Chapter 3: Problem 26
What is the sum of the following four vectors in (a) unitvector notation, and as (b) a magnitude and (c) an angle? $$\begin{array}{ll}\vec{A}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(3.00 \mathrm{~m}) \hat{\mathrm{j}} & \vec{B}: 4.00 \mathrm{~m}, \text { at }+65.0^{\circ} \\ \vec{C}=(-4.00 \mathrm{~m}) \hat{\mathrm{i}}+(-6.00 \mathrm{~m}) \hat{\mathrm{j}} & \vec{D}: 5.00 \mathrm{~m}, \text { at }-235^{\circ}\end{array}$$
Short Answer
Step by step solution
Convert Vector B to Unit Vector Notation
Convert Vector D to Unit Vector Notation
Add All Vectors in Unit Vector Notation
Calculate the Magnitude of the Resultant Vector
Determine the Angle of the Resultant Vector
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Vector Notation
- The \( \hat{i} \) component corresponds to the x-axis.
- The \( \hat{j} \) component corresponds to the y-axis.
- For the x-component: Multiply the magnitude by the cosine of the angle.
- For the y-component: Multiply the magnitude by the sine of the angle.
- an \( \hat{i} \) component of \( 4.00 \cos(65^{\circ}) \)
- and a \( \hat{j} \) component of \( 4.00 \sin(65^{\circ}) \)
Magnitude of a Vector
In a two-dimensional space (2D), you can calculate the magnitude of a vector \( \vec{R} = a\hat{i} + b\hat{j} \) using the Pythagorean theorem:\[R = \sqrt{a^2 + b^2}\] This formula is derived from considering the vector's components as part of a right triangle, where the magnitude is the hypotenuse. For example, if our resultant vector \( \vec{R} \) has components \( -3.1776 \) along \( \hat{i} \) and \( -3.4708 \) along \( \hat{j} \), then:\[R = \sqrt{(-3.1776)^2 + (-3.4708)^2} = 4.705 \text{ m}\] The magnitude provides us with essential information regarding the vector's overall size, independent of its direction, useful in various applications from physics to engineering.
Angle of a Vector
To find the angle \( \theta \) of a vector with components \( R_x \) and \( R_y \), use the tangent function:\[\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)\] This inverse tangent function calculates the angle opposite one of the vector's components.
For the vector \( \vec{R} \) with \( R_x = -3.1776 \) and \( R_y = -3.4708 \), the angle is:\[\theta = \tan^{-1}\left(\frac{-3.4708}{-3.1776}\right) = 47.77^{\circ}\] Since both components are negative, the vector lies in the third quadrant. This requires an adjustment to find the correct angle:\[\theta = 180^{\circ} + 47.77^{\circ} = 227.77^{\circ}\] Recognizing the vector's quadrant ensures accurate direction, which is vital in applications like navigation and vector field analysis.
Quadrant Identification
- **First Quadrant**: Both \( x \) and \( y \) components are positive.
- **Second Quadrant**: The \( x \) component is negative, and \( y \) is positive.
- **Third Quadrant**: Both \( x \) and \( y \) components are negative.
- **Fourth Quadrant**: \( x \) is positive, and \( y \) is negative.
This quadrant consideration is essential when calculating vectors' angles, as it influences the direction measure from the positive \( x \)-axis. Knowing the quadrant ensures that any angle calculations are adjusted accordingly, making the vector representation full and correct, which is crucial in fields like cartography and robotics.