Problem 96 - Algebra and Trigonometry Real Mathematics, Real People

Question

A system of pulleys is loaded with 128 -pound and 32 -pound weights (see figure). The tensions \(t_{1}\) and \(t_{2}\) in the ropes and the acceleration \(a\) of the 32 -pound weight are modeled by the following system, where \(t_{1}\) and \(t_{2}\) are measured in pounds and \(a\) is in feet per second squared. Solve the system. $$\left\\{\begin{aligned} t_{1}-2 t_{2} &=0 \\ t_{1} &-2 a=128 \\ t_{2}+& a=32 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution of the system is \( t_{1} = 0 \) pounds, \( t_{2} = 0 \) pounds, and \( a = 32 \) feet per second squared.

Step 1: Setting up the System

The 3 equations given are: \n\n1) \( t_{1} - 2t_{2} = 0 \)\n\n2) \( t_{1} - 2a = 128 \)\n\n3) \( t_{2} + a = 32 \)

Step 2: Solving the First Equation

Rearranging equation 1) gives \( t_{1} = 2t_{2} \).

Step 3: Substitution

Substitute \( t_{1} = 2t_{2} \) from equation 1) into equation 2) and equation 3) \n\nFrom equation 2), this gives \( 2t_{2} - 2a = 128 \), which further simplifies to \( t_{2} - a = 64 \).\n\nSubstituting into the third equation then gives \( t_{2} = 32 - a \).

Step 4: Finding Values

We can set the above two equations equal to each other: \( t_{2} - a = 32 - a \). Solving for \( a \), we find \( a = 64 - 32 = 32 \) feet per second squared (ft/s²). \n\nSubstituting \( a = 32 \) back into the equation \( t_{2} = 32 - a \), we find \( t_{2} = 32 - 32 = 0 \) pounds. \n\nFinally, substituting \( t_{2} = 0 \) into equation 1), we find \( t_{1} = 2 * 0 = 0 \) pounds.

Key Concepts

Linear Algebra

Linear algebra is a fundamental area of mathematics that deals with vectors, vector spaces, linear mappings, and the systems of linear equations. It is the cornerstone of many fields such as engineering, physics, computer science, and data analysis. In the context of our exercise, linear algebra comes into play as we deal with a system of equations that need to be solved simultaneously to determine the tensions in the ropes, represented by the variables t_1 and t_2, and the acceleration, represented by a. Each equation in the system models a physical relationship between these variables, reflecting the forces and movements in a system of pulleys.

Using linear algebra techniques, such as matrix operations and vector spaces, we can systematically approach and solve these equations. However, in this particular problem, we use more straightforward methods by direct manipulation and substitution, which are often taught as introductory steps before delving into the more abstract concepts of linear algebra.

Simultaneous Equations

Simultaneous equations are a set of equations containing multiple variables that are solved together, at the same time, hence the term simultaneous. These equations are intertwined in such a way that the solution to one equation depends on the others. Our exercise presents three simultaneous equations, and our goal is to find values for t_1, t_2, and a that satisfy all three equations at once.

Why Solve Simultaneously?

Solving simultaneous equations gives us a fuller picture of the scenario being modeled. In practical applications, such as when calculating forces in mechanical systems or currents in electrical circuits, the interdependency of variables is crucial for an accurate analysis. By solving the simultaneous equations, we ensure that all physical constraints and relationships are respected. These equations are a typical application of linear algebra and often appear in real-world problem-solving scenarios.

Substitution Method

The substitution method is a technique used to solve systems of simultaneous equations. This method involves solving one of the equations for one variable in terms of the others and then substituting this expression into the remaining equations. This process reduces the number of equations and variables step by step until all variables are isolated and can be solved. In our exercise, after setting up the three equations, we solved the first equation for t_1, finding t_1 = 2t_2. This expression for t_1 was then substituted into the second and third equations to find the values of t_2 and a.

Advantages of the Substitution Method

• It is simple to understand and apply, especially for systems with two or three equations.
• It works well when one of the equations is easily solvable for one of the variables.
• It can be applied to any system of equations, whether or not the coefficients are integers.

By reducing the number of variables incrementally, the substitution method makes it easier to see the relationship between different parts of the equation and ultimately to find the solution for all variables within the system.