Question

Vector \(\overrightarrow{A}\) is 2.80 cm long and is 60.0\(^{\circ}\) above the \(x\)-axis in the first quadrant. Vector \(\overrightarrow{B}\) is 1.90 cm long and is 60.0\(^{\circ}\) below the x-axis in the fourth quadrant (Fig. E1.35).Use components to find the magnitude and direction of (a) \(\overrightarrow{A}\) \(+\) \(\overrightarrow{B}\); (b) \(\overrightarrow{A}\) \(-\) \(\overrightarrow{B}\); (c) \(\overrightarrow{B}\) \(-\) \(\overrightarrow{A}\). In each case, sketch the vector addition or subtraction and show that your numerical answers are in qualitative agreement with your sketch.

Short answer

(a) 2.48 cm at 18.0°, (b) 4.09 cm at 85.9°, (c) 4.09 cm at 265.9°.

Step-by-step solution

Understand Vectors in Terms of Components

Vector \(\overrightarrow{A}\) is 2.80 cm at an angle of 60.0° above the x-axis. The components are:- \(A_x = 2.80 \cos(60°)\)- \(A_y = 2.80 \sin(60°)\)Vector \(\overrightarrow{B}\) is 1.90 cm at an angle of 60.0° below the x-axis. The components are:- \(B_x = 1.90 \cos(60°)\)- \(B_y = -1.90 \sin(60°)\) (negative because it is below the x-axis)

Calculate Components for Each Vector

Calculate the components:For \(\overrightarrow{A}\):\[A_x = 2.80 \times 0.5 = 1.40\, \text{cm}\]\[A_y = 2.80 \times \frac{\sqrt{3}}{2} = 2.42\, \text{cm}\]For \(\overrightarrow{B}\):\[B_x = 1.90 \times 0.5 = 0.95\, \text{cm}\]\[B_y = 1.90 \times -\frac{\sqrt{3}}{2} = -1.65\, \text{cm}\]

Add Vector Components for \(\overrightarrow{A} + \overrightarrow{B}\)

To find \(\overrightarrow{A} + \overrightarrow{B}\):\[ (A_x + B_x, A_y + B_y) = (1.40 + 0.95, 2.42 - 1.65) \]So, \(\overrightarrow{A} + \overrightarrow{B} = (2.35, 0.77)\, \text{cm}\).Find magnitude and angle:\[ \text{magnitude} = \sqrt{(2.35)^2 + (0.77)^2} = 2.48\, \text{cm} \]\[ \text{angle} = \tan^{-1}\left(\frac{0.77}{2.35}\right) \approx 18.0°\] above the x-axis

Subtract Vector Components for \(\overrightarrow{A} - \overrightarrow{B}\)

For \(\overrightarrow{A} - \overrightarrow{B}\):\[ (A_x - B_x, A_y - B_y) = (1.40 - 0.95, 2.42 + 1.65) \]So, \(\overrightarrow{A} - \overrightarrow{B} = (0.45, 4.07)\, \text{cm}\).Find magnitude and angle:\[ \text{magnitude} = \sqrt{(0.45)^2 + (4.07)^2} = 4.09\, \text{cm} \]\[ \text{angle} = \tan^{-1}\left(\frac{4.07}{0.45}\right) \approx 85.9°\] above the x-axis

Find Components for \(\overrightarrow{B} - \overrightarrow{A}\)

For \(\overrightarrow{B} - \overrightarrow{A}\):\[ (B_x - A_x, B_y - A_y) = (0.95 - 1.40, -1.65 - 2.42) \]So, \(\overrightarrow{B} - \overrightarrow{A} = (-0.45, -4.07)\, \text{cm}\).Find magnitude and angle:\[ \text{magnitude} = \sqrt{(-0.45)^2 + (-4.07)^2} = 4.09\, \text{cm} \]\[ \text{angle} = \tan^{-1}\left(\frac{-4.07}{-0.45}\right) \approx 265.9°\] counterclockwise from the positive x-axis

Summarize Results and Sketch

For \(\overrightarrow{A} + \overrightarrow{B}\):- Magnitude: 2.48 cm, Direction: 18.0° above x-axisFor \(\overrightarrow{A} - \overrightarrow{B}\):- Magnitude: 4.09 cm, Direction: 85.9° above x-axisFor \(\overrightarrow{B} - \overrightarrow{A}\):- Magnitude: 4.09 cm, Direction: 265.9°Sketch the vectors accordingly to visualize these results.

Key concepts

Vector Components

When discussing vectors, it is crucial to understand their components. A vector can be broken down into its horizontal and vertical components along the x and y axes, respectively. This helps in simplifying vector addition, subtraction, and other calculations.
Consider Vector \(\overrightarrow{A}\) that is 2.80 cm long, making an angle of 60° above the x-axis. To find its components, you can use trigonometry:

• Horizontal component \(A_x = 2.80 \cos(60°)\)
• Vertical component \(A_y = 2.80 \sin(60°)\)

Solving these gives \(A_x = 1.40\, \text{cm}\) and \(A_y = 2.42\, \text{cm}\).
Similarly, for Vector \(\overrightarrow{B}\) which is 1.90 cm and 60° below the x-axis, its components are:

• \(B_x = 1.90 \cos(60°)\)
• \(B_y = -1.90 \sin(60°)\), where the negative sign indicates a direction below the x-axis.

Calculating these yields \(B_x = 0.95\, \text{cm}\) and \(B_y = -1.65\, \text{cm}\). Understanding these components is foundational for further vector operations.

Magnitude and Direction

The magnitude and direction of a vector describe its length and the angle it makes with a reference line, typically the x-axis. Knowing these helps in both physics and engineering when you're interested in precisely knowing a vector's effect in a system.
Once you have the components:

• The magnitude can be found using the Pythagorean theorem: \[ \text{magnitude} = \sqrt{(A_x + B_x)^2 + (A_y + B_y)^2} \]
• The direction or angle \(\theta\) is found using: \[ \theta = \tan^{-1}\left(\frac{A_y + B_y}{A_x + B_x}\right) \]

For instance, when adding vectors \(\overrightarrow{A} + \overrightarrow{B}\), the resultant magnitude is computed as 2.48 cm, with the direction being 18.0° above the x-axis.
These calculations are not just numbers; they illustrate how these vectors affect physical quantities like displacement or force.

Vector Subtraction

Vector subtraction is essentially adding a vector with the negative of another vector. This is useful when you need to find a difference in effects, often seen in relative velocities or changes in forces.
To subtract vectors, calculate their components. For instance, in \(\overrightarrow{A} - \overrightarrow{B}\):

• Subtract their x-components: \(A_x - B_x = 1.40 - 0.95 = 0.45\, \text{cm}\)
• Subtract their y-components: \(A_y - B_y = 2.42 + 1.65 = 4.07\, \text{cm}\)

The result is a vector \( (0.45, 4.07)\, \text{cm}\). Similarly, for \(\overrightarrow{B} - \overrightarrow{A}\), the resultant vector is \((-0.45, -4.07)\, \text{cm}\).
The magnitude and direction are found with the same methods as addition: \[ \text{magnitude} = \sqrt{(0.45)^2 + (4.07)^2} = 4.09\, \text{cm} \] \[ \theta = \tan^{-1}\left(\frac{4.07}{0.45}\right) \approx 85.9° \]\Direction can be tricky with subtraction, especially when negative angles arise, so always visualize your vectors to better understand their orientation.