Problem 45 - College Physics

Question

Vector \(A\) has components \(A_{x}=1.30 \mathrm{cm}, \quad A_{y}=\) \(2.25 \mathrm{cm} ; \quad\) vector \(\vec{B}\) has components \(B_{x}=4.10 \mathrm{cm}\) \(B_{y}=-3.75 \mathrm{cm} .\) Find \((\mathrm{a})\) the components of the vector sum \(A+B ;\) (b) the magnitude and direction of \(\vec{A}+\vec{B}\) (c) the components of the vector difference \(\vec{B}-\vec{A}\) (d) the magnitude and direction of \(\vec{\boldsymbol{B}}-\vec{\boldsymbol{A}}\)

Step-by-Step Solution

Answer

(a) \( \vec{A} + \vec{B} = (5.40\,\text{cm}, -1.50\,\text{cm}) \). (b) Magnitude \( 5.60\,\text{cm}, \) direction \( -15.5^\circ \). (c) \( \vec{B} - \vec{A} = (2.80\,\text{cm}, -6.00\,\text{cm}) \). (d) Magnitude \( 6.62\,\text{cm}, \) direction \( -64.9^\circ \).

Step 1: Find components of vector sum A+B

To find the components of the vector sum \( \vec{A} + \vec{B} \), we simply add the corresponding components of vectors \( \vec{A} \) and \( \vec{B} \). \[ (\vec{A} + \vec{B})_x = A_x + B_x = 1.30\,\text{cm} + 4.10\,\text{cm} = 5.40\,\text{cm} \] \[ (\vec{A} + \vec{B})_y = A_y + B_y = 2.25\,\text{cm} - 3.75\,\text{cm} = -1.50\,\text{cm} \] Thus, the components of vector \( \vec{A} + \vec{B} \) are \( 5.40\,\text{cm} \) and \( -1.50\,\text{cm} \).

Step 2: Calculate magnitude of vector A+B

The magnitude of the vector \( \vec{A} + \vec{B} \) can be found using the Pythagorean theorem. \[ |\vec{A} + \vec{B}| = \sqrt{(\vec{A} + \vec{B})_x^2 + (\vec{A} + \vec{B})_y^2} \] Substituting the values, we get \[ |\vec{A} + \vec{B}| = \sqrt{(5.40)^2 + (-1.50)^2} = \sqrt{29.16 + 2.25} = \sqrt{31.41} \approx 5.60\,\text{cm} \]

Step 3: Determine direction of vector A+B

The direction angle \( \theta \) of the vector \( \vec{A} + \vec{B} \) is found using the tangent function: \[ \theta = \tan^{-1}\left(\frac{(\vec{A} + \vec{B})_y}{(\vec{A} + \vec{B})_x}\right) = \tan^{-1}\left(\frac{-1.50}{5.40}\right) \] Using a calculator, we find \[ \theta \approx \tan^{-1}(-0.278) \approx -15.5^\circ \] Since the angle is negative in the standard position (counter-clockwise from the positive x-axis), it actually points in the fourth quadrant.

Step 4: Find components of vector difference B-A

Similarly, for the vector difference \( \vec{B} - \vec{A} \), we subtract the components of \( \vec{A} \) from \( \vec{B} \). \[ (\vec{B} - \vec{A})_x = B_x - A_x = 4.10\,\text{cm} - 1.30\,\text{cm} = 2.80\,\text{cm} \] \[ (\vec{B} - \vec{A})_y = B_y - A_y = -3.75\,\text{cm} - 2.25\,\text{cm} = -6.00\,\text{cm} \] Hence, the components of vector \( \vec{B} - \vec{A} \) are \( 2.80\,\text{cm} \) and \( -6.00\,\text{cm} \).

Step 5: Calculate magnitude of vector B-A

The magnitude of the vector \( \vec{B} - \vec{A} \) is calculated similarly: \[ |\vec{B} - \vec{A}| = \sqrt{(2.80)^2 + (-6.00)^2} = \sqrt{7.84 + 36.00} = \sqrt{43.84} \approx 6.62\,\text{cm} \]

Step 6: Determine direction of vector B-A

For vector \( \vec{B} - \vec{A} \), the direction \( \theta \) is \[ \theta = \tan^{-1}\left(\frac{(\vec{B} - \vec{A})_y}{(\vec{B} - \vec{A})_x}\right) = \tan^{-1}\left(\frac{-6.00}{2.80}\right) \] Calculating, we find \[ \theta \approx \tan^{-1}(-2.14) \approx -64.9^\circ \] This angle is in the fourth quadrant, as the \( x \)-component is positive and the \( y \)-component is negative.

Key Concepts

Vector Addition

When we add two vectors, we combine their respective components. This process involves adding each corresponding component to form a new vector. For instance, when vectors \( \vec{A} \) and \( \vec{B} \) are added together:

• We find the x-component of the vector sum by adding \( A_x \) and \( B_x \).
• Similarly, the y-component is the sum of \( A_y \) and \( B_y \).

As an example, given \( A_x = 1.30 \, \text{cm} \) and \( B_x = 4.10 \, \text{cm} \), their sum becomes:\[(\vec{A} + \vec{B})_x = 1.30 + 4.10 = 5.40 \, \text{cm}\]For the y-components, with \( A_y = 2.25 \, \text{cm} \) and \( B_y = -3.75 \, \text{cm} \), we have:\[(\vec{A} + \vec{B})_y = 2.25 - 3.75 = -1.50 \, \text{cm}\]The result gives us a new vector with components \( (5.40, -1.50) \). This process is straightforward and essential for understanding how multiple directions play together.

Vector Subtraction

Subtracting one vector from another is similar to addition but involves reversing the direction of the second vector and then performing vector addition. For vectors \( \vec{B} \) and \( \vec{A} \):

• Subtract the x-component of \( \vec{A} \) from \( \vec{B} \).
• Subtract the y-component of \( \vec{A} \) from \( \vec{B} \).

The operation looks like this for our given vectors:\[(\vec{B} - \vec{A})_x = 4.10 - 1.30 = 2.80 \, \text{cm}\]For the y-component:\[(\vec{B} - \vec{A})_y = -3.75 - 2.25 = -6.00 \, \text{cm}\]This method helps clarify direction changes and can be visualized as moving forces or displacements in opposite directions, giving us the new vector \((2.80, -6.00)\). It's a crucial skill to understand how to find resultant effects or changes effectively.

Magnitude Calculation

Finding the magnitude of a vector is about measuring its length or size using its components. This is done through the Pythagorean theorem for the given components of a vector.For a vector \((x, y)\), the magnitude \(|\vec{V}|\) is calculated as:\[|\vec{V}| = \sqrt{x^2 + y^2}\]Applying this to our vector \( \vec{A} + \vec{B} \) with components \( 5.40 \, \text{cm} \) and \( -1.50 \, \text{cm} \), we derive:\[|\vec{A} + \vec{B}| = \sqrt{5.40^2 + (-1.50)^2} = \sqrt{29.16 + 2.25} = \sqrt{31.41} \approx 5.60 \, \text{cm}\]This process is vital for practical applications where knowing the exact length or size, such as in physics for determining force magnitudes, is necessary.

Direction Calculation

Calculating the direction of a vector involves finding the angle it makes with the horizontal axis. This is typically done using trigonometric functions—primarily the tangent function, which relates the opposite and adjacent sides of a right triangle.For the direction \( \theta \) of a vector with components \((x, y)\), we use:\[\theta = \tan^{-1}\left(\frac{y}{x}\right)\]Applying this to our vector difference \( \vec{B} - \vec{A} \) with \( x = 2.80 \, \text{cm} \) and \( y = -6.00 \, \text{cm} \), we calculate:\[\theta = \tan^{-1}\left(\frac{-6.00}{2.80}\right) \approx \tan^{-1}(-2.14) \approx -64.9^\circ\]Because the angle is negative, the vector points in the fourth quadrant. Understanding this helps in making accurate predictions about the behavior of vectors, such as flight paths or motion trajectories, in various fields.