Problem 41 - Precalculus : Building Concepts and Connections

Question

In this set of exercises, you will use vectors and dot products to study real- world problems. \- Work A parent pulling a wagon in which her child is riding along level ground exerts a force of 20 pounds on the handle. The handle makes an angle of \(45^{\circ}\) with the horizontal. How much work is done in pulling the wagon 100 feet, to the nearest foot-pound?

Step-by-Step Solution

Answer

The work done in pulling the wagon 100 feet is approximately \(1000 \sqrt{2}\) foot-pounds.

Step 1: Identify given values

The force \(F\) exerted on the wagon handle is 20 pounds, the angle \(\Theta\) between the force and the horizontal is \(45^{\circ}\) and the distance \(d\) the wagon is pulled is 100 feet.

Step 2: Convert the angle from degrees to radians

As trigonometric functions in the formula for work are usually operated in radians, we need to convert \(\Theta\) from degrees to radians. \(\Theta = 45^{\circ} = \frac{\pi}{4}\) radians.

Step 3: Calculate work done

The formula for work done \(W\) is the product of force, distance and the cosine of the angle between them. So, \(W = F \cdot d \cdot cos(\Theta)\). Substituting the given values, \(W = 20 \cdot 100 \cdot cos(\frac{\pi}{4}) = 2000 \cdot \frac{\sqrt{2}}{2} = 1000 \sqrt{2}\) foot-pounds.

Key Concepts

Dot Product

The dot product is an essential concept in vector calculus, especially when dealing with vectors such as forces or directions in physics. The dot product of two vectors gives a scalar quantity, not another vector. This scalar is calculated by multiplying corresponding components of the vectors and summing those products. If vectors \(\mathbf{A}\) and \(\mathbf{B}\) are represented as \(\langle A_x, A_y \rangle\) and \(\langle B_x, B_y \rangle\), respectively, then their dot product is calculated as:

• \(\mathbf{A} \cdot \mathbf{B} = A_x \cdot B_x + A_y \cdot B_y\)

The dot product also can be expressed in terms of the angle \(\Theta\) between the two vectors:

• \(\mathbf{A} \cdot \mathbf{B} = \|\mathbf{A}\| \cdot \|\mathbf{B}\| \cdot \cos(\Theta)\)

This formula is particularly useful when calculating work, as it links force, distance, and the angle between them. Understanding the dot product's role helps in analyzing how different force components contribute to the work done.

Work Done

Work is a fundamental concept in physics, and its calculation often involves vectors. When you calculate work done, you are essentially measuring the amount of energy transferred by a force moving an object a certain distance. For work done involving vectors, the formula is as follows:

• \(W = F \cdot d \cdot \cos(\Theta)\)

Here, \(W\) is the work done, \(F\) is the force applied, \(d\) is the distance over which the force is applied, and \(\Theta\) is the angle between the force vector and the direction of movement.

The function \(\cos(\Theta)\) is crucial because it measures how much of the force is acting in the direction of the movement. If the force is applied directly in the line of movement, the full force contributes to the work done. However, if the force is at an angle, only the component of the force in the direction of the movement does work.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are foundational in math and applicable in vector calculus for understanding angles in relation to horizontal and vertical planes. For the calculation of work done in real-world problems, the function \(\cos(\Theta)\) is used, where \(\Theta\) is the angle of the force in relation to the horizontal direction.

It's useful to remember that the angle in trigonometric functions typically needs to be in radians for calculations. Converting degrees to radians:

• The formula for conversion: \(1^{\circ} = \frac{\pi}{180}\) radians.
• So, \(45^{\circ} = \frac{\pi}{4}\) radians.

In work calculations, the cosine value determines the effective component of force doing actual work. At \(45^{\circ}\), \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\), representing an adequate compromise between vertical and horizontal components of force.

Real-World Problems

Problem-solving with vector calculus in everyday scenarios often involves understanding how vectors represent real-world quantities like force and direction. The exercise of a parent pulling a wagon is a practical example. In this scenario:

• The force exerted (20 pounds) acting on the handle is one vector.
• The movement of the wagon (100 feet) is another direction vector.
• The angle of application (\(45^{\circ}\)) affects how efficiently the work is done.

By considering the angle, we incorporate real-world variables such as the height or efficiency of pulling, which vary depending on how the handle is held. This example demonstrates how theoretical concepts like dot products and trigonometric conversions solve practical issues, showing you how such calculations are woven into everyday tasks, making physics less abstract and more relevant.