/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 The components of vector \(\over... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The components of vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}\) and \(A_{y}\) (both positive), and the angle that it makes with respect to the positive \(x\) axis is \(\theta\). (a) Does increasing the component \(A_{x}\) (while holding \(A_{y}\) constant) increase or decrease the angle \(\theta\) ? (b) Does increasing the component \(A_{v}\) (while holding \(A_{r}\) constant) increase or decrease the angle \theta? Account for your answers. Problem (a) The components of displacement vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}=12 \mathrm{~m}\) and \(A_{y}=12 \mathrm{~m}\) Find \(\theta\). (b) The components of displacement vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}=17 \mathrm{~m}\) and \(A_{y}=12 \mathrm{~m}\) Find \(\theta\). (c) The components of displacement vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}=12 \mathrm{~m}\) and \(A_{y}=17 \mathrm{~m}\) Find \(\theta\). Be sure that your answers are consistent with your answers to the Concept Questions.

Short Answer

Expert verified
(a) Increasing \(A_x\) decreases \(\theta\); \(\theta\) = 45° when \(A_x = 12\,m\). (b) Increasing \(A_y\) increases \(\theta\); \(\theta\) = 35.26° when \(A_x = 17\,m\); \(\theta\) = 54.74° when \(A_y = 17\,m\).

Step by step solution

01

Understanding the vector components

The vector \(\overrightarrow{\mathrm{A}}\) can be described using its horizontal and vertical components, \(A_x\) and \(A_y\), which corresponds to movement in the \(x\) and \(y\) directions respectively. The angle \(\theta\) is the angle between the vector and the positive \(x\)-axis. \(\theta\) can be calculated using the tangent function: \(\tan \theta = \frac{A_y}{A_x}\).
02

Solving for \(\theta\) when \(A_x = 12\, \mathrm{m}\) and \(A_y = 12\, \mathrm{m}\)

In this case, apply the formula \(\tan \theta = \frac{A_y}{A_x}\). Thus, \(\tan \theta = \frac{12}{12} = 1\). To find \(\theta\), we take the arctangent of both sides: \(\theta = \tan^{-1}(1) = 45^\circ\).
03

Analyzing the effect of increasing \(A_x\) (Problem a)

If we increase the \(A_x\) component while keeping \(A_y\) constant, the ratio \(\frac{A_y}{A_x}\) becomes smaller, indicating that \(\tan \theta\) decreases. Therefore, \(\theta\) decreases.
04

Solving for \(\theta\) when \(A_x = 17\, \mathrm{m}\) and \(A_y = 12\, \mathrm{m}\)

Use the formula \(\tan \theta = \frac{A_y}{A_x}\). Now, \(\tan \theta = \frac{12}{17}\). Therefore, \(\theta = \tan^{-1}\left(\frac{12}{17}\right) \approx 35.26^\circ\).
05

Analyzing the effect of increasing \(A_y\) (Problem b)

If we increase the \(A_y\) component while keeping \(A_x\) constant, the ratio \(\frac{A_y}{A_x}\) becomes larger, indicating that \(\tan \theta\) increases. Therefore, \(\theta\) increases.
06

Solving for \(\theta\) when \(A_x = 12\, \mathrm{m}\) and \(A_y = 17\, \mathrm{m}\)

Using the formula again, \(\tan \theta = \frac{17}{12}\). Hence, \(\theta = \tan^{-1}\left(\frac{17}{12}\right) \approx 54.74^\circ\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement Vector
A displacement vector is a mathematical quantity used to represent a change in position. It has both magnitude and direction, allowing us to determine where exactly an object has moved from its initial position. When working with displacement vectors, they are often represented in terms of their components in the Cartesian coordinate system. This means we break down the vector into two parts: one that points along the x-axis (horizontal) and another along the y-axis (vertical). These components are denoted as \( A_x \) and \( A_y \) respectively.
The overall direction of the displacement vector is essential. It guides us in understanding the vector's influence on movement. In simple terms, you could imagine it as an arrow pointing in the direction in which displacement occurs. Both the magnitude, which determines the length of the arrow, and its orientation are crucial when examining any motion problem involving displacement vector. Keeping these aspects in mind when dealing with vector analysis makes solving related problems much more straightforward.
Angle Calculation
Calculating the angle \( \theta \) of a displacement vector is essential as it determines the vector's orientation with respect to a reference direction, usually the positive x-axis. To find this angle, we make use of trigonometric functions, specifically the tangent function.
The key formula used in angle calculation is \( \tan \theta = \frac{A_y}{A_x} \), where \( A_y \) is the vertical component and \( A_x \) is the horizontal component. By rearranging this equation, we can solve for \( \theta \) using the inverse tangent function, which is \( \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \).
It is crucial to understand the underlying relationship: the angle \( \theta \) will vary depending on the ratio of \( A_y \) to \( A_x \). A larger \( A_y \) implies a steeper angle, while a larger \( A_x \) results in a shallower angle. Mastering this concept aids in quickly estimating the range or direction of a vector when troubleshooting or confirming calculations.
Tangent Function
The tangent function is a fundamental tool in trigonometry that helps in linking the ratio of the opposite to the adjacent side of a right triangle to an angle. In the context of vectors, we use the tangent function to find the angle \( \theta \) between a displacement vector and a reference axis.
In mathematical terms, for a given angle \( \theta \) in a right triangle, the tangent is expressed as \( \tan \theta = \frac{A_y}{A_x} \), where \( A_y \) is the vertical component and \( A_x \) is the horizontal component.
The significance of the tangent function in vector analysis comes from its ability to help deduce \( \theta \) from known vector components. This is done by calculating \( \theta \) through the inverse operation: \( \theta = \tan^{-1} \left( \frac{A_y}{A_x} \right) \). Understanding how to manipulate this function—whether to increase or decrease the angle by adjusting \(A_x\) or \(A_y\)—is essential to mastering vector component analysis.
Component Analysis
Component analysis is the process of breaking down vectors into their horizontal and vertical parts. This approach simplifies complex vector calculations into manageable pieces. Each vector is defined by its components \( A_x \) and \( A_y \), and these components determine its overall direction and length.
When analyzing vector components, our goal is often to determine how changes in these components (either \( A_x \) or \( A_y \)) influence the vector properties such as magnitude and direction. For instance, an increase in the horizontal component \( A_x \) while holding \( A_y \) constant decreases the angle \( \theta \) the vector makes with the horizontal axis.
Conversely, if we increase the vertical component \( A_y \) while keeping \( A_x \) constant, the angle \( \theta \) increases. Component analysis thus provides valuable insights into not only vector properties but also the dynamic interaction between these components in physics and engineering applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A spring is hanging down from the ceiling, and an object of mass \(m\) is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time \(T\) required for one complete up-and-down oscillation is given by the equation \(T=2 \pi \sqrt{m / k},\) where \(k\) is known as the spring constant. What must be the dimension of \(k\) for this equation to be dimensionally correct?

You are driving into St. Louis, Missouri, and in the distance you see the famous Gateway-to-the-West arch. This monument rises to a height of \(192 \mathrm{~m}\). You estimate your line of sight with the top of the arch to be \(2.0^{\circ}\) above the horizontal. Approximately how far (in kilometers) are you from the base of the arch?

(a) Considering the fact that \(3.28 \mathrm{ft}=1 \mathrm{~m}\), which is the larger unit for measuring area, \(1 \mathrm{ft}^{2}\) or \(1 \mathrm{~m}^{2} ?\) (b) Consider a \(1330-\mathrm{ft}^{2}\) apartment. With your answer to part (a) in mind and without doing any calculations, decide whether this apartment has an area that is greater than or less than \(1330 \mathrm{~m}^{2}\). In a \(1330-\mathrm{ft}^{2}\) apartment, how many square meters of area are there? Be sure that your answer is consistent with your answers to the Concept Questions.

A circus performer begins his act by walking out along a nearly horizontal high wire. He slips and falls to the safety net, \(25.0 \mathrm{ft}\) below. The magnitude of his displacement from the beginning of the walk to the net is \(26.7 \mathrm{ft}\). (a) How far out along the high wire did he walk? (b) Find the angle that his displacement vector makes below the horizontal.

The two hot-air balloons in the drawing are \(48.2\) and \(61.0 \mathrm{~m}\) above the ground. A person in the left balloon observes that the right balloon is \(13.3^{\circ}\) above the horizontal. What is the horizontal distance \(x\) between the two balloons?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.