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Chapter 3: Problem 56

Find the sum of the following four vectors in (a) unit-vector notation, and as (b) a magnitude and (c) an angle relative to \(+x\). \(\vec{P}: 10.0 \mathrm{~m}\), at \(25.0^{\circ}\) counterclockwise from \(+x\) \(\vec{Q}: 12.0 \mathrm{~m}\), at \(10.0^{\circ}\) counterclockwise from \(+y\) \(\vec{R}: 8.00 \mathrm{~m}\), at \(20.0^{\circ}\) clockwise from \(-y\) \(\vec{S}: 9.00 \mathrm{~m}\), at \(40.0^{\circ}\) counterclockwise from \(-y\)

Short Answer

Expert verified
a) \(\vec{R}_{\text{sum}} = 1.50 \hat{i} + 14.31 \hat{j} \; \mathrm{m}\), b) Magnitude: 14.38 m, c) Angle: 84.0掳 from +x.

Step by step solution

01

Resolve Vector P into Components

Vector \(\vec{P}\) has a magnitude of 10.0 m and is oriented \(25.0^{\circ}\) counterclockwise from the positive x-axis. To resolve it into components:\[ \begin{align*} P_x &= 10.0 \cos(25.0^{\circ}) \ P_y &= 10.0 \sin(25.0^{\circ}) \end{align*} \] Calculating these, we find \(P_x \approx 9.06\) m and \(P_y \approx 4.23\) m.
02

Resolve Vector Q into Components

Vector \(\vec{Q}\) has a magnitude of 12.0 m and is \(10.0^{\circ}\) counterclockwise from the positive y-axis, which is equivalent to \(80.0^{\circ}\) from the positive x-axis. To resolve it into components:\[ \begin{align*} Q_x &= 12.0 \cos(80.0^{\circ}) \ Q_y &= 12.0 \sin(80.0^{\circ}) \end{align*} \] Calculating these, we find \(Q_x \approx 2.08\) m and \(Q_y \approx 11.81\) m.
03

Resolve Vector R into Components

Vector \(\vec{R}\) has a magnitude of 8.00 m and is \(20.0^{\circ}\) clockwise from the negative y-axis, equivalent to \(250.0^{\circ}\) from the positive x-axis. To resolve it into components:\[ \begin{align*} R_x &= 8.00 \cos(250.0^{\circ}) \ R_y &= 8.00 \sin(250.0^{\circ}) \end{align*} \] Computing these, we get \(R_x \approx -2.74\) m and \(R_y \approx -7.52\) m.
04

Resolve Vector S into Components

Vector \(\vec{S}\) has a magnitude of 9.00 m and is oriented \(40.0^{\circ}\) counterclockwise from the negative y-axis, equivalent to \(140.0^{\circ}\) from the positive x-axis. To resolve it into components: \[ \begin{align*} S_x &= 9.00 \cos(140.0^{\circ}) \ S_y &= 9.00 \sin(140.0^{\circ}) \end{align*} \] Calculating these results in \(S_x \approx -6.90\) m and \(S_y \approx 5.79\) m.
05

Sum the Components to Find the Resultant Vector

To find the resultant vector \(\vec{R}_{\text{sum}}\) in unit-vector notation, add up all the x-components and y-components separately: \[ \begin{align*} R_{\text{sum},x} &= P_x + Q_x + R_x + S_x = 9.06 + 2.08 - 2.74 - 6.90 \approx 1.50 \; \mathrm{m} \ R_{\text{sum},y} &= P_y + Q_y + R_y + S_y = 4.23 + 11.81 - 7.52 + 5.79 \approx 14.31 \; \mathrm{m} \end{align*} \] So, \(\vec{R}_{\text{sum}} = 1.50 \hat{i} + 14.31 \hat{j} \; \mathrm{m}\).
06

Calculate the Magnitude of the Resultant Vector

The magnitude \(|\vec{R}_{\text{sum}}|\) is given by: \[ |\vec{R}_{\text{sum}}| = \sqrt{(R_{\text{sum},x})^2 + (R_{\text{sum},y})^2} = \sqrt{(1.50)^2 + (14.31)^2} \] Upon calculation, \(|\vec{R}_{\text{sum}}| \approx 14.38 \; \mathrm{m}\).
07

Determine the Angle Relative to the Positive X-axis

The angle \(\theta\) can be found using the inverse tangent function: \[ \theta = \tan^{-1}\left(\frac{R_{\text{sum},y}}{R_{\text{sum},x}}\right) = \tan^{-1}\left(\frac{14.31}{1.50}\right) \] Solving gives \(\theta \approx 84.0^{\circ}\) from the positive x-axis.

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Key Concepts

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

Unit-Vector Notation
Unit-vector notation is a concise way to express vectors in terms of their x and y components. When working in two dimensions, we use the unit vectors \(\hat{i}\) and \(\hat{j}\) to represent the x and y directions, respectively. This notation is extremely helpful because it not only provides the magnitude of the vector along each axis but also indicates its direction.
For instance, in the problem we are solving, the resultant vector \(\vec{R}_{\text{sum}}\) is expressed as \(1.50 \hat{i} + 14.31 \hat{j}\; \mathrm{m}\). This tells us that the vector points 1.50 meters in the positive x-direction and 14.31 meters in the positive y-direction.
  • \(\hat{i}\) and \(\hat{j}\) are unit vectors, meaning they have a magnitude of 1 but are directed along the x and y axes.
  • Unit-vector notation simplifies the addition and subtraction of vectors by focusing separately on each dimension.
  • This form of notation is essential for understanding complex vector operations in physics and engineering.
Magnitude Calculation
Calculating the magnitude of a vector is one of the fundamental tasks in vector analysis. It tells us "how much" of a quantity we have in without referring to its direction. This is achieved using the Pythagorean theorem.
For a vector decomposed into x and y components, its magnitude is given by \(|\vec{V}| = \sqrt{V_x^2 + V_y^2}\). You essentially square the components, sum them, and take the square root to find the length of the vector.
In the exercise, the resultant vector's magnitude \(|\vec{R}_{\text{sum}}|\) is calculated as \(\sqrt{(1.50)^2 + (14.31)^2}\), yielding approximately 14.38 meters.
  • The magnitude provides a scalar value representing the overall size of the vector.
  • While direction is crucial, the magnitude alone is often used to gauge kinetic energy, force, and other scalar quantities in physics.
  • Magnitude is always a non-negative number since it represents a distance or size.
Vector Resolution
Vector resolution is the process of breaking a vector down into its perpendicular components, typically along the x and y axes. This involves using trigonometric functions such as sine and cosine to determine how much of the vector is directed along each axis.
Consider Vector \(\vec{P}\) from the exercise. It is resolved into components as follows:
\[\begin{align*} P_x &= 10.0 \cos(25.0^\circ) \ P_y &= 10.0 \sin(25.0^\circ) \end{align*}\]
This tells us exactly how much of \(\vec{P}\) is directed along the x-axis and y-axis.
  • Resolution is particularly useful when dealing with forces, velocity, and acceleration in more complex systems.
  • It allows for simple vector addition by focusing on each component separately.
  • This method is crucial for converting between polar and rectangular coordinates.
Overall, vector resolution simplifies many problems by dividing them into manageable parts.

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Most popular questions from this chapter

Here are three vectors in meters: $$ \begin{array}{l} \vec{d}_{1}=-3.0 \hat{\mathrm{i}}+3.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}} \\\ \vec{d}_{2}=-2.0 \hat{\mathrm{i}}-4.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{k}} \\\ \vec{d}_{3}=2.0 \hat{\mathrm{i}}+3.0 \hat{\mathrm{j}}+1.0 \hat{\mathrm{k}} \end{array} $$ What results from (a) \(\vec{d}_{1} \cdot\left(\vec{d}_{2}+\vec{d}_{3}\right)\), (b) \(\vec{d}_{1} \cdot\left(\vec{d}_{2} \times \vec{d}_{3}\right)\), and (c) \(\vec{d}_{1} \times\left(\vec{d}_{2}+\vec{d}_{3}\right) ?\)

If \(\vec{d}_{1}+\vec{d}_{2}=5 \vec{d}_{3}, \vec{d}_{1}-\vec{d}_{2}=3 \vec{d}_{3}\), and \(\vec{d}_{3}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}\), then what are, in unit-vector notation, (a) \(\vec{d}_{1}\) and (b) \(\vec{d}_{2}\) ?

A sailboat sets out from the U.S. side of Lake Erie for a point on the Canadian side, \(90.0 \mathrm{~km}\) due north. The sailor, however, ends up \(50.0 \mathrm{~km}\) due east of the starting point. (a) How far and (b) in what direction must the sailor now sail to reach the original destination?

In the sum \(\vec{A}+\vec{B}=\vec{C}\), vector \(\vec{A}\) has a magnitude of \(12.0 \mathrm{~m}\) and is angled \(40.0^{\circ}\) counterclockwise from the \(+x\) direction, and vector \(\vec{C}\) has a magnitude of \(15.0 \mathrm{~m}\) and is angled \(20.0^{\circ}\) counterclockwise from the \(-x\) direction. What are (a) the magnitude and (b) the angle (relative to \(+x)\) of \(\vec{B}\) ?

Three vectors \(\vec{a}, \vec{b}\), 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 (c) the magnitude and (d) the angle of \(\vec{a}-\vec{b}+\vec{c} ?\) What are the (e) magnitude and (f) angle of a fourth vector \(\vec{d}\) such that \((\vec{a}+\vec{b})-(\vec{c}+\vec{d})=0 ?\)
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