Problem 15 - Calculus: Early Transcendentals

Question

A force of 15 pounds is applied to a wagon, directed at an angle of \(45^{\circ} .\) 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

Both vertical and horizontal components are approximately 10.61 pounds.

Step 1: Understanding the Problem

We have a force of 15 pounds applied at an angle of \( 45^{\circ} \). We need to find the force components in the vertical (up) direction and the horizontal direction.

Step 2: Identify Components of the Force

The force can be split into two components: one vertical (\( F_v \)) and one horizontal (\( F_h \)). These components are perpendicular to each other, forming a right triangle with the hypotenuse as the original force.

Step 3: Calculate Vertical Component

The vertical component of the force can be found using sine: \[ F_v = F \cdot \sin(\theta) \] where \( F = 15 \) pounds and \( \theta = 45^{\circ} \). Therefore, \[ F_v = 15 \cdot \sin(45^{\circ}) = 15 \cdot \frac{\sqrt{2}}{2} \approx 10.61 \] pounds.

Step 4: Calculate Horizontal Component

The horizontal component of the force can be found using cosine: \[ F_h = F \cdot \cos(\theta) \] where \( F = 15 \) pounds and \( \theta = 45^{\circ} \). Therefore, \[ F_h = 15 \cdot \cos(45^{\circ}) = 15 \cdot \frac{\sqrt{2}}{2} \approx 10.61 \] pounds.

Step 5: Conclude the Result

Since both components involve a \( 45^{\circ} \) angle, they are of equal magnitude. Both the vertical and horizontal components of the force are approximately 10.61 pounds.

Key Concepts

Trigonometry in Physics

Trigonometry is an essential tool in physics. It helps us understand how forces act in different directions. Whenever a force acts at an angle, trigonometry allows us to split this force into two parts, or components. This is particularly useful because these components can be analyzed separately in physics problems.

When a force is applied at a specific angle, using trigonometric functions like sine and cosine helps in determining the exact influence of that force in different directions. For instance:

• Sine (\(\sin(\theta)\)) is used to find the component of the force that acts vertically.
• Cosine (\(\cos(\theta)\)) is used to determine the component acting horizontally.

These mathematical functions are vital in translating between the force's angle and its effect in single directions. Understanding how to use these functions in physics makes it easier to tackle various real-world problems such as the motion of objects, tension, and equilibrium.

Vector Decomposition

Vector decomposition is the process of breaking down a vector into simpler parts. In force-related problems, the original force is the vector, which we often decompose into horizontal and vertical parts. This technique makes it easier to analyze and solve physical problems, including those involving motion and balance.

To decompose a vector, we use its magnitude and direction. The direction is typically given by an angle, as in the case of our example where a 15-pound force acts at a 45-degree angle. Knowing this angle, we can use trigonometric functions to find:

• The vertical component (\(F_v\)
• The horizontal component (\(F_h\)

By doing this, we transform a single force vector into two orthogonal (right-angled) components. These components are much easier to work with because they align with the axes of a coordinate system, allowing us to apply laws of motion and other physical principles directly to them.

Right Triangle Components

When we decompose a force into components, we often visualize a right triangle. In physics, the original force is usually the hypotenuse, and the components are the other two sides. This triangle helps us see how each component relates to the original force.

For our example with a 15-pound force acting at a 45-degree angle, the process involves understanding:

• The hypotenuse represents the total force applied (15 pounds).
• The vertical side represents the upward force component (\(F_v\): 15 \on(45^\circ)).
• The horizontal side represents the force component along the ground (\(F_h\): 15 \oc(45^\circ)).

The properties of the right triangle make it clear that, at a 45-degree angle, both components are equal in size. This balance is unique to the 45-degree angle in a right triangle, notably simplifying calculations. Recognizing the setup as a right triangle helps you apply trigonometry with confidence, making the problem more approachable.