Problem 76 - Calculus with Concepts in Calculus

Question

A weight \(W\) is suspended from a ceiling by three strings of length \(L\) so that the points of attachment form an equilateral triangle of side \(S\) (Figure 4.30). If the weights of the various strings are negligible, then the tension \(T\) in the three strings attached to the ceiling is given by $$ T=\frac{W}{3 \sqrt{1-S^{2} / 3 L^{2}}} $$ a. Show that for fixed values of \(S\) and \(L, T\) is an increasing function of \(W\). b. Show that for fixed values of \(W\) and \(L, T\) is an increasing function of \(S\). c. Show that for fixed values of \(S\) and \(W, T\) is a decreasing function of \(L\).

Step-by-Step Solution

Answer

T increases with W and S, and decreases with L.

Step 1: Understand the Given Formula

The given problem involves the tension \( T \) in the strings holding a weight \( W \). The formula provided is:\[ T = \frac{W}{3 \sqrt{1-\frac{S^{2}}{3 L^{2}}}} \]Here, \( S \) is the side of the equilateral triangle formed by the attachment points, and \( L \) is the length of each string.

Step 2: Show T Increases With W

To show \( T \) is an increasing function of \( W \) when \( S \) and \( L \) are fixed, differentiate \( T \) with respect to \( W \):\[\frac{dT}{dW} = \frac{1}{3 \sqrt{1-\frac{S^{2}}{3 L^{2}}}}\]This derivative is positive, indicating that \( T \) increases as \( W \) increases.

Step 3: Show T Increases With S

To demonstrate \( T \) increases with \( S \) for fixed \( W \) and \( L \):Consider the expression inside the square root, \( 1 - \frac{S^{2}}{3L^{2}} \). As \( S \) increases, \( \frac{S^{2}}{3L^{2}} \) increases, making \( \sqrt{1-\frac{S^{2}}{3L^{2}}} \) smaller. Consequently, \( T = \frac{W}{3 \sqrt{1-\frac{S^{2}}{3 L^{2}}}} \) increases since the denominator decreases.

Step 4: Show T Decreases With L

For showing \( T \) decreases with \( L \) when \( S \) and \( W \) are constant:In the term \( \frac{S^{2}}{3L^{2}} \), increasing \( L \) decreases \( \frac{S^{2}}{3L^{2}} \), making \( \sqrt{1-\frac{S^{2}}{3L^{2}}} \) larger. Thus, \( T = \frac{W}{3 \sqrt{1-\frac{S^{2}}{3 L^{2}}}} \) decreases, as the denominator increases.

Key Concepts

Differentiation

Differentiation is a fundamental tool in calculus that helps us understand how a function changes with respect to one of its variables. It's like peeling back the layers of an onion to see how different changes affect the outcome.
In the context of the exercise, differentiation allows us to see how the tension \( T \) varies as the weight \( W \), side length \( S \), and string length \( L \) change. To do this, you calculate the derivative, which shows the rate of change.

• When we differentiated \( T \) with respect to \( W \), we found that the derivative \( \frac{dT}{dW} \) is positive, meaning tension increases as the weight increases.
• Similarly, changes in \( S \) affect \( T \) through the term \( \sqrt{1 - \frac{S^{2}}{3L^{2}}} \). As \( S \) grows, this expression decreases, thereby increasing \( T \).
• The influence of \( L \) is opposite. Increasing \( L \) increases the value under the square root, leading to a decrease in tension.

Differentiation helps us predict these behaviors accurately, making it an essential skill.

Functions

Functions are the backbone of expressing relationships between variables in mathematics. They help transform complex physical relationships into equations that we can analyze and understand.
In this exercise, the formula for tension \( T \) is a function that depends on several variables: the weight \( W \), the side of the triangle \( S \), and the length of the string \( L \). This function reflects how changes in these parameters affect the tension in each string.

• The input to our function includes \( W, S, \) and \( L \), while the output is the calculated tension.
• By controlling for two variables, we can analyze the impact of the third. This process forms the basis for each part of the problem.
• This function is a great example because it brings physical intuition about tension and balance into a precise mathematical format.

Understanding functions is essential for solving real-life problems involving multiple changing quantities.

Equilateral Triangle

An equilateral triangle is a geometric shape where all three sides are of equal length, and its angles are each \( 60^{\circ} \). It is a symmetrical form that appears frequently in both geometry and physical applications.
In this exercise, the points of attachment of the strings form an equilateral triangle. Each side of this triangle is denoted by \( S \). This setup ensures uniform symmetry, making it an useful scenario for calculating tension.

• The equal side lengths simplify calculations by ensuring that the tension in each string is initially assumed to be equal, given the weight is directly beneath the center.
• The symmetry of an equilateral triangle contributes to even distribution of forces, which is why the setup is stable under weight.
• This characteristic is pivotal when exploring the role of changing \( S \) in the tension formula because it provides the base geometry.

Recognizing the properties of an equilateral triangle offers crucial insights into how and why tension changes affect each side evenly.