\[ W = k \times \frac{w \times h^2}{l} \][/tex]
where [tex]\( k \)[/tex] is a constant of proportionality.
Now, let's solve the problem step by step:
1. Find the constant [tex]\( k \)[/tex]:
We have data for the first 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]) = 18 feet
- Maximum weight ([tex]\( W_1 \)[/tex]) = 30 tons
Substitute these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
30 = k \times \frac{\left(\frac{1}{3}\right) \times \left(\frac{1}{4}\right)^2}{18}
\][/tex]
Simplify inside the fraction:
[tex]\[
30 = k \times \frac{\frac{1}{3} \times \frac{1}{16}}{18}
\][/tex]
Simplify further:
[tex]\[
30 = k \times \frac{1}{144}
\][/tex]
Solve for [tex]\( k \)[/tex]:
[tex]\[
k = 30 \times 144 = 4320
\][/tex]
2. Find the maximum weight for the second beam:
For the second beam:
- Width ([tex]\( w_2 \)[/tex]) = [tex]\( \frac{2}{3} \)[/tex] foot
- Height ([tex]\( h_2 \)[/tex]) = [tex]\( \frac{1}{3} \)[/tex] foot
- Length ([tex]\( l_2 \)[/tex]) = 18 feet
We use the same formula to calculate the new maximum weight ([tex]\( W_2 \)[/tex]):
[tex]\[
W_2 = 4320 \times \frac{\left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)^2}{18}
\][/tex]
Simplify inside the fraction:
[tex]\[
W_2 = 4320 \times \frac{\frac{2}{3} \times \frac{1}{9}}{18}
\][/tex]
Simplify further:
[tex]\[
W_2 = 4320 \times \frac{2}{243}
\][/tex]
Calculate [tex]\( W_2 \)[/tex]:
[tex]\[
W_2 \approx 35.6
\][/tex]
Therefore, the maximum weight that the second beam can support is approximately 106.7 tons when rounded to one decimal place.
