Answer :
To solve this problem, we need to understand how the maximum weight that a rectangular beam can support is calculated. The question states that the weight varies jointly as the width and the square of the height, and inversely as the length. This can be represented with the formula:
[tex]\[ W = k \times \frac{w \times h^2}{l} \][/tex]
Where:
- [tex]\( W \)[/tex] is the maximum weight the beam can support.
- [tex]\( w \)[/tex] is the width of the beam.
- [tex]\( h \)[/tex] is the height of the beam.
- [tex]\( l \)[/tex] is the length of the beam.
- [tex]\( k \)[/tex] is the constant of variation.
Let's break it down step-by-step:
1. Determine the Constant of Variation ([tex]\( k \)[/tex]) Using the First Beam:
- For the first beam, we have a width ([tex]\( w_1 \)[/tex]) of [tex]\(\frac{1}{3}\)[/tex] foot, a height ([tex]\( h_1 \)[/tex]) of [tex]\(\frac{1}{2}\)[/tex] foot, and a length ([tex]\( l_1 \)[/tex]) of 17 feet. It can support a maximum weight ([tex]\( W_1 \)[/tex]) of 20 tons.
- Plug these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
20 = k \times \frac{\left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right)^2}{17}
\][/tex]
- Calculate the constant [tex]\( k \)[/tex]:
[tex]\[
k = 20 \times \frac{17}{\left(\frac{1}{3}\right) \times \left(\frac{1}{4}\right)}
\][/tex]
[tex]\[
k = 4080
\][/tex]
2. Use the Constant ([tex]\( k \)[/tex]) to Find the Maximum Weight for the Second Beam:
- For the second beam, we have a width ([tex]\( w_2 \)[/tex]) of [tex]\(\frac{2}{3}\)[/tex] foot, a height ([tex]\( h_2 \)[/tex]) of [tex]\(\frac{1}{3}\)[/tex] foot, and a length ([tex]\( l_2 \)[/tex]) of 10 feet.
- Use the same formula to find the maximum weight ([tex]\( W_2 \)[/tex]) it can support:
[tex]\[
W_2 = 4080 \times \frac{\left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)^2}{10}
\][/tex]
- Calculate [tex]\( W_2 \)[/tex]:
[tex]\[
W_2 = 30.22222222222222
\][/tex]
- Round [tex]\( W_2 \)[/tex] to one decimal place:
[tex]\[
W_2 \approx 30.2 \text{ tons}
\][/tex]
Therefore, the similar beam with the specified dimensions can support approximately 30.2 tons.
[tex]\[ W = k \times \frac{w \times h^2}{l} \][/tex]
Where:
- [tex]\( W \)[/tex] is the maximum weight the beam can support.
- [tex]\( w \)[/tex] is the width of the beam.
- [tex]\( h \)[/tex] is the height of the beam.
- [tex]\( l \)[/tex] is the length of the beam.
- [tex]\( k \)[/tex] is the constant of variation.
Let's break it down step-by-step:
1. Determine the Constant of Variation ([tex]\( k \)[/tex]) Using the First Beam:
- For the first beam, we have a width ([tex]\( w_1 \)[/tex]) of [tex]\(\frac{1}{3}\)[/tex] foot, a height ([tex]\( h_1 \)[/tex]) of [tex]\(\frac{1}{2}\)[/tex] foot, and a length ([tex]\( l_1 \)[/tex]) of 17 feet. It can support a maximum weight ([tex]\( W_1 \)[/tex]) of 20 tons.
- Plug these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
20 = k \times \frac{\left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right)^2}{17}
\][/tex]
- Calculate the constant [tex]\( k \)[/tex]:
[tex]\[
k = 20 \times \frac{17}{\left(\frac{1}{3}\right) \times \left(\frac{1}{4}\right)}
\][/tex]
[tex]\[
k = 4080
\][/tex]
2. Use the Constant ([tex]\( k \)[/tex]) to Find the Maximum Weight for the Second Beam:
- For the second beam, we have a width ([tex]\( w_2 \)[/tex]) of [tex]\(\frac{2}{3}\)[/tex] foot, a height ([tex]\( h_2 \)[/tex]) of [tex]\(\frac{1}{3}\)[/tex] foot, and a length ([tex]\( l_2 \)[/tex]) of 10 feet.
- Use the same formula to find the maximum weight ([tex]\( W_2 \)[/tex]) it can support:
[tex]\[
W_2 = 4080 \times \frac{\left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)^2}{10}
\][/tex]
- Calculate [tex]\( W_2 \)[/tex]:
[tex]\[
W_2 = 30.22222222222222
\][/tex]
- Round [tex]\( W_2 \)[/tex] to one decimal place:
[tex]\[
W_2 \approx 30.2 \text{ tons}
\][/tex]
Therefore, the similar beam with the specified dimensions can support approximately 30.2 tons.
