Question

The maximum safe load for a horizontal rectangular beam varies directly with the width of the beam and the square of the thickness of the beam and inversely with its length. If an 8 -foot beam will support up to 750 pounds when the beam is 4 inches wide and 2 inches thick, what is the maximum safe load in a similar beam 10 feet long, 6 inches wide, and 2 inches thick?

Short answer

The maximum safe load for the new beam is 900 pounds.

Step-by-step solution

- Understand the Relationship

The maximum safe load (L) varies directly with the width (w) and the square of the thickness (t) and inversely with the length (l). This relationship can be expressed as: \[ L = k \frac{w t^2}{l} \]where k is a constant of proportionality.

- Determine the Constant of Proportionality (k)

Given that for an 8-foot beam, 4 inches wide and 2 inches thick, the load is 750 pounds, we can substitute these values into the equation to find k:\[ 750 = k \frac{4 \times 2^2}{8} \]Solving this:\[ 750 = k \frac{4 \times 4}{8} \]\[ 750 = k \frac{16}{8} \]\[ 750 = 2k \]\[ k = 375 \]

- Find the Maximum Safe Load for New Beam

Now, substitute the values for the new beam (10 feet long, 6 inches wide, 2 inches thick) into the equation along with k = 375:\[ L = 375 \frac{6 \times 2^2}{10} \]\[ L = 375 \frac{6 \times 4}{10} \]\[ L = 375 \frac{24}{10} \]\[ L = 375 \times 2.4 \]\[ L = 900 \]

Key concepts

Proportionality Constant

In mathematics and science, a proportionality constant (k) helps us understand how different variables in an equation are related. In our example, the formula \( L = k \frac{w t^2}{l} \) shows that the maximum safe load (L) is directly proportional to the width (w) and the square of the thickness (t), but inversely proportional to its length (l). This common proportionality equation helps in different fields such as physics and engineering. To find k, we used given values for a specific scenario: a beam 8 feet long, 4 inches wide, and 2 inches thick supporting 750 pounds. By substituting these known values into our equation, we solved for the proportionality constant:

• \( 750 = k \frac{4 \times 2^2}{8} \)
• \( 750 = k \frac{16}{8} \)
• \( 750 = 2k \)
• \( k = 375 \)

By understanding and finding this constant, we can predict the behavior of similar systems under different conditions. This is crucial for tasks like beam load calculation where safety and accuracy are paramount.

Beam Load Calculation

Calculating the load a beam can safely support involves understanding direct and inverse variations. Using our formula \( L = k \frac{w t^2}{l} \), where:

• \( L \) is the maximum safe load
• \( w \) is the width
• \( t \) is the thickness
• \( l \) is the length
• \( k \) is the proportionality constant

We initially calculated k for a given beam and then used the same constant to find the load capacity for a new beam. For the example: Given an 8-foot long, 4-inch wide, and 2-inch thick beam that supports 750 pounds, we first found k to be 375.Next, we used this k value to calculate the maximum safe load for a 10-foot, 6-inch wide, and 2-inch thick beam:

• \( L = 375 \frac{6 \times 2^2}{10} \)
• \( L = 375 \frac{24}{10} \)
• \( L = 900 \)

Following these steps allows us to determine that the new beam can safely support up to 900 pounds. This method is reliable, ensuring beams are used within their safe loading limits.

Variation Equations

Variation equations help describe how one quantity changes in relation to another. In direct variation, when one variable increases, the other does too. For inverse variation, when one variable increases, the other decreases. In our example, the formula \( L = k \frac{w t^2}{l} \) shows mixed variation, combining both direct and inverse variations.

• \( L \propto w t^2 \) (Direct Variation)
• \( L \propto \frac{1}{l} \) (Inverse Variation)

Understanding these equations helps in many areas such as physics, engineering, and economics. They clarify how and why certain changes occur. In our beam problem, the maximum safe load depends on three factors at once: width, thickness, and length, highlighting the usefulness of variation equations. By mastering these concepts, students can solve similar problems more easily and apply these methods in complex real-world scenarios.