Problem 19 - College Physics

Question

(a) What is the sum of \(\overrightarrow{\mathbf{A}}=3.0 \hat{\mathbf{x}}+5.0 \hat{\mathbf{y}}\) and \(\overrightarrow{\mathbf{B}}=1.0 \hat{\mathbf{x}}-3.0 \hat{\mathbf{y}}\) ? (b) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) ?

Step-by-Step Solution

Answer

(a) The sum is \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \). (b) Magnitude: \( 2\sqrt{5} \); Direction: \( 26.57^\circ \).

Step 1: Finding the Sum of Vectors

To find the sum of two vectors, we add the corresponding components. The vectors given are \( \overrightarrow{\mathbf{A}}=3.0 \hat{\mathbf{x}}+5.0 \hat{\mathbf{y}} \) and \( \overrightarrow{\mathbf{B}}=1.0 \hat{\mathbf{x}}-3.0 \hat{\mathbf{y}} \). Add the \( \hat{\mathbf{x}} \) components: \( 3.0 + 1.0 = 4.0 \). Add the \( \hat{\mathbf{y}} \) components: \( 5.0 - 3.0 = 2.0 \). Therefore, the sum is \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} = 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \).

Step 2: Calculating the Magnitude of the Sum

The magnitude of a vector \( \overrightarrow{\mathbf{C}} = a \hat{\mathbf{x}} + b \hat{\mathbf{y}} \) is given by \( \sqrt{a^2 + b^2} \). For \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} \), the sum is \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \). Therefore, the magnitude is \( \sqrt{(4.0)^2 + (2.0)^2} = \sqrt{16 + 4} = \sqrt{20} \). Therefore, the magnitude is \( 2\sqrt{5} \).

Step 3: Determining the Direction of the Sum

The direction \( \theta \) of a vector \( \overrightarrow{\mathbf{C}} \) can be found using \( \tan \theta = \frac{b}{a} \), where \( b \) is the \( \hat{\mathbf{y}} \) component and \( a \) is the \( \hat{\mathbf{x}} \) component. For the sum \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \), we calculate \( \tan \theta = \frac{2.0}{4.0} = \frac{1}{2} \). Therefore, \( \theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.57^\circ \) above the positive \( x \)-axis.

Key Concepts

Vector Components

Vector components are a way to express a vector using its projections along the coordinate axes. Vectors can be broken down into two parts: one along the x-axis and one along the y-axis. These are called vector components. Each component of a vector is scaled by a unit vector (like \( \hat{\mathbf{x}} \) and \( \hat{\mathbf{y}} \)), which shows the direction along the respective axis.

• The x-component represents how much the vector moves in the x-direction, while
• The y-component signifies the motion in the y-direction.

In the exercise, we added the components of vectors \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) to find their sum. By adding the x-components (3.0 + 1.0) and y-components (5.0 - 3.0), we effectively combined two vectors into one new vector: \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \). This method simplifies vector addition and enables us to work in a more component-wise straightforward manner.

Magnitude of a Vector

The magnitude of a vector reveals its size or length. It is a measure of how far the vector extends, regardless of its direction, denoted often simply as \(|\overrightarrow{\mathbf{C}}|\). It's determined using the Pythagorean theorem for the components, reminiscent of finding the length of the hypotenuse of a right triangle. To calculate the magnitude of a vector \(\overrightarrow{\mathbf{C}} = a \hat{\mathbf{x}} + b \hat{\mathbf{y}}\), use the formula:\[\sqrt{a^2 + b^2}\]For our sum vector \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \), its magnitude can be calculated as:\[\sqrt{(4.0)^2 + (2.0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\]This result, \(2\sqrt{5}\), demonstrates the length of the resulting vector. Magnitude helps us understand the vector's influence or effect without considering its direction.

Direction of a Vector

The direction of a vector describes where the vector points in the coordinate system. Unlike magnitude, which is purely about size, the direction tells you in which direction the vector is acting. This is usually measured as the angle \(\theta\) with respect to the positive x-axis. To find the direction, trigonometry is often employed with the tangent function:\[\tan \theta = \frac{b}{a}\]where \(b\) is the y-component and \(a\) is the x-component of the vector.Applying this to our result, we have:\[\tan \theta = \frac{2.0}{4.0} = \frac{1}{2}\]Thus, the angle is calculated as:\[\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.57^\circ\]This angle tells us that the resultant vector \(4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}}\) is directed approximately 26.57 degrees above the positive x-axis. Knowing the direction is crucial for fully understanding vector behavior in physics and engineering contexts.