Answer :
To solve the problem, we need to utilize the relationship given in the question about how the maximum weight a beam can support varies. Let's go through the process step-by-step.
### Step-by-Step Solution:
1. Understand the Relationship:
The maximum weight [tex]\( W \)[/tex] that a beam can support varies:
- Jointly as its width [tex]\( w \)[/tex], meaning [tex]\( W \propto w \)[/tex].
- Jointly as the square of its height [tex]\( h^2 \)[/tex], meaning [tex]\( W \propto h^2 \)[/tex].
- Inversely as its length [tex]\( l \)[/tex], meaning [tex]\( W \propto \frac{1}{l} \)[/tex].
Combining these, we can express the relationship as:
[tex]\[
W = k \cdot \frac{w \cdot h^2}{l}
\][/tex]
where [tex]\( k \)[/tex] is a constant.
2. Use the Given Data to Find [tex]\( k \)[/tex]:
For the first beam:
- Width [tex]\( w_1 = \frac{1}{4} \)[/tex] foot
- Height [tex]\( h_1 = \frac{1}{3} \)[/tex] foot
- Length [tex]\( l_1 = 13 \)[/tex] feet
- Maximum weight supported [tex]\( W_1 = 30 \)[/tex] tons
Substitute these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
30 = k \cdot \frac{\frac{1}{4} \cdot \left(\frac{1}{3}\right)^2}{13}
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
30 = k \cdot \frac{\frac{1}{4} \cdot \frac{1}{9}}{13}
\][/tex]
[tex]\[
30 = k \cdot \frac{1}{36 \cdot 13}
\][/tex]
[tex]\[
k = 30 \cdot 468 = 14040
\][/tex]
3. Calculate the Weight for the New Beam:
For the second beam:
- Width [tex]\( w_2 = \frac{2}{3} \)[/tex] foot
- Height [tex]\( h_2 = \frac{1}{4} \)[/tex] foot
- Length [tex]\( l_2 = 10 \)[/tex] feet
Use the formula with the new dimensions:
[tex]\[
W_2 = 14040 \cdot \frac{\frac{2}{3} \cdot \left(\frac{1}{4}\right)^2}{10}
\][/tex]
[tex]\[
W_2 = 14040 \cdot \frac{\frac{2}{3} \cdot \frac{1}{16}}{10}
\][/tex]
[tex]\[
W_2 = 14040 \cdot \frac{1}{240}
\][/tex]
[tex]\[
W_2 = \frac{14040}{240} = 58.5
\][/tex]
Thus, the maximum weight that the new beam can support is [tex]\( 58.5 \)[/tex] tons.
### Step-by-Step Solution:
1. Understand the Relationship:
The maximum weight [tex]\( W \)[/tex] that a beam can support varies:
- Jointly as its width [tex]\( w \)[/tex], meaning [tex]\( W \propto w \)[/tex].
- Jointly as the square of its height [tex]\( h^2 \)[/tex], meaning [tex]\( W \propto h^2 \)[/tex].
- Inversely as its length [tex]\( l \)[/tex], meaning [tex]\( W \propto \frac{1}{l} \)[/tex].
Combining these, we can express the relationship as:
[tex]\[
W = k \cdot \frac{w \cdot h^2}{l}
\][/tex]
where [tex]\( k \)[/tex] is a constant.
2. Use the Given Data to Find [tex]\( k \)[/tex]:
For the first beam:
- Width [tex]\( w_1 = \frac{1}{4} \)[/tex] foot
- Height [tex]\( h_1 = \frac{1}{3} \)[/tex] foot
- Length [tex]\( l_1 = 13 \)[/tex] feet
- Maximum weight supported [tex]\( W_1 = 30 \)[/tex] tons
Substitute these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
30 = k \cdot \frac{\frac{1}{4} \cdot \left(\frac{1}{3}\right)^2}{13}
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
30 = k \cdot \frac{\frac{1}{4} \cdot \frac{1}{9}}{13}
\][/tex]
[tex]\[
30 = k \cdot \frac{1}{36 \cdot 13}
\][/tex]
[tex]\[
k = 30 \cdot 468 = 14040
\][/tex]
3. Calculate the Weight for the New Beam:
For the second beam:
- Width [tex]\( w_2 = \frac{2}{3} \)[/tex] foot
- Height [tex]\( h_2 = \frac{1}{4} \)[/tex] foot
- Length [tex]\( l_2 = 10 \)[/tex] feet
Use the formula with the new dimensions:
[tex]\[
W_2 = 14040 \cdot \frac{\frac{2}{3} \cdot \left(\frac{1}{4}\right)^2}{10}
\][/tex]
[tex]\[
W_2 = 14040 \cdot \frac{\frac{2}{3} \cdot \frac{1}{16}}{10}
\][/tex]
[tex]\[
W_2 = 14040 \cdot \frac{1}{240}
\][/tex]
[tex]\[
W_2 = \frac{14040}{240} = 58.5
\][/tex]
Thus, the maximum weight that the new beam can support is [tex]\( 58.5 \)[/tex] tons.
