Problem 14 - Calculus: Early Transcendentals

Question

A force of 10 pounds is applied to a wagon, directed at an angle of \(30^{\circ} .\) Find the component of this force pulling the wagon straight up, and the component pulling it horizontally along the ground.

Step-by-Step Solution

Answer

Vertical component is 5 pounds; horizontal component is approximately 8.66 pounds.

Step 1: Component Definitions

In this exercise, we need to find two components of the force: the vertical component (which pulls upwards) and the horizontal component (which pulls along the ground).

Step 2: Convert Angle to Radians

Since trigonometric calculations are generally done in radians, we first convert the angle from degrees to radians. We use the formula: \[ \text{Radians} = \text{Degrees} \times \frac{\pi}{180} \]For \( 30^{\circ} \), this gives us:\[ 30 \times \frac{\pi}{180} = \frac{\pi}{6} \text{ radians} \]

Step 3: Calculate Vertical Component

Use the sine function to find the vertical component of the force. For a force \( F \) at angle \( \theta \), the vertical component \( F_v \) is given by:\[ F_v = F \cdot \sin(\theta) \]Substituting the given values:\[ F_v = 10 \cdot \sin\left(\frac{\pi}{6}\right) = 10 \cdot \frac{1}{2} = 5 \text{ pounds} \]

Step 4: Calculate Horizontal Component

Use the cosine function to find the horizontal component of the force. For a force \( F \) at angle \( \theta \), the horizontal component \( F_h \) is given by:\[ F_h = F \cdot \cos(\theta) \]Substituting the given values:\[ F_h = 10 \cdot \cos\left(\frac{\pi}{6}\right) = 10 \cdot \frac{\sqrt{3}}{2} \approx 8.66 \text{ pounds} \]

Key Concepts

Trigonometric Functions

Trigonometric functions are essential in understanding how angles and sides relate in right triangles.
In physics and engineering, they help break forces into perpendicular components that are much easier to work with.
There are two primary trigonometric functions used in vector decomposition: sine and cosine.

• The sine function is calculated as the ratio of the opposite side to the hypotenuse in a right triangle. It is used to determine the vertical component of a force.
• The cosine function is calculated as the ratio of the adjacent side to the hypotenuse. It is used for finding the horizontal component.

When you have a force acting at an angle, by using these functions, you can resolve it into components:

• The vertical component uses the sine of the angle: \( F_v = F \cdot \sin(\theta) \)
• The horizontal component uses the cosine: \( F_h = F \cdot \cos(\theta) \)

Simple, straightforward application of these functions allows for efficient solution of complex vector problems.

Angle Conversion to Radians

Angles can be measured in different units, among which degrees and radians are the most common.
When working with trigonometric functions, especially in mathematical and scientific contexts, radians are usually preferred.
This is because they relate more naturally to other mathematical concepts.

• A radian is based on the radius of a circle. One full revolution of a circle is \( 2\pi \) radians, which equals 360 degrees.
• To convert an angle in degrees to radians, use the formula: \[ \text{Radians} = \text{Degrees} \times \frac{\pi}{180} \]

In our example, converting 30 degrees to radians involves:

• \( 30 \times \frac{\pi}{180} = \frac{\pi}{6} \) radians.

Understanding how to convert angles to radians ensures accurate calculations in trigonometry.

Vertical and Horizontal Components

Once the angle conversion to radians is completed, we apply trigonometric functions to find the vector components.
This process is a crucial aspect of vector decomposition which involves breaking down a vector into its perpendicular parts.

• For the vertical component, we use the sine function. It describes how much of the force acts in the upward direction:
  \( F_v = F \cdot \sin(\theta) \). In our example: \( 10 \times \sin(\frac{\pi}{6}) = 5 \) pounds.
• For the horizontal component, the cosine function tells us how much of the force acts along the ground or horizontal direction:
  \( F_h = F \cdot \cos(\theta) \). In our example: \( 10 \times \cos(\frac{\pi}{6}) \approx 8.66 \) pounds.

This method is vital in physics to resolve forces or velocities acting in a non-linear direction into simpler, one-dimensional problems.