Problem 40 - University Physics with Modern Physics

Question

Find the magnitude and direction of the vector represented by the following pairs of components: (a) \(A_{x}=-8.60 \mathrm{cm}\), \(A_{y}=5.20 \mathrm{cm}\) (b) \(A_{x}=-9.70 \mathrm{m}, A_{y}=-2.45 \mathrm{m};\) (b) \(A_{x}=-9.70 \mathrm{m}, A_{y}=-2.45 \mathrm{m};\) (c) \(A_{x}=7.75 \mathrm{km}\), \(A_{y}=-2.70 \mathrm{km}\).

Step-by-Step Solution

Answer

(a) Magnitude: 10.05 cm, Direction: 148.9°; (b) Magnitude: 10.00 m, Direction: 194.1°; (c) Magnitude: 8.21 km, Direction: 340.8°.

Step 1: Calculate the Magnitude for Part a

We calculate the magnitude of vector \( A \) using the Pythagorean theorem: \[ |A| = \sqrt{A_x^2 + A_y^2} \]Substituting the given values, we have:\[ |A| = \sqrt{(-8.60)^2 + (5.20)^2} = \sqrt{73.96 + 27.04} = \sqrt{101} \approx 10.05 \text{ cm} \]

Step 2: Calculate the Direction for Part a

The direction \( \theta \) of the vector is calculated using the inverse tangent function:\[ \theta = \arctan\left(\frac{A_y}{A_x}\right) \]Substitute the given values:\[ \theta = \arctan\left(\frac{5.20}{-8.60}\right) \approx \arctan(-0.6047) \]This results in a direction of approximately \[ \theta \approx -31.1^\circ \] However, since it's in the second quadrant, add 180°:\[ \theta = 180^\circ - 31.1^\circ = 148.9^\circ \]

Step 3: Calculate the Magnitude for Part b

We use the Pythagorean theorem again:\[ |A| = \sqrt{(-9.70)^2 + (-2.45)^2} = \sqrt{94.09 + 6.0025} = \sqrt{100.0925} \approx 10.00 \text{ m} \]

Step 4: Calculate the Direction for Part b

Calculate the angle using:\[ \theta = \arctan\left(\frac{-2.45}{-9.70}\right) \approx \arctan(0.2526) \approx 14.1^\circ \] Since both \( A_x \) and \( A_y \) are negative, the vector is in the third quadrant. Thus, \[ \theta = 180^\circ + 14.1^\circ = 194.1^\circ \]

Step 5: Calculate the Magnitude for Part c

Using the Pythagorean theorem:\[ |A| = \sqrt{(7.75)^2 + (-2.70)^2} = \sqrt{60.0625 + 7.29} = \sqrt{67.3525} \approx 8.21 \text{ km} \]

Step 6: Calculate the Direction for Part c

Calculate the angle:\[ \theta = \arctan\left(\frac{-2.70}{7.75}\right) \approx \arctan(-0.3484) \approx -19.2^\circ \] Since only \( A_y \) is negative, the vector is in the fourth quadrant, thus \[ \theta = 360^\circ - 19.2^\circ = 340.8^\circ \]

Key Concepts

Magnitude of a Vector

The magnitude of a vector is a crucial aspect of vector calculation, representing its size or length. If you picture a vector as an arrow, the magnitude is simply how long that arrow is. To find this, we use the components of the vector, usually given in the x and y directions.

If you're given a vector with components \(A_x\) and \(A_y\), the magnitude \(|A|\) is determined using the Pythagorean theorem, which relates the square of the hypotenuse of a right triangle to the sum of the squares of its other two sides. The formula is:

• \(|A| = \sqrt{A_x^2 + A_y^2}\)

This formula ensures that regardless of the direction of the vector, its magnitude is always a non-negative number, providing a complete description of its size.

Direction of a Vector

Understanding the direction of a vector is just as important as knowing its magnitude. Direction tells you where the vector is pointing. It's typically measured as an angle from the positive x-axis, and can range from \(0^\circ\) to \(360^\circ\).

To find the direction \(\theta\) of a vector with components \(A_x\) and \(A_y\), we use the arctangent function, denoted as \(\arctan\). The formula is:

• \(\theta = \arctan\left(\frac{A_y}{A_x}\right)\)

However, the angle provided by \(\arctan\) might not always reflect the true directional angle. It can result in negative angles or miss the correct quadrant of the vector. Therefore, it's important to adjust the angle based on which quadrant the vector resides in.

Pythagorean Theorem

The Pythagorean theorem is a foundational principle in mathematics used extensively in vector calculations. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides. This theorem provides the foundation for calculating the magnitude of a vector.

For instance, if you have a vector that can be expressed by the components along the x-axis and y-axis, you essentially form a right triangle. The vector itself serves as the hypotenuse, and the relationship as stated by the Pythagorean theorem is:

• \(c^2 = a^2 + b^2\)
• Where \(c\) is the vector (hypotenuse), and \(a\) and \(b\) are its components (legs of the triangle).

This principle allows us to find the magnitude of any vector regardless of its direction.

Quadrant Analysis

Quadrant analysis is an essential step in vector calculations as it helps interpret the directional angle correctly. The coordinate plane is divided into four quadrants, each characterized by signs of the x and y components:

• First Quadrant: \((+x, +y)\)
• Second Quadrant: \((-x, +y)\)
• Third Quadrant: \((-x, -y)\)
• Fourth Quadrant: \((+x, -y)\)

Using quadrant analysis allows us to adjust the direction \(\theta\) if needed. Angles must be adapted based on the specific quadrant the vector lies in. For example, if \(\arctan\) gives a negative angle or an angle less than \(0^\circ\), add \(180^\circ\) or \(360^\circ\) depending on the quadrant. Understanding this ensures the angle accurately corresponds to the vector's actual direction in the plane.