College

The maximum weight that a rectangular beam can support varies jointly as its width and the square of its height, and inversely as its length.

If a beam [tex]\frac{1}{3}[/tex] foot wide, [tex]\frac{1}{4}[/tex] foot high, and 15 feet long can support 14 tons, find how much a similar beam can support if the beam is [tex]\frac{1}{4}[/tex] foot wide, [tex]\frac{1}{2}[/tex] foot high, and 20 feet long.

Answer :

To solve this problem, we need to understand how the weight that a beam can support varies with its dimensions. The maximum weight that a rectangular beam can support is determined by the following relationships:

- It varies jointly as the width of the beam.
- It varies as the square of the beam's height.
- It varies inversely with the length of the beam.

Let's denote:
- [tex]\( W \)[/tex] as the weight the beam can support,
- [tex]\( w \)[/tex] as the width of the beam,
- [tex]\( h \)[/tex] as the height of the beam,
- [tex]\( l \)[/tex] as the length of the beam.

The mathematical expression of these relationships is:
[tex]\[ W = k \cdot w \cdot h^2 \cdot \frac{1}{l} \][/tex]
where [tex]\( k \)[/tex] is the constant of proportionality.

Step 1: Determine the constant of proportionality, [tex]\( k \)[/tex], using the first beam's data.
- Width, [tex]\( w_1 = \frac{1}{3} \)[/tex] foot
- Height, [tex]\( h_1 = \frac{1}{4} \)[/tex] foot
- Length, [tex]\( l_1 = 15 \)[/tex] feet
- Weight supported, [tex]\( W_1 = 14 \)[/tex] tons

Plug these values into the equation:
[tex]\[ 14 = k \cdot \frac{1}{3} \cdot \left(\frac{1}{4}\right)^2 \cdot \frac{1}{15} \][/tex]

Simplify:
[tex]\[ 14 = k \cdot \frac{1}{3} \cdot \frac{1}{16} \cdot \frac{1}{15} \][/tex]

[tex]\[ 14 = k \cdot \frac{1}{720} \][/tex]

To find the constant [tex]\( k \)[/tex]:
[tex]\[ k = 14 \times 720 = 10080 \][/tex]

Step 2: Calculate the weight that the second beam can support using the same formula.

For the second beam:
- Width, [tex]\( w_2 = \frac{1}{4} \)[/tex] foot
- Height, [tex]\( h_2 = \frac{1}{2} \)[/tex] foot
- Length, [tex]\( l_2 = 20 \)[/tex] feet

We now know that [tex]\( k = 10080 \)[/tex], so:
[tex]\[ W_2 = 10080 \cdot \frac{1}{4} \cdot \left(\frac{1}{2}\right)^2 \cdot \frac{1}{20} \][/tex]

Calculate the weight:
[tex]\[ W_2 = 10080 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{20} \][/tex]

[tex]\[ W_2 = 10080 \cdot \frac{1}{320} \][/tex]

[tex]\[ W_2 = 31.5 \][/tex]

Therefore, the second beam can support 31.5 tons.