Question

Use a variation model to solve for the unknown value. Round your answer to the nearest whole number.
The strength of a wooden beam varies jointly as the width of the beam and the square of the thickness of the beam, and inversely as the length of the beam. A beam that is 72 in . long, 6 in . wide, and 2 in. thick can support a load of 405 lb.
Find the maximum load that can be safely supported by a board that is 12 in . wide, 144 in . long, and 4 in . thick.?

Answer 1

To solve the problem, we will use the formula for variation based on the information given. The strength \( S \) of a wooden beam varies jointly as the width \( w \) of the beam and the square of the thickness \( t \) of the beam, and inversely as the length \( l \) of the beam. This can be expressed mathematically as:

\[
S = k \frac{w t^2}{l}
\]

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

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

We know from the problem that a beam with the following specifications can support a load of 405 lb:
- Length \( l = 72 \) in
- Width \( w = 6 \) in
- Thickness \( t = 2 \) in

Substituting these values into the equation, we have:

\[
405 = k \frac{6 \cdot 2^2}{72}
\]

Calculating \( 2^2 \):

\[
405 = k \frac{6 \cdot 4}{72}
\]

This simplifies to:

\[
405 = k \frac{24}{72}
\]

Now, simplifying \( \frac{24}{72} = \frac{1}{3} \):

\[
405 = k \cdot \frac{1}{3}
\]

To find \( k \), multiply both sides by 3:

\[
k = 405 \cdot 3 = 1215
\]

**Step 2: Find the maximum load for the new beam**

Now we want to find the maximum load \( S \) that can be supported by a beam with the following specifications:
- Length \( l = 144 \) in
- Width \( w = 12 \) in
- Thickness \( t = 4 \) in

Using the formula for strength using the found value of \( k \):

\[
S = 1215 \frac{12 \cdot 4^2}{144}
\]

Calculating \( 4^2 = 16 \):

\[
S = 1215 \frac{12 \cdot 16}{144}
\]

Calculating \( 12 \cdot 16 = 192 \):

\[
S = 1215 \frac{192}{144}
\]

Now, simplifying \( \frac{192}{144} \):

\[
\frac{192}{144} = \frac{4}{3}
\]

So we have:

\[
S = 1215 \cdot \frac{4}{3}
\]

Calculating \( 1215 \cdot \frac{4}{3} \):

\[
S = 1215 \cdot 1.3333 \approx 1620
\]

Thus, rounding our answer to the nearest whole number, the maximum load that can be safely supported by the new board is

\[
\boxed{1620} \text{ lb.}
\]