Problem 10 - Fundamentals of Physics

Question

Find the (a) \(x\), (b) \(y\), and (c) \(z\) components of the sum \(\vec{r}\) of the displacements \(\vec{c}\) and \(\vec{d}\) whose components in meters are \(c_{x}=7.4, c_{y}=-3.8, c_{z}=-6.1 ; d_{x}=4.4, d_{y}=-2.0, d_{z}=3.3\) \(-11\) ssM (a) In unit-vector notation, what is the sum \(\vec{a}+\vec{b}\) if \(\vec{a}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}+(3.0 \mathrm{~m}) \hat{\mathrm{j}}\) and \(\vec{b}=(-13.0 \mathrm{~m}) \hat{\mathrm{i}}+(7.0 \mathrm{~m}) \hat{\mathrm{j}} ?\) What are the (b) magnitude and (c) direction of \(\vec{a}+\vec{b}\) ?

Step-by-Step Solution

Answer

(a) \(\vec{r} = (11.8 \, \mathrm{m}, -5.8 \, \mathrm{m}, -2.8 \, \mathrm{m})\); (b) Magnitude is approximately 13.45 m; (c) Direction adjusted angle from the x-axis.

Step 1: Find x-component of vector sum \(\vec{r}\)

The x-component of \(\vec{r}\) is the sum of the x-components of vectors \(\vec{c}\) and \(\vec{d}\). Calculate: \[ r_x = c_x + d_x \] \[ r_x = 7.4 + 4.4 = 11.8 \]

Step 2: Find y-component of vector sum \(\vec{r}\)

The y-component of \(\vec{r}\) is the sum of the y-components of vectors \(\vec{c}\) and \(\vec{d}\). Calculate: \[ r_y = c_y + d_y \] \[ r_y = -3.8 + (-2.0) = -5.8 \]

Step 3: Find z-component of vector sum \(\vec{r}\)

The z-component of \(\vec{r}\) is the sum of the z-components of vectors \(\vec{c}\) and \(\vec{d}\). Calculate: \[ r_z = c_z + d_z \] \[ r_z = -6.1 + 3.3 = -2.8 \]

Step 4: Sum of vectors \(\vec{a}\) and \(\vec{b}\) in unit-vector notation

Compute the sum of vectors \(\vec{a}\) and \(\vec{b}\) in unit-vector notation. \[ \vec{a} = (4.0 \, \mathrm{m}) \hat{\mathrm{i}} + (3.0 \, \mathrm{m}) \hat{\mathrm{j}} \]\[ \vec{b} = (-13.0 \, \mathrm{m}) \hat{\mathrm{i}} + (7.0 \, \mathrm{m}) \hat{\mathrm{j}} \]Adding them gives: \[ \vec{a} + \vec{b} = (4.0 - 13.0) \hat{\mathrm{i}} + (3.0 + 7.0) \hat{\mathrm{j}} \]\[ \vec{a} + \vec{b} = (-9.0 \, \mathrm{m}) \hat{\mathrm{i}} + (10.0 \, \mathrm{m}) \hat{\mathrm{j}} \]

Step 5: Calculate magnitude of \(\vec{a} + \vec{b}\)

The magnitude is calculated using the Pythagorean theorem: \[ |\vec{a} + \vec{b}| = \sqrt{(-9.0)^2 + (10.0)^2} \]\[ |\vec{a} + \vec{b}| = \sqrt{81 + 100} \]\[ |\vec{a} + \vec{b}| = \sqrt{181} \approx 13.45 \, \mathrm{m} \]

Step 6: Determine the direction of \(\vec{a} + \vec{b}\)

The direction is given by the angle \(\theta\) with respect to the negative x-axis (as the x-component is negative): \[ \theta = \tan^{-1}\left(\frac{10.0}{-9.0}\right) \]\[ \theta = \tan^{-1}(-1.11) \]The original result is in the second quadrant, thus adjust the angle by adding 180 degrees to the calculated angle.

Key Concepts

Components of Vectors

Understanding how vectors are made up of components is key in vector addition. A vector, like a displacement, has both a magnitude and a direction. To simplify mathematical operations, each vector is typically broken down into its orthogonal components. For three-dimensional vectors, these components are typically represented along the x, y, and z axes.

• The x-component refers to the part of the vector that runs parallel to the x-axis.
• The y-component aligns with the y-axis.
• The z-component, similarly, aligns with the z-axis.

When adding or subtracting vectors, you handle each component separately. For example, when adding two vectors, you would sum the individual x-components together, the y-components, and the z-components. This results in a new vector whose x, y, and z components are the sums of the respective components of the original vectors.
The exercise you are working with illustrates this beautifully by taking vectors \( \vec{c} \) and \( \vec{d} \) and summing their components to find \( \vec{r} \). This method allows us to handle complex vector operations more efficiently.

Magnitude and Direction

Once you have the components of a vector, you can determine its magnitude and direction to understand its overall effect. The magnitude of a vector (\(|\vec{v}|\)) is essentially its length, often found using the Pythagorean Theorem. This theorem is pretty handy and works here—they are like spatial distances in mathematics.
For a two-dimensional vector with components \((a, b)\), the magnitude is found using: \ |\vec{v}| = \sqrt{a^2 + b^2} \ Conversely, in three-dimensional space, the formula extends to include the z-component: \ |\vec{v}| = \sqrt{a^2 + b^2 + c^2} \ Magnitude provides a scalar value, which simply means a single value rather than a vector with direction. Direction, on the other hand, reveals exactly where this displacement is pointing. It is often expressed as an angle. In the scenario given, angles can be calculated using trigonometric functions like arctangent. The combination of magnitude and direction fully describes the vector's effect, akin to saying not just how far, but where to go.

Unit-Vector Notation

Unit-vector notation is a more straightforward form of representing vectors. This method uses the fundamental unit vectors \( \hat{\mathrm{i}}, \hat{\mathrm{j}}, \hat{\mathrm{k}} \), which point in the directions of the x, y, and z axes, respectively. They each have a magnitude of one but provide the direction for the vector.
In unit-vector notation, a vector \( \vec{v} \) is expressed as:

• \( \vec{v} = v_x \hat{\mathrm{i}} + v_y \hat{\mathrm{j}} + v_z \hat{\mathrm{k}} \)

where \( v_x, v_y, \) and \( v_z \) are the components of the vector. This corresponds to the contribution along each axis.
Unit vectors make vector arithmetic, such as addition and subtraction, very manageable. When working with vectors \( \vec{a} \) and \( \vec{b} \) in unit-vector form, you simply add their corresponding coefficients. The result remains in unit-vector form, preserving both magnitude (through individual components) and direction (through the unit vectors). Thus, unit-vector notation simplifies the calculation and provides a clear visualizable way of understanding vector components.