Problem 37 - College Physics

Question

In each of the cases that follow, the magnitude of a vector is given along with the counterclockwise angle it makes with the \(+x\) axis. Use trigonometry to find the \(x\) and \(y\) components of the vector. Also, sketch each vector approximately to scale to see if your calculated answers seem reasonable. (a) \(50.0 \mathrm{~N}\) at \(60.0^{\circ},\) (b) \(75 \mathrm{~m} / \mathrm{s}\) at \(5 \pi / 6 \mathrm{rad},\) (c) \(254 \mathrm{lb}\) at \(325^{\circ},\) (d) \(69 \mathrm{~km}\) at \(1.1 \pi\) rad.

Step-by-Step Solution

Answer

The components for each vector are: (a) 25 N, 43.3 N; (b) -64.95 m/s, 37.5 m/s; (c) 208.9 lb, -134.6 lb; (d) -67.6 km, -21.4 km.

Step 1: Understanding the Problem

We have four vectors described by their magnitudes and angles with the positive x-axis. The task is to find the x and y components of each. We'll use trigonometry: the x-component is given by \( x = |v| \cos \theta \) and the y-component by \( y = |v| \sin \theta \).

Step 2: Calculate Components for (a)

Given: Magnitude \( |v| = 50.0 \) N and angle \( \theta = 60^{\circ} \).Convert angle to radians: \( \theta = \left( \frac{\pi}{180} \right) 60 \).Thus, \( \theta \approx 1.047 \) radians.\[ x = 50.0 \cos(1.047) \approx 25 \text{ N} \]\[ y = 50.0 \sin(1.047) \approx 43.3 \text{ N} \].

Step 3: Calculate Components for (b)

Given: Magnitude \( |v| = 75 \) m/s and angle \( \theta = \frac{5\pi}{6} \) rad.\[ x = 75 \cos\left(\frac{5\pi}{6}\right) \approx -64.95 \text{ m/s} \]\[ y = 75 \sin\left(\frac{5\pi}{6}\right) \approx 37.5 \text{ m/s} \].

Step 4: Calculate Components for (c)

Given: Magnitude \( |v| = 254 \) lb and angle \( \theta = 325^{\circ} \).Convert angle to radians: \( \theta = \left( \frac{\pi}{180} \right) 325 \).Thus, \( \theta \approx 5.672 \) radians.\[ x = 254 \cos(5.672) \approx 208.9 \text{ lb} \]\[ y = 254 \sin(5.672) \approx -134.6 \text{ lb} \].

Step 5: Calculate Components for (d)

Given: Magnitude \( |v| = 69 \) km and angle \( \theta = 1.1 \pi \) rad.\[ x = 69 \cos(1.1\pi) = -69 \cos(0.1\pi) \approx -67.6 \text{ km} \]\[ y = 69 \sin(1.1\pi) = -69 \sin(0.1\pi) \approx -21.4 \text{ km} \].

Step 6: Sketch and Verify

Create a rough sketch for each vector based on the calculated x and y components, making sure the vectors point in the expected direction and have the approximate expected length. For example, vector (a) should point more in the positive y-direction, (b) should point mostly in the negative x-direction and slightly in the positive y-direction, (c) in the positive x and negative y direction, and (d) mostly in the negative x and slightly in the negative y direction.

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. It is crucial in understanding vector components because it allows us to decompose vectors into their horizontal (x-axis) and vertical (y-axis) components.

To find these vector components, we use two primary trigonometric functions:

• Cosine: This function helps determine the x-component of a vector. Given a vector with magnitude \( |v| \) and angle \( \theta \,\) the x-component is calculated as \( x = |v| \cos \theta \).
• Sine: This function determines the y-component. The y-component is found by \( y = |v| \sin \theta \).

By using these trigonometric relationships, we can translate the magnitude and direction of the vector into usable components for various applications. This understanding is foundational for solving physics problems involving forces, velocities, and other vector quantities.

Vector Magnitude

The magnitude of a vector is a measure of its length or size and is often represented by the absolute value symbols \( |v| \). In physical terms, this could be the force in newtons, displacement in meters, speed in meters per second, or any other measurable quantity.

Magnitudes are always positive, as they represent the size of something rather than its direction.

When working with vectors:

• Keep in mind that magnitude describes how much – it could be the strength of a force or the total distance of a journey.
• Be cautious to distinguish between a vector's magnitude and its direction, which are two separate attributes of the vector.

In contexts like physics and engineering, understanding vector magnitudes alongside their direction forms the basis of analyzing many types of motion and forces.

Angle Conversion to Radians

Angles can be measured in degrees or radians, and each has its use in different contexts. Radians are often used in mathematics and physics because they offer a direct way to calculate angular relationships.

A full circle is equal to \( 360^{\circ} \) or \( 2\pi \) radians. To convert degrees to radians, the formula is \( \theta_{\text{rad}} = \frac{\pi}{180} \times \theta_{\text{degrees}} \).

Here’s how you convert angles:

• For a degree measure, multiply the degree value by \( \frac{\pi}{180} \).
• The resulting value will be in radians, suitable for use in trigonometric calculations.

Using radians simplifies many mathematical expressions, making them more concise and directly related to the properties of circular motion and oscillations.

Vector Sketching

Sketching vectors is an essential step in visualizing how they operate in space. It helps ensure your calculations align with reality by providing a visual representation of the vectors and their components.

Here’s how to effectively sketch vectors:

• Firstly, draw an axis with an x and y reference.
• Plot your vector according to its calculated x and y components.
• Ensure the arrow's length and direction match the vector's magnitude and overall direction.
• Draw and label the angle the vector makes with the x-axis.

This sketch serves as a verification step and can quickly show if a vector is pointing in the right direction and seems reasonably scaled compared to other vectors in the same system. It’s a great strategy for checking your work and understanding vector relations.