Problem 38 - College Physics

Question

In each of the cases that follow, the components of a vector \(A\) are given. 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) \(A_{x}=-3.0 \mathrm{~km}, A_{y}=-6.0 \mathrm{~km}\) (c) \(A_{x}=9.0 \mathrm{~m} / \mathrm{s}, A_{y}=-17 \mathrm{~m} / \mathrm{s}\) (d) \(A_{x}=-8.0 \mathrm{~N}, A_{y}=12 \mathrm{~N}\)

Step-by-Step Solution

Answer

Magnitudes are approximately: (a) 6.40 m, (b) 6.71 km, (c) 19.23 m/s, (d) 14.42 N. Angles: (a) 51.34°, (b) 243.43°, (c) 297.27°, (d) 123.69°.

Step 1: Compute Magnitude for Vector (a)

The components of vector \( A \) are given as \( A_x = 4.0 \, \text{m} \) and \( A_y = 5.0 \, \text{m} \). To find the magnitude \( |\vec{A}| \), use the formula: \(|\vec{A}| = \sqrt{A_x^2 + A_y^2}\). Substitute the values: \( |\vec{A}| = \sqrt{4.0^2 + 5.0^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.40 \, \text{m} \).

Step 2: Calculate Angle for Vector (a)

The angle \( \theta \) the vector makes with the +x axis can be found using the tangent function: \( \tan \theta = \frac{A_y}{A_x} \). Thus, \( \theta = \tan^{-1}\left(\frac{5.0}{4.0}\right) \). Calculate \( \theta \): \( \theta \approx \tan^{-1}(1.25) \approx 51.34^\circ \).

Step 3: Magnitude and Angle for Vector (b)

For vector (b), \( A_x = -3.0 \, \text{km} \) and \( A_y = -6.0 \, \text{km} \). The magnitude is \(|\vec{A}| = \sqrt{(-3.0)^2 + (-6.0)^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.71 \, \text{km} \). The angle is \( \theta = \tan^{-1}\left(\frac{-6.0}{-3.0}\right) = \tan^{-1}(2) \approx 63.43^\circ \). Because both components are negative, the vector is in the third quadrant, so adjust: \( \theta \approx 180^\circ + 63.43^\circ = 243.43^\circ \).

Step 4: Magnitude and Angle for Vector (c)

For vector (c), \( A_x = 9.0 \, \text{m/s} \) and \( A_y = -17 \, \text{m/s} \). Calculate the magnitude: \( |\vec{A}| = \sqrt{9.0^2 + (-17)^2} = \sqrt{81 + 289} = \sqrt{370} \approx 19.23 \, \text{m/s} \). The angle is \( \theta = \tan^{-1}\left(\frac{-17}{9.0}\right) \approx \tan^{-1}(-1.89) \approx -62.73^\circ \). Adjust for the fourth quadrant by adding 360°: \( \theta \approx 297.27^\circ \).

Step 5: Magnitude and Angle for Vector (d)

For vector (d), \( A_x = -8.0 \, \text{N} \) and \( A_y = 12.0 \, \text{N} \). Compute the magnitude: \( |\vec{A}| = \sqrt{(-8.0)^2 + 12.0^2} = \sqrt{64 + 144} = \sqrt{208} \approx 14.42 \, \text{N} \). The angle is \( \theta = \tan^{-1}\left(\frac{12.0}{-8.0}\right) \approx \tan^{-1}(-1.5) \approx -56.31^\circ \). Adjust for the second quadrant: \( \theta \approx 180^\circ - 56.31^\circ = 123.69^\circ \).

Step 6: Sketch and Verify

Draw vectors (a), (b), (c), and (d) with their respective angles and magnitudes. Confirm that the vectors align within the appropriate quadrants: (a) in the first quadrant, (b) in the third, (c) in the fourth, and (d) in the second. This ensures the calculated angles are reasonable.

Key Concepts

Trigonometry in Physics

Trigonometry plays a crucial role in physics, especially when dealing with vectors. Vectors are mathematical entities with both magnitude and direction. In physics, they help represent various quantities such as force, velocity, and acceleration. To solve problems involving vectors, understanding basic trigonometric concepts like sine, cosine, and tangent is essential.

When given the components of a vector, trigonometry allows us to find both the length (magnitude) and the direction relative to a reference axis, often the x-axis. The magnitude of a vector can be calculated using the Pythagorean theorem:

• Magnitude: \(|\vec{A}| = \sqrt{A_x^2 + A_y^2}\)
• Direction: The angle \(\theta\) with the x-axis is found using the tangent function: \(\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)\)

By applying these trigonometric concepts, physics students can navigate complex vector problems with more ease and precision.

Vector Components

Vectors can be broken down into components, which simplifies analysis and calculations. A vector in a 2D plane can be expressed in terms of its horizontal and vertical components, typically denoted as \(A_x\) and \(A_y\). These components represent the effect of the vector in each respective direction along the x-axis and y-axis.

Understanding vector components is essential for physics problem-solving because it allows the decomposition of a vector into parts that can be individually analyzed using algebraic methods. Each component can be used to determine the vector's overall effect, making it easier to handle situations involving forces, motion, and other vector quantities.

Working with components involves converting between vector form and component form. To do this, you need to use:

• Coordinate axes as references
• Trigonometric identities to resolve the vector into its components: \( A_x = |\vec{A}|\cos(\theta) \) and \( A_y = |\vec{A}|\sin(\theta) \)

This methodical breakdown makes vectors more manageable and empowers students to solve physics problems with greater clarity and accuracy.

Analytical Geometry

Analytical geometry, or coordinate geometry, merges algebra with geometry and is vital for understanding the properties of shapes and their relations in space. When applied to vectors, it offers a graphical and algebraic way to view and manipulate vector operations.

This branch of mathematics allows us to visualize vectors as arrows in a coordinate system, with direction from one point to another and a fixed magnitude. In this space:

• Vectors are represented as points or positions determined by their endpoints.
• Their components are the differences in the x and y coordinates of these points.

Using analytical geometry, vectors can be added or subtracted visually or through algebraic methods, helping to confirm solutions and increase comprehension through graphical representations. This approach simplifies complex systems into geometric interpretations, which can be a more intuitive way to understand vector relationships and transformations.

Physics Problem-Solving

Solving physics problems requires a structured approach that combines mathematical tools and physical insight. With vectors involved, the process is even more nuanced, demanding not just mathematical computations but also spatial visualization skills.

Effective physics problem-solving involves:

• Identifying what quantities are vectorial and determining their components.
• Using trigonometry to find magnitudes and angles, applying \(|\vec{A}| = \sqrt{A_x^2 + A_y^2}\) and \(\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)\), recognizing quadrant placements for accurate angle adjustments.
• Checking results through sketches and ensuring consistency in quadrant placement and expected vector directions.

Approaching physics problems with a clear method not only ensures accuracy but also builds intuition about the physical world. With practice, this structured reasoning becomes second nature, aiding in tackling more complex challenges with confidence.