Problem 26 - Fundamentals of Physics

Question

What is the sum of the following four vectors in (a) unitvector notation, and as (b) a magnitude and (c) an angle? $$\begin{array}{ll}\vec{A}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(3.00 \mathrm{~m}) \hat{\mathrm{j}} & \vec{B}: 4.00 \mathrm{~m}, \text { at }+65.0^{\circ} \\ \vec{C}=(-4.00 \mathrm{~m}) \hat{\mathrm{i}}+(-6.00 \mathrm{~m}) \hat{\mathrm{j}} & \vec{D}: 5.00 \mathrm{~m}, \text { at }-235^{\circ}\end{array}$$

Step-by-Step Solution

Answer

In unit vector notation: \(-3.1776\hat{i} - 3.4708\hat{j}\); magnitude: 4.705 m; angle: 227.77°.

Step 1: Convert Vector B to Unit Vector Notation

Vector \( \vec{B} \) has a magnitude of 4.00 m and an angle of +65.0 degrees. To convert into unit vector notation:\( \vec{B} = (4.00 \cos(65.0^{\circ}))\hat{i} + (4.00 \sin(65.0^{\circ}))\hat{j} \).Calculating these components:\( \vec{B} = (4.00 \times 0.4226)\hat{i} + (4.00 \times 0.9063)\hat{j} = 1.6904\hat{i} + 3.6252\hat{j} \).

Step 2: Convert Vector D to Unit Vector Notation

Vector \( \vec{D} \) has a magnitude of 5.00 m and an angle of -235 degrees. To convert it into unit vector notation:\( \vec{D} = (5.00 \cos(-235^{\circ}))\hat{i} + (5.00 \sin(-235^{\circ}))\hat{j} \).Calculating the components:\( \vec{D} = (5.00 \times -0.5736)\hat{i} + (5.00 \times -0.8192)\hat{j} = -2.8680\hat{i} + -4.0960\hat{j} \).

Step 3: Add All Vectors in Unit Vector Notation

Now, add all vectors, \( \vec{A}, \vec{B}, \vec{C}, \) and \( \vec{D} \) in unit vector notation:\[ \vec{A} + \vec{B} + \vec{C} + \vec{D} = (2.00\hat{i} + 3.00\hat{j}) + (1.6904\hat{i} + 3.6252\hat{j}) + (-4.00\hat{i} - 6.00\hat{j}) + (-2.8680\hat{i} - 4.0960\hat{j}) \].Simplify by combining like terms:\[ \vec{R} = (2.00 + 1.6904 - 4.00 - 2.8680)\hat{i} + (3.00 + 3.6252 - 6.00 - 4.0960)\hat{j} \]\[ \vec{R} = (-3.1776)\hat{i} + (-3.4708)\hat{j} \].

Step 4: Calculate the Magnitude of the Resultant Vector

The magnitude \( R \) of the resultant vector \( \vec{R} \) is given by the formula:\[ R = \sqrt{(R_x)^2 + (R_y)^2} \]\[ R = \sqrt{(-3.1776)^2 + (-3.4708)^2} \]\[ R = \sqrt{10.0984 + 12.0475} = \sqrt{22.1459} = 4.705 \text{ m} \].

Step 5: Determine the Angle of the Resultant Vector

The angle \( \theta \) of the resultant vector from the positive x-axis is:\[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \]Using \( R_x = -3.1776 \) and \( R_y = -3.4708 \):\[ \theta = \tan^{-1}\left(\frac{-3.4708}{-3.1776}\right) = \tan^{-1}(1.0923) \]\[ \theta = 47.77^{\circ} \].Since both \( R_x \) and \( R_y \) are negative, the vector is in the third quadrant, so adjust the angle:\[ \theta = 180^{\circ} + 47.77^{\circ} = 227.77^{\circ} \].

Key Concepts

Unit Vector Notation

Unit vector notation is a convenient way to express vectors, especially in two or three dimensions. It involves breaking down a vector into its horizontal and vertical components along the x and y axes, respectively. These components are represented using the unit vectors \( \hat{i} \) and \( \hat{j} \).

• The \( \hat{i} \) component corresponds to the x-axis.
• The \( \hat{j} \) component corresponds to the y-axis.

To convert a vector expressed with a certain magnitude and angle into unit vector notation, you calculate the horizontal and vertical components:

1. For the x-component: Multiply the magnitude by the cosine of the angle.
2. For the y-component: Multiply the magnitude by the sine of the angle.

For example, a vector \( \vec{B} \) with a magnitude of 4.00 meters at an angle of 65 degrees would have:

• an \( \hat{i} \) component of \( 4.00 \cos(65^{\circ}) \)
• and a \( \hat{j} \) component of \( 4.00 \sin(65^{\circ}) \)

Hence, in unit vector notation, such a vector can be expressed as: \( \vec{B} = 1.6904 \hat{i} + 3.6252 \hat{j} \). This method allows for easy addition and subtraction of vectors, acting like simple algebraic quantities.

Magnitude of a Vector

The magnitude of a vector is a measure of its length, providing a scalar quantity that describes how large or small a vector is. Magnitude is always a positive value and is represented by the symbol \( |\vec{V}| \) or simply \( V \).
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

The angle of a vector is crucial as it provides direction in relation to a known reference line, typically the positive x-axis. Understanding the angle helps in graphically representing the vector's orientation.
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

Quadrant identification is an essential skill when working with vectors, as it helps determine the vector's direction based on its components. The coordinate plane is divided into four quadrants:

• **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.

Identifying the correct quadrant helps in defining the vector's angle more precisely. For instance, if a vector has negative \( x \) and \( y \) components, it means the vector points toward the third quadrant.
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.