Problem 55 - Fundamentals of Physics

Question

A particle undergoes three successive displacements in plane, as follows: \(\vec{d}_{1}, 4.00 \mathrm{~m}\) southwest; then \(\vec{d}_{2}, 5.00 \mathrm{~m}\) east; an finally \(\vec{d}_{3}, 6.00 \mathrm{~m}\) in a direction \(60.0^{\circ}\) north of east. Choose a coon dinate system with the \(y\) axis pointing north and the \(x\) axis pointin east. What are (a) the \(x\) component and (b) the \(y\) component of \(\vec{d}_{1}\) What are (c) the \(x\) component and (d) the \(y\) component of \(\vec{d}_{2}\) What are (e) the \(x\) component and (f) the \(y\) component of \(\vec{d}_{3}\) Next, consider the net displacement of the particle for the thre successive displacements. What are (g) the \(x\) component, (h) the component, (i) the magnitude, and (j) the direction of the net dis placement? If the particle is to return directly to the starting poin (k) how far and (1) in what direction should it move?

Step-by-Step Solution

Answer

Net displacement is 5.68 m at 25.1° north of east. Return by moving 5.68 m at 25.1° south of west.

Step 1: Identify Coordinates for Displacement 1

Since southwest is at an angle of 45 degrees away from both south and west, the components of \(\vec{d}_1\) can be calculated using trigonometry. Both components will be negative because southwest is in the negative regions of the coordinate system.\(d_{1x} = -4.00 \cos(45^\circ) = -2.83 \text{ m}\) and \(d_{1y} = -4.00 \sin(45^\circ) = -2.83 \text{ m}\).

Step 2: Calculate Coordinates for Displacement 2

Since \(\vec{d}_2\) is directed entirely along the east (positive x-axis), it has no y-component.\(d_{2x} = 5.00 \text{ m}\) and \(d_{2y} = 0 \text{ m}\).

Step 3: Determine Coordinates for Displacement 3

For \(\vec{d}_3\), resolve the vector using the given direction angle of 60 degrees north of east.\(d_{3x} = 6.00 \cos(60^\circ) = 3.00 \text{ m}\) and \(d_{3y} = 6.00 \sin(60^\circ) = 5.20 \text{ m}\).

Step 4: Calculate Net Displacement (x-component)

Add up all the x-components of the displacements to get the net x-component.\(d_{\text{net}, x} = d_{1x} + d_{2x} + d_{3x} = -2.83 + 5.00 + 3.00 = 5.17 \text{ m}\).

Step 5: Calculate Net Displacement (y-component)

Add up all the y-components of the displacements to get the net y-component.\(d_{\text{net}, y} = d_{1y} + d_{2y} + d_{3y} = -2.83 + 0 + 5.20 = 2.37 \text{ m}\).

Step 6: Calculate Magnitude of Net Displacement

Use the Pythagorean theorem to calculate the magnitude of the net displacement.\(d_{\text{net}} = \sqrt{(d_{\text{net}, x})^2 + (d_{\text{net}, y})^2} = \sqrt{(5.17)^2 + (2.37)^2} = 5.68 \text{ m}\).

Step 7: Determine Direction of Net Displacement

Find the direction of the net displacement by calculating the angle with respect to the x-axis.\(\theta = \tan^{-1}\left(\frac{d_{\text{net}, y}}{d_{\text{net}, x}}\right) = \tan^{-1}\left(\frac{2.37}{5.17}\right) = 25.1^\circ \text{ north of east}\).

Step 8: Calculate Return Displacement

To return to the starting point, the particle must move opposite the net displacement. The magnitude remains the same, 5.68 m, and the direction becomes \(180^\circ + 25.1^\circ = 205.1^\circ\) from east, or equivalently, it moves \(25.1^\circ\) south of west.

Key Concepts

Displacement Components

When a particle moves in a plane, each movement or displacement can be represented by a vector. A vector has both a magnitude and a direction. To analyze these displacements in a coordinate system, we break them down into two components, one along the x-axis and the other along the y-axis. This process is known as resolving a vector into its components.

For example, if a movement is described as southwest, it will form a right triangle with equal x and y components since it is at a 45-degree angle between the south and west directions. Therefore, both components will be negative because they point in the opposite direction of the positive x and y axes.

The mathematical tool we often use to resolve a vector is trigonometry, which helps us find these components using the cosine and sine of the angle with respect to the axis. In summary,

• The x-component is found using cosine: \(d_x = d \cos(\theta)\).
• The y-component is found using sine: \(d_y = d \sin(\theta)\).

where \(d\) is the magnitude of the vector and \(\theta\) is the angle of the vector.

Coordinate System

A coordinate system provides a framework for analyzing the position and movement of objects in space. It consists of two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). In this exercise, the x-axis was chosen to point east and the y-axis north. This setup allows us to easily assign directions and calculate movements.

When determining displacements, it is essential to maintain consistency with the coordinate system. For instance:

- Moving east corresponds to a positive x-component.
- Moving north corresponds to a positive y-component.
- Conversely, moving west or south accounts for negative components accordingly.
Thus, any vector can be translated into numerical values that reflect its direction and distance within this system. Establishing a clear coordinate system helps efficiently solve complex problems involving multiple movements.

Trigonometry

Trigonometry is a mathematical tool used to solve problems involving angles and distances. In the context of vector addition, it is crucial for breaking down vectors into their respective components. Trigonometry allows you to calculate distances using the relationships between the angles and the lengths of triangles.

Key trigonometric functions involved include:

• Cosine (\(\cos\)): Relates the adjacent side to the hypotenuse in a right-angled triangle.
• Sine (\(\sin\)): Relates the opposite side to the hypotenuse.
• Tangent (\(\tan\)): Relates the opposite side to the adjacent side.

Using the given angle in a displacement helps determine precise measurements for each component. For example, the eastward component of a vector pointing north of east is calculated using cosine, whereas the northward component uses sine.

Net Displacement

Net displacement is the overall change in position of an object after multiple movements. In vector addition, it is the sum of all individual displacements. To find the net displacement, you must add all x-components together and all y-components together.

Once the components are summed, the Pythagorean theorem helps calculate the magnitude of the net displacement, providing the total distance from the starting point to the ending location.

The formula for magnitude is:

• \[d_{\text{net}} = \sqrt{(d_{\text{net}, x})^2 + (d_{\text{net}, y})^2}\]

Moreover, to find the net direction, the arctangent function (\(\tan^{-1}\)) is employed to relate the y-component to the x-component. Thus, net displacement isn't just a distance, but a holistic description of an object's journey, encompassing both magnitude and direction. It's essential to understand this concept when analyzing complex movements.