Problem 34 - University Physics with Modern Physics

Question

Use a scale drawing to find the \(x\) - and \(y\) -components of the following vectors. For each vector the numbers given are the magnitude of the vector and the angle, measured in the sense from the \(+x\) -axis toward the \(+y\) -axis, that it makes with the \(+x\) -axis: (a) magnitude \(9.30 \mathrm{m},\) angle \(60.0^{\circ} ;\)(b) magnitude \(22.0 \mathrm{km},\) angle \(135^{\circ} ;\) (c) magnitude \(6.35 \mathrm{cm},\) angle \(307^{\circ} .\)

Step-by-Step Solution

Answer

(a) x = 4.65 m, y = 8.06 m; (b) x = -15.56 km, y = 15.56 km; (c) x = 3.81 cm, y = -5.07 cm.

Step 1: Understanding the Problem

We have three vectors with given magnitudes and angles. Our task is to find the components along the x-axis and y-axis using a scale drawing and understanding of trigonometry.

Step 2: Recall the Formulas

The x-component of a vector can be found using the equation: \( x = r \cos \theta \), and the y-component can be found using: \( y = r \sin \theta \), where \( r \) is the magnitude of the vector and \( \theta \) is the angle from the positive x-axis.

Step 3: Calculate Components for Vector (a)

For vector (a) with magnitude \( 9.30 \mathrm{m} \) and angle \( 60.0^{\circ} \):- The x-component is \( x = 9.30 \cos 60.0^{\circ} = 9.30 \times 0.5 = 4.65 \, \mathrm{m} \).- The y-component is \( y = 9.30 \sin 60.0^{\circ} = 9.30 \times \frac{\sqrt{3}}{2} \approx 8.06 \, \mathrm{m} \).

Step 4: Calculate Components for Vector (b)

For vector (b) with magnitude \( 22.0 \, \mathrm{km} \) and angle \( 135^{\circ} \):- The x-component is \( x = 22.0 \cos 135^{\circ} = 22.0 \times (-\frac{\sqrt{2}}{2}) \approx -15.56 \, \mathrm{km} \).- The y-component is \( y = 22.0 \sin 135^{\circ} = 22.0 \times \frac{\sqrt{2}}{2} \approx 15.56 \, \mathrm{km} \).

Step 5: Calculate Components for Vector (c)

For vector (c) with magnitude \( 6.35 \, \mathrm{cm} \) and angle \( 307^{\circ} \):- The x-component is \( x = 6.35 \cos 307^{\circ} = 6.35 \cos (360^{\circ} - 53^{\circ}) = 6.35 \cos 53^{\circ} = 6.35 \times 0.6 \approx 3.81 \, \mathrm{cm} \).- The y-component is \( y = 6.35 \sin 307^{\circ} = 6.35 \sin (360^{\circ} - 53^{\circ}) = 6.35 (-\sin 53^{\circ}) \approx -5.07 \, \mathrm{cm} \).

Key Concepts

Trigonometry Basics in Vector Decomposition

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. When we work with vectors, which are quantities having both magnitude and direction, trigonometry becomes essential for breaking them down into their components. A key aspect of this method is to think of a vector as the hypotenuse of a right triangle, where we need to find the lengths of the sides (the components along the x and y axes).
To express a vector in terms of its components, we employ basic trigonometric functions:

• The cosine function for the x-component: \(x = r \cos \theta\), where \( r \) is the magnitude and \( \theta \) is the angle with the positive x-axis.
• The sine function for the y-component: \(y = r \sin \theta\).

This understanding enables us to convert polar coordinates (magnitude and angle) into rectangular coordinates (x and y components), which are much easier to manage in physics and engineering problems.

Understanding Magnitude and Direction

When dealing with vectors, two primary aspects define them: magnitude and direction. The magnitude of a vector represents how strong or long the vector is. This could be a force, velocity, or any other vector quantity depending on the context.
The direction is equally significant, as it points from which angle the vector is acting. In our exercise, angles are measured from the positive x-axis, moving counterclockwise towards the positive y-axis. It’s crucial to properly understand this as angles can affect the sign of vector components. For instance:

• A vector pointing directly north (90°) would have its entire magnitude in the y-component.
• A vector pointing e directly east (0°) would have its entire magnitude in the x-component.
• For vectors directed in the second quadrant (between 90° and 180°), the x component becomes negative, as seen in our vector (b) example, where the angle is 135°.

This concept helps map directional influence of any physical quantity more precisely.

Performing Vector Calculations

In vector analysis, finding the x and y components using vector calculations is a routine task, requiring the application of trigonometric functions. Here’s how you can carry out such calculations, as seen in the exercise.
Consider a vector defined by its magnitude and direction. First, identify or measure the angle it makes with the positive x-axis. Then employ the trigonometric formulas:

• Calculate the x-component using \(x = r \cos \theta\). This will tell you how much of the vector is pointing in the horizontal direction.
• Calculate the y-component using \(y = r \sin \theta\). This represents the vector’s influence in the vertical direction.

Applying these calculations gives insight into how different vectors combine and interact with each other in complex systems. Practice using these vector calculations with a variety of magnitudes and angles to master the art of decomposition.