Problem 68 - Precalculus with Limits

Question

Pulley System 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 found by solving the system of equations $$\left\\{ \begin{aligned} t _ { 1 } - 2 t _ { 2 } & = 0 \\ t _ { 1 } & \- 2 a = 128 \\ t _ { 2 } + a & = 32 \end{aligned} \right.$$ where \(t _ { 1 }\) and \(t _ { 2 }\) are in pounds and \(a\) is in feet per second squared. Solve this system.

Step-by-Step Solution

Answer

The solution to the system is \( t _ { 1 } = 96 \) lbs, \( t _ { 2 } = 48 \) lbs, and \( a = -16 \) ft/s².

Step 1: Simplify the equations

The given system of equations is: \[\begin{aligned} t _ { 1 } - 2 t _ { 2 } & = 0 \ t _ { 1 } - 2 a & = 128 \ t _ { 2 } + a & = 32 \end{aligned} \]From the first equation, we can express \(t _ { 1 }\) in terms of \(t _ { 2 }\): \(t _ { 1 } = 2 t _ { 2 }\). This would be our simplified equation (1).

Step 2: Substitution

Substitute the expression for \(t _ { 1 }\) from equation (1) into the other two equations:For equation 2, \((2 t _ { 2 }) - 2 a = 128\), which simplifies to \(t _ { 2 } = a + 64\).For equation 3, \(t _ { 2 } + a = 32\), substituting for \(t _ { 2 }\) from what we got above, we get \(a + 64 + a = 32\), which gives us \(a = -16\) ft/s².

Step 3: Solve for \( t _ { 2 } \) and \( t _ { 1 }\)

Substitute \(a = -16\) into equation (2) to find \( t _ { 2 }\):\( t _ { 2 }= (-16) + 64 = 48\) lbs.Substitute \( t _ { 2 } = 48\) into equation (1) to find \( t _ { 1 }\):\( t _ { 1 } = 2 * 48 = 96 \) lbs.

Key Concepts

Pulley System

Pulley systems are an essential part of mechanical systems, used extensively to change the direction and magnitude of forces. They consist of wheels and ropes that help move weight with less effort. Understanding how they function can assist in solving complex physics problems, as they rely on principles of mechanical advantage.

• Each pulley in a system contributes to redirecting the applied force.
• The tension in the ropes is crucial, as it helps determine the system's equilibrium and acceleration.
• In problems, recognizing the relation between tensions and forces can simplify understanding pulley mechanics.

In our exercise, we are looking at two weights, a 128-pound and a 32-pound, connected through ropes over pulleys. The ultimate goal is to solve the tensions in the ropes and the acceleration of the lighter weight.

Substitution Method

The substitution method is a basic algebraic technique used to solve systems of equations. It involves replacing a variable from one equation to simplify the system, allowing us to solve for one variable at a time. This method is particularly useful when one variable can easily be expressed in terms of another.

• Identify an equation where a variable can be isolated easily.
• Substitute the expression of the isolated variable into the other equations to reduce complexity.
• Solve the simplified equation to find the value of one variable, then backtrack to find others.

In the given pulley problem, the substitution method allowed us to find expressions for the tensions and acceleration directly, leading to a solution that is easier to handle and interpret.

Physics Applications in Math

Physics problems often require mathematical techniques to solve complex environments like pulleys or objects in motion. By setting up equations based on physical laws, we can tap into math to find quantitative solutions.

• Equations of motion or force can lead to a system of linear equations.
• Pulley systems fundamentally rely on balance of forces tensile and gravitational.
• Using algebra, we convert real-world physics phenomena into solvable equations.

This exercise blends physics concepts like tension and acceleration with algebraic techniques, showcasing how real-world problems are often translated into mathematical models.

Algebraic Manipulation

Algebraic manipulation involves arranging, simplifying, and solving equations, which is essential for finding precise solutions to complex problems. It involves operations such as addition, subtraction, multiplication, and substitution.

• Recognizing equivalent expressions helps simplify and solve equations.
• Careful rearrangement of terms is crucial to find variable values.
• Simplifying equations by eliminating variables can unveil the relationships between different elements in a system.

In our problem, algebraic manipulation was used to break down the tension and acceleration equations into simpler components. Navigating through algebraic manipulations lets us find the exact values needed, such as pounds for tension and feet per second squared for acceleration.