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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
Sum the components to find unit-vector notation; calculate magnitude with Pythagorean theorem; find angle using arctangent.

Step by step solution

01

Convert Each Vector to Component Form

To express each vector in component form, we'll use trigonometry for conversion. For vector \( \vec{P} \):- \( \vec{P}_x = 10.0 \cos(25.0^\circ) \)- \( \vec{P}_y = 10.0 \sin(25.0^\circ) \)For vector \( \vec{Q} \):- The angle relative to \(+x\) is \( 80.0^\circ \) because it is 10.0° from \(+y\).- \( \vec{Q}_x = 12.0 \cos(80.0^\circ) \)- \( \vec{Q}_y = 12.0 \sin(80.0^\circ) \)For vector \( \vec{R} \):- The angle from \(+x\) is \(250.0^\circ\) as it is 20° clockwise from \(-y\).- \( \vec{R}_x = 8.0 \cos(250.0^\circ) \)- \( \vec{R}_y = 8.0 \sin(250.0^\circ) \)For vector \( \vec{S} \):- The angle from \(+x\) is \(130.0^\circ\) as it is 40.0° CCW from \(-y\).- \( \vec{S}_x = 9.0 \cos(130.0^\circ) \)- \( \vec{S}_y = 9.0 \sin(130.0^\circ) \)

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Key Concepts

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

Unit-Vector Notation
In physics and engineering, it's common to describe vectors using unit-vector notation.This can help simplify calculations and make the understanding of a vector's direction more intuitive.Unit-vector notation involves expressing a vector in terms of its components along the x- and y-axes, typically using the unit vectors \( \hat{i} \) and \( \hat{j} \).
For example, a vector \( \vec{V} \) can be represented as \( \vec{V} = V_x \hat{i} + V_y \hat{j} \), where \( V_x \) and \( V_y \) are the magnitudes of the vector along the x and y directions, respectively.
  • \( \hat{i} \) is a unit vector pointing in the direction of the x-axis.
  • \( \hat{j} \) is a unit vector pointing in the direction of the y-axis.
In practical problems, converting vectors to unit-vector notation is the first step.This allows us to perform operations such as addition more efficiently.
Trigonometry
Trigonometry is a branch of mathematics focusing on the relationships between angles and sides in triangles.When dealing with vector components, trigonometry is essential.It helps us break down vectors that are given in terms of magnitude and direction into their x- and y-components.
Using trigonometric functions like sine and cosine:
  • The cosine function (\( \cos \) ) can be used to find the component of the vector along the x-axis. This is calculated as \( V_x = V \cos(\theta) \), where \( V \) is the magnitude of the vector and \( \theta \) is the angle it makes with the horizontal axis.
  • The sine function (\( \sin \) ) can be used to find the component along the y-axis, \( V_y = V \sin(\theta) \).
These formulas are the backbone of transforming a vector's magnitude and angle into its corresponding coordinate components.
Vector Components
The components of a vector are the projections of that vector along the coordinate axes, commonly the x- and y-axes.They are crucial for simplifying complex vector problems.By dealing with each component separately, operations such as addition and subtraction become more straightforward.
In a 2-dimensional plane, a vector \( \vec{V} \) is expressed in terms of its components as \( \vec{V} = V_x \hat{i} + V_y \hat{j} \).Here, \( V_x \) represents the horizontal component, and \( V_y \) is the vertical component.
  • The horizontal component is computed by \( V_x = V \cos(\theta) \).
  • The vertical component is computed by \( V_y = V \sin(\theta) \).
By converting vectors into components, we can easily add or subtract multiple vectors by simply adding or subtracting their respective components.
Magnitude and Angle Calculation
Once vectors are expressed in terms of their components, calculating their resultant vector becomes manageable.The magnitude of a vector is essentially its length, calculated from its components using the Pythagorean theorem.
The formula to find the magnitude \( R \) of a vector \( \vec{R} \) from its components \( R_x \) and \( R_y \) is:
\[ R = \sqrt{R_x^2 + R_y^2} \]
Besides magnitude, the direction is equally important.The angle \( \theta \) of the resultant vector relative to the positive x-axis can be determined using the tangent function:
\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \]
This gives a complete understanding of both the length and direction of the resultant vector.This complete picture can then be used for further applications or interpretations in physics or engineering contexts.

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Most popular questions from this chapter

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 \(-2.0 \vec{d} ?\)

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

A ship sets out to sail to a point \(120 \mathrm{~km}\) due north. An unexpected storm blows the ship to a point \(100 \mathrm{~km}\) due east of its starting point. (a) How far and (b) in what direction must it now sail to reach its original destination?

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

Displacement \(\vec{d}_{1}\) is in the \(y z\) plane \(63.0^{\circ}\) from the positive direction of the \(y\) axis, has a positive \(z\) component, and has a magnitude of \(4.50 \mathrm{~m} .\) Displacement \(\vec{d}_{2}\) is in the \(x z\) plane \(30.0^{\circ}\) from the positive direction of the \(x\) axis, has a positive \(z\) component, and has magnitude \(1.40 \mathrm{~m} .\) What are \((\mathrm{a}) \vec{d}_{1} \cdot \vec{d}_{2},(\mathrm{~b}) \vec{d}_{1} \times \vec{d}_{2},\) and \((\mathrm{c})\) the angle between \(\vec{d}_{1}\) and \(\vec{d}_{2} ?\)

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