Problem 100 - A Graphical Approach to College Algebra

Question

Solve each problem. See Example 9. The strength \(S\) of a rectangular beam varies directly with its width \(W\) and the square of its thickness \(T,\) and inversely with its length \(L\). A beam that is 2 inches wide, 6 inches thick, and 96 inches long can support a load of 375 pounds. Determine how much a similar beam that is 3.5 inches wide, 8 inches thick, and 128 inches long can support.

Step-by-Step Solution

Answer

The new beam can support 875 pounds.

Step 1: Write the formula for the beam strength

The strength of the beam, \( S \), can be expressed as \( S = k \cdot \frac{W \cdot T^2}{L} \), where \( k \) is the proportionality constant, \( W \) is the width, \( T \) is the thickness, and \( L \) is the length.

Step 2: Calculate the constant of proportionality

Use the given dimensions and load of the first beam to find \( k \). Substituting \( W = 2 \), \( T = 6 \), \( L = 96 \), and \( S = 375 \), we get: \\( 375 = k \cdot \frac{2 \cdot 6^2}{96} \). \Simplify this to find \( k \): \\( 375 = k \cdot \frac{72}{96} \) or \( 375 = k \cdot 0.75 \). \Solving for \( k \) gives \( k = \frac{375}{0.75} = 500 \).

Step 3: Calculate the strength for the new beam

Use the found \( k \) value to calculate the strength of the new beam with different dimensions: \( W = 3.5 \), \( T = 8 \), \( L = 128 \). \Using the formula, \( S = 500 \cdot \frac{3.5 \cdot 8^2}{128} \). \Plug in the numbers: \\( S = 500 \cdot \frac{3.5 \cdot 64}{128} = 500 \cdot \frac{224}{128} \). \Simplify the expression: \\( S = 500 \cdot 1.75 = 875 \).

Key Concepts

Proportionality Constant

To solve problems involving direct and inverse variations, determining the proportionality constant is essential. The proportionality constant \( k \) is a unique value that relates the variables involved in the relationship. In our beam problem, the strength \( S \) varies directly with the width \( W \) and the square of the thickness \( T^2 \), and inversely with the length \( L \). This means that mathematically, the relation can be expressed as: \[ S = k \cdot \frac{W \cdot T^2}{L} \] Here, \( k \) serves as a bridge between the physical measurements and the strength they ultimately determine. To find \( k \), we use known values. By inserting the dimensions of the initial beam and its strength, we solve the equation to find:

• Width \( W = 2 \)
• Thickness \( T = 6 \)
• Length \( L = 96 \)
• Strength \( S = 375 \)

Substituting these values, we calculate \( k = 500 \). Once established, this constant allows us to calculate the strength of other similar beams by substituting their respective dimensions.

Rectangular Beam Strength

Rectangular beam strength is influenced by geometric and material properties that dictate how it can withstand loads. Strength depends critically on three factors:

• The **width** \( W \) of the beam, contributing linearly to strength.
• The **thickness** \( T \), which enhances strength exponentially as \( T^2 \).
• The **length** \( L \), which inversely affects strength: longer beams are generally weaker.

Combining these variables, the equation \( S = k \cdot \frac{W \cdot T^2}{L} \) provides a concise expression for beam strength. In our example, the first beam supported 375 pounds, and by inputting its dimensions, we factored in all the affecting elements. For the new beam, with its larger width and thickness but increased length, each metric modifies the overall capacity. The adjustments in width and thickness enhance this new beam's capacity, which allows it to support an increased load of 875 pounds despite its longer length.

Algebraic Formulas

Algebraic formulas are at the heart of solving variation problems, enabling us to systematically understand complex relationships involving multiple variables. In this context, the general formula for beam strength \( S = k \cdot \frac{W \cdot T^2}{L} \) allows us to bridge concrete physical dimensions with abstract mathematical reasoning. Converting the physical problem into an algebraic form offers several advantages:

• Clarity in math allows easy manipulation to solve for variables like \( k \) or \( S \).
• It provides a repeatable method applicable to various scenarios by simply switching input values.
• Facilitates predictions for similar problems, enhancing understanding and saving time.

By isolating and solving for the unknowns, students can apply these algebraic principles to other similar structures and materials, thus deepening their grasp of direct and inverse relationships.