/*! 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 45 A vector \(\vec{B}\), with a mag... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 45

A vector \(\vec{B}\), with a magnitude of \(8.0 \mathrm{~m}\), is added to a vector \(\vec{A}\), which lies along an \(x\) axis. The sum of these two vectors is a third vector that lies along the \(y\) axis and has a magnitude that is twice the magnitude of \(\vec{A}\). What is the magnitude of \(\vec{A}\) ?

Short Answer

Expert verified
The magnitude of \( \vec{A} \) is 4 meters.

Step by step solution

01

Understand the problem

We need to find the magnitude of vector \(\vec{A}\) given the following information: \(\vec{B}\) has a magnitude of 8.0 m and the sum of \(\vec{A}\) and \(\vec{B}\) lies along the y-axis with a magnitude twice that of \(\vec{A}\). This implies that the resultant vector is orthogonal to vector \(\vec{A}\).
02

Express vectors algebraically

Let \(\vec{A} = A\hat{i}\) where A is the magnitude of vector \(\vec{A}\) along the x-axis. Vector \(\vec{B}\) can be expressed as \(\vec{B} = 8\hat{j}\), as it lies along the y-axis with magnitude 8.0 m.
03

Find the resultant vector

The resultant vector \(\vec{R}\) is given by the sum of \(\vec{A}\) and \(\vec{B}\). So \(\vec{R} = \vec{A} + \vec{B} = A\hat{i} + 8\hat{j}\).
04

Magnitude condition

According to the problem, the resultant vector lies along the y-axis, which means its x-component must be zero. Also, its magnitude is twice that of \(\vec{A}\), so \(|\vec{R}| = 2A\).
05

Set up the equation

Since \(\vec{R}\) lies on the y-axis, the x-component A must be zero. Hence, \(\vec{A}\)'s contribution to the vector sum must be exactly cancelled by some component of \(\vec{B}\). But since \(\vec{B}\) contributes nothing to the x-axis, the problem simplifies to finding when \(|8| = 2A\).
06

Solve for A

From \(|8| = 2A\), we get \(|8| = 2A\), which simplifies to \(8 = 2A\). Solving for \(\ A \), we get \(A = 4 \).

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 91影视!

Key Concepts

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

Vector Magnitude
Understanding the magnitude of a vector is crucial in physics. The magnitude of a vector represents its length or size and is always a non-negative value. For a vector \(\backslash \backslashvec{B}\) with components in the Cartesian coordinate system, its magnitude can be found using the formula: \[\backslashleft|\vec{B}\backslashright| = \backslashsqrt{B_x^2 + B_y^2}\]. Here, \(B_x\) and \(B_y\) are the components of \(\backslashvec{B}\) along the x and y axes, respectively.
In our example exercise, one of the vectors, \(\backslash \backslashvec{B}\), has a given magnitude of 8.0 m. Magnitude is also sometimes called the length or norm of the vector. This simplifies our calculations as we directly use the value given.
Resultant Vector
The resultant vector is the vector obtained when two or more vectors are added together. It effectively represents the combination of the direction and magnitude of these vectors.
The resultant vector, \(\backslash \backslashvec{R}\), can be found by vector addition. For vectors \(\backslash \backslashvec{A}\) and \(\backslash \backslashvec{B}\), the resultant vector is \(\backslashvec{R} = \backslashvec{A} + \backslashvec{B}\). In our exercise example, since the given condition is that the resultant vector lies along the y-axis and has a magnitude twice that of \(\backslash \backslashvec{A}\), this indicates specific behavior along coordinate axes.
Orthogonal Vectors
Orthogonal vectors are vectors that are perpendicular to each other. In two dimensions, this means they form a right angle (90 degrees) with each other. One key property of orthogonal vectors \(\backslash \backslashvec{A}\) and \(\backslash \backslashvec{B}\) is that their dot product is zero: \(\backslashvec{A} \backslashcdot \backslashvec{B} = 0\).
In our example, \(\backslash \backslashvec{A}\) lies along the x-axis and \(\backslash \backslashvec{B}\) lies along the y-axis. Since the resultant vector lies along the y-axis and the magnitude condition implies orthogonality. This tells us that we are dealing with perpendicular vectors.
Vector Components
Understanding vector components is essential in solving vector addition problems. Vector components split a vector into parts along the coordinate axes. For a vector \(\backslash \backslashvec{A}\), its x-component \(A_x\) and y-component \(A_y\) can be described as: \(\backslash \backslashvec{A} = A_x \backslashhat{i} + A_y \backslashhat{j}\).
In the given exercise, \(\backslash \backslashvec{A}\) lies along the x-axis, so it can be expressed as \(\backslash \backslashvec{A} = A\backslashhat{i}\). Meanwhile, \(\backslash \backslashvec{B}\) lies along the y-axis and is written as \(\backslash \backslashvec{B} = 8\backslashhat{j}\). This clear breakdown of vector components helps in forming the equation to find the magnitude of \(\backslash \backslashvec{A}\).

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 golfer takes three putts to get the ball into the hole. The first putt displaces the ball \(3.66 \mathrm{~m}\) north, the second \(1.83 \mathrm{~m}\) southeast, and the third \(0.91 \mathrm{~m}\) southwest. What are (a) the magnitude and (b) the direction of the displacement needed to get the ball into the hole on the first putt?

A vector \(\vec{d}\) has a magnitude \(3.0 \mathrm{~m}\) and is directed south. What are (a) the magnitude and (b) the direction of the vector \(5.0 \vec{d} ?\) What are (c) the magnitude and (d) the direction of the vector \(-2.0 \vec{d}\) ?

Three vectors \(\vec{a}\), and \(\vec{b}\), and \(\vec{c}\) each have a magnitude of \(50 \mathrm{~m}\) and lie in an \(x y\) plane. Their directions relative to the positive direction of the \(x\) axis are \(30^{\circ}, 195^{\circ}\), and \(315^{\circ}\), respectively. What are (a) the magnitude and (b) the angle of the vector \(\vec{a}+\vec{b}+\) \(\vec{c}\), and \((\mathrm{c})\) the magnitude and \((\mathrm{d})\) the angle of \(\vec{a}-\vec{b}+\vec{c} ?\) What are (e) the magnitude and (f) the angle of a fourth vector \(\vec{d}\) such that \((\vec{a}+\vec{b})-(\vec{c}+\vec{d})=0\)

A pea leaves a pea shooter at a speed of \(5.4 \mathrm{~m} / \mathrm{s}\). It makes an angle of \(+30^{\circ}\) with respect to the horizontal. (a) Calculate the \(x\) -component of the pea's initial velocity. (b) Calculate the \(y\) -component of the pea's initial velocity. (c) Write an expression for the pea's velocity, \(\vec{v}\), using unit vectors for the \(x\) direction and the \(y\) direction.

In each case below, sketch the velocity vector. Find the magnitude and direction of motion with respect to the \(x\) axis of the coordinate system: (a) \(\vec{v}=(2.45 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.67 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) (b) \(\vec{v}=(-2.45 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(5.20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Electrostatics

Read Explanation

Electricity and Magnetism

Read Explanation

Force

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Modern Physics

Read Explanation
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.