Problem 41 - College Physics

Question

\(\cdot\) In each of the cases that follow, the components of a vector \(\vec{A}\) are given. Use trigonometry to find the magnitude of that vector and the counterclockwise angle it makes with the \(+x\) axis. Also, sketch each vector approximately to scale to see if your calculated answers seem reasonable. (a) \(A_{x}=4.0 \mathrm{m}, A_{y}=5.0 \mathrm{m},\) (b) \(A_{x}=-3.0 \mathrm{km}, A_{y}=-6.0 \mathrm{km},(\mathrm{c}) A_{x}=9.0 \mathrm{m} / \mathrm{s}, A_{y}=\) \(-17 \mathrm{m} / \mathrm{s},(\mathrm{d}) A_{x}=-8.0 \mathrm{N}, A_{\mathrm{y}}=12 \mathrm{N}\)

Step-by-Step Solution

Answer

(a) Magnitude: 6.4 m, Angle: 51.3°; (b) Magnitude: 6.7 km, Angle: 153.4°; (c) Magnitude: 19.2 m/s, Angle: 297.8°; (d) Magnitude: 14.4 N, Angle: 123.7°.

Step 1: Understanding the Problem

We are given vector components \( A_x \) and \( A_y \), and we need to find the magnitude and angle (counterclockwise from the positive x-axis) of vector \( \vec{A} \) for four different cases. We will also sketch each vector to check the results.

Step 2: Calculate Magnitude

The magnitude of a vector \( \vec{A} \) with components \( A_x \) and \( A_y \) is given by the Pythagorean theorem: \( |\vec{A}| = \sqrt{A_x^2 + A_y^2} \). We will apply this formula to each case.

Step 3: Finding the Angle

The angle \( \theta \) can be found using the arctangent function: \( \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \). We will find this angle for each case, adjusting for the correct quadrant if necessary.

Step 4: Calculate for Case (a)

For \( A_x = 4.0 \) m and \( A_y = 5.0 \) m:- Magnitude: \( |\vec{A}| = \sqrt{4.0^2 + 5.0^2} = \sqrt{16 + 25} = 6.4 \) m.- Angle: \( \theta = \tan^{-1}\left(\frac{5.0}{4.0}\right) \approx 51.3^\circ \). This is in the first quadrant.

Step 5: Calculate for Case (b)

For \( A_x = -3.0 \) km and \( A_y = -6.0 \) km:- Magnitude: \( |\vec{A}| = \sqrt{(-3.0)^2 + (-6.0)^2} = \sqrt{9 + 36} = 6.7 \) km.- Angle: \( \theta = \tan^{-1}\left(\frac{-6.0}{-3.0}\right) = \tan^{-1}(2) \approx 116.6^\circ \). Since both components are negative, this is in the third quadrant, so adjust to \( 180^\circ + \theta = 153.4^\circ \).

Step 6: Calculate for Case (c)

For \( A_x = 9.0 \) m/s and \( A_y = -17.0 \) m/s:- Magnitude: \( |\vec{A}| = \sqrt{9.0^2 + (-17.0)^2} = \sqrt{81 + 289} = 19.2 \) m/s.- Angle: \( \theta = \tan^{-1}\left(\frac{-17.0}{9.0}\right) \approx -62.2^\circ \). This is in the fourth quadrant, so adjust to \( 360^\circ - 62.2^\circ = 297.8^\circ \).

Step 7: Calculate for Case (d)

For \( A_x = -8.0 \) N and \( A_y = 12.0 \) N:- Magnitude: \( |\vec{A}| = \sqrt{(-8.0)^2 + 12.0^2} = \sqrt{64 + 144} = 14.4 \) N.- Angle: \( \theta = \tan^{-1}\left(\frac{12.0}{-8.0}\right) \approx -56.3^\circ \). Since \( A_x \) is negative and \( A_y \) is positive, this is in the second quadrant, so adjust to \( 180^\circ + (-56.3^\circ) = 123.7^\circ \).

Step 8: Sketch and Review Results

Sketch each vector based on the calculated magnitude and angle to check if results seem reasonable visually. Ensure vectors in the correct quadrant as per the angle adjustments done in Step 7.

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that deals with triangles, specifically right triangles, and the relationships between their angles and side lengths. It plays a crucial role in vector analysis, as it allows us to determine various properties of vectors, such as direction and magnitude.
In the exercise, we apply trigonometric functions such as the tangent. The tangent function is particularly useful because in a right triangle formed by the vector components and the resultant vector, the tangent of the angle gives us the ratio of the opposite side to the adjacent side.

• The formula used is: \( \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \).

This formula helps us find the angle that the vector makes with the positive x-axis. Understanding the triangle formed by the vector components helps in visualizing and solving the vector problems more effectively.

Magnitude Calculation

Magnitude refers to the length or size of a vector. It tells us how much of a quantity the vector represents, ignoring its direction.
To find the magnitude of a vector from its components, we utilize the Pythagorean theorem. This formula calculates the hypotenuse of a right triangle formed by the components of the vector.
The standard formula is:

• \( |\vec{A}| = \sqrt{A_x^2 + A_y^2} \)

This formula gives us a scalar value that represents the vector's magnitude. By applying this calculation to each case, we can accurately determine how strong or significant the vector is in terms of its representation in physical space. For instance, in the exercise, case (a) had a result of 6.4 m, indicating the precise size of the vector.

Angle Finding

Finding the angle in vector analysis is crucial to understand the vector's direction relative to a reference axis. The angle gives insight into where the vector is pointing and is vital for various applications, like navigation or physics problems.
In the exercise, we use the arctangent function, which provides the angle \( \theta \). The formula is:

• \( \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \)

The result from this function helps to pinpoint the angle relative to the positive x-axis. However, it's important to note that this angle initially provided by the arctangent function could have ambiguities in terms of quadrant placement. Consequently, further steps may be needed to adjust and ensure it's accurately placed on the coordinate system.

Quadrant Adjustments

Correctly identifying which quadrant a vector lies in is essential for accurately descri bing its direction. The initial computation of the angle using arctan assumes the vector lies in the first or fourth quadrant, often necessitating adjustments.
Depending on the signs of the components \( A_x \) and \( A_y \):

• If both components are positive, the vector is in the first quadrant, and no adjustment is needed.
• If both components are negative, the vector is in the third quadrant. You need to add \(180^\circ\) to the angle.
• If \( A_x \) is negative and \( A_y \) is positive, the vector is in the second quadrant. Adjust the angle by adding \(180^\circ\) to a negative result.
• If \( A_x \) is positive and \( A_y \) is negative, it is in the fourth quadrant. Adjust the angle by subtracting from \(360^\circ\).

Adjustments are crucial to ensure the vector's direction is represented accurately, aligning well with the standard coordinate system expectations.