The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, and the square of the depth, d, and inversely as the length, l. A wooden beam 7in.
7
⁢
in.
wide, 3in.
3
⁢
in.
deep, and 9ft
9
⁢
ft
long holds up 49419lb
49419
⁢
lb
. What load would a beam 3in.
3
⁢
in.
wide, 8in.
8
⁢
in.
deep, and 12ft
12
⁢
ft
. long, of the same material, support? Round your answer to the nearest integer if necessary.

The problem states that the safe load, \(L\), of a wooden beam varies jointly as the width (\(w\)) and the square of the depth (\(d^2\)), and inversely as the length (\(l\)).

This means we can express \(L\) as:

\[
L = k \cdot \frac{w \cdot d^2}{l}
\]

where \(k\) is a constant of proportionality.

### Step 1: Find the constant \(k\)

We are given the following for the first beam:
- Width: \( w_1 = 7 \, \text{in} \)
- Depth: \( d_1 = 3 \, \text{in} \)
- Length: \( l_1 = 9 \, \text{ft} = 108 \, \text{in} \) (converted to inches)
- Load: \( L_1 = 49419 \, \text{lb} \)

Now, substitute these values into the equation for \(L\):

\[
49419 = k \cdot \frac{7 \cdot 3^2}{108}
\]

Simplify:

\[
49419 = k \cdot \frac{7 \cdot 9}{108}
\]

\[
49419 = k \cdot \frac{63}{108}
\]

\[
49419 = k \cdot 0.5833
\]

Solve for \(k\):

\[
k = \frac{49419}{0.5833} \approx 84734
\]

### Step 2: Use \(k\) to find the load for the second beam

We are given the following for the second beam:
- Width: \( w_2 = 3 \, \text{in} \)
- Depth: \( d_2 = 8 \, \text{in} \)
- Length: \( l_2 = 12 \, \text{ft} = 144 \, \text{in} \) (converted to inches)

Substitute these values into the formula:

\[
L_2 = 84734 \cdot \frac{3 \cdot 8^2}{144}
\]

Simplify:

\[
L_2 = 84734 \cdot \frac{3 \cdot 64}{144}
\]

\[
L_2 = 84734 \cdot \frac{192}{144}
\]

\[
L_2 = 84734 \cdot 1.3333
\]

\[
L_2 \approx 113000 \, \text{lb}
\]

### Final Answer:
The load that the second beam would support is approximately **113,000 lb**.

Would you like further details or have any questions?

---

Here are five related questions to extend your understanding:
1. How would the load change if the depth of the second beam were halved?
2. If the length of the second beam were increased to 18 feet, what would the new load capacity be?
3. How does the safe load vary if both the width and depth are doubled, but the length remains the same?
4. What is the effect on the load capacity if only the length of the beam changes, with the other variables held constant?
5. How would the safe load change if a beam had the same dimensions but was made of a different material with a different constant \(k\)?

**Tip**: When solving joint variation problems, it's crucial to convert all units to be consistent (e.g., inches for all measurements of length) to avoid errors.

The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, and the square of the depth, d, and inversely as the length, l. A wooden beam 7 inches wide, 3 inches deep, and 9 feet long holds up 49,419 pounds. What load would a beam 3 inches wide, 8 inches deep, and 12 feet long, of the same material, support?

Discover how to calculate the safe load of a wooden beam that varies jointly with width and the square of depth, and inversely with length. This problem walks through a detailed step-by-step solution for beams of varying dimensions, suitable for students in Grades 10-12.

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