Answer :
To solve this problem, we need to understand how the maximum weight that a rectangular beam can support is calculated. The weight varies jointly as the width and the square of its height, and inversely as its length. We can express this relationship with the formula:
[tex]\[ W = k \cdot \frac{w \cdot h^2}{l} \][/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 a constant of proportionality.
### Step 1: Find the constant [tex]\( k \)[/tex]
We're given that a beam with dimensions [tex]\( w = \frac{1}{3} \)[/tex] foot, [tex]\( h = \frac{1}{4} \)[/tex] foot, and [tex]\( l = 12 \)[/tex] feet can support 13 tons. We can use this information to find [tex]\( k \)[/tex].
First, substitute the known values into the formula:
[tex]\[ 13 = k \cdot \frac{\left(\frac{1}{3}\right) \cdot \left(\frac{1}{4}\right)^2}{12} \][/tex]
Solving for [tex]\( k \)[/tex], we get:
1. Calculate [tex]\( \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)[/tex].
2. Plug the values back into the formula:
[tex]\[ 13 = k \cdot \frac{\left(\frac{1}{3}\right) \cdot \frac{1}{16}}{12} \][/tex]
3. Simplify:
[tex]\[ 13 = k \cdot \frac{1}{576} \][/tex]
4. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = 13 \cdot 576 = 7488 \][/tex]
### Step 2: Find the maximum weight for the new beam dimensions
Now we calculate the maximum weight for a beam with dimensions [tex]\( w = \frac{1}{3} \)[/tex] foot, [tex]\( h = \frac{1}{2} \)[/tex] foot, and [tex]\( l = 15 \)[/tex] feet using the same constant [tex]\( k = 7488 \)[/tex].
Substitute these values into the formula:
[tex]\[ W = 7488 \cdot \frac{\left(\frac{1}{3}\right) \cdot \left(\frac{1}{2}\right)^2}{15} \][/tex]
1. Calculate [tex]\( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)[/tex].
2. Substitute into the formula:
[tex]\[ W = 7488 \cdot \frac{\left(\frac{1}{3}\right) \cdot \frac{1}{4}}{15} \][/tex]
3. Simplify:
[tex]\[ W = 7488 \cdot \frac{1}{180} \][/tex]
4. Calculate:
[tex]\[ W = 41.6 \][/tex]
Therefore, the maximum weight that the new beam can support is 41.6 tons.
[tex]\[ W = k \cdot \frac{w \cdot h^2}{l} \][/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 a constant of proportionality.
### Step 1: Find the constant [tex]\( k \)[/tex]
We're given that a beam with dimensions [tex]\( w = \frac{1}{3} \)[/tex] foot, [tex]\( h = \frac{1}{4} \)[/tex] foot, and [tex]\( l = 12 \)[/tex] feet can support 13 tons. We can use this information to find [tex]\( k \)[/tex].
First, substitute the known values into the formula:
[tex]\[ 13 = k \cdot \frac{\left(\frac{1}{3}\right) \cdot \left(\frac{1}{4}\right)^2}{12} \][/tex]
Solving for [tex]\( k \)[/tex], we get:
1. Calculate [tex]\( \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)[/tex].
2. Plug the values back into the formula:
[tex]\[ 13 = k \cdot \frac{\left(\frac{1}{3}\right) \cdot \frac{1}{16}}{12} \][/tex]
3. Simplify:
[tex]\[ 13 = k \cdot \frac{1}{576} \][/tex]
4. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = 13 \cdot 576 = 7488 \][/tex]
### Step 2: Find the maximum weight for the new beam dimensions
Now we calculate the maximum weight for a beam with dimensions [tex]\( w = \frac{1}{3} \)[/tex] foot, [tex]\( h = \frac{1}{2} \)[/tex] foot, and [tex]\( l = 15 \)[/tex] feet using the same constant [tex]\( k = 7488 \)[/tex].
Substitute these values into the formula:
[tex]\[ W = 7488 \cdot \frac{\left(\frac{1}{3}\right) \cdot \left(\frac{1}{2}\right)^2}{15} \][/tex]
1. Calculate [tex]\( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)[/tex].
2. Substitute into the formula:
[tex]\[ W = 7488 \cdot \frac{\left(\frac{1}{3}\right) \cdot \frac{1}{4}}{15} \][/tex]
3. Simplify:
[tex]\[ W = 7488 \cdot \frac{1}{180} \][/tex]
4. Calculate:
[tex]\[ W = 41.6 \][/tex]
Therefore, the maximum weight that the new beam can support is 41.6 tons.