Problem 73 - University Physics with Modern Physics

Question

As noted in Exercise \(1.33,\) a spelunker is surveying a cave. She follows a passage 180 \(\mathrm{m}\) straight west, then 210 \(\mathrm{m}\) in a direction \(45^{\circ}\) east of south, and then 280 \(\mathrm{m}\) at \(30^{\circ}\) east of north. After a fourth unmeasured displacement she finds herself back where she started. Use the method of components to determine the magnitude and direction of the fourth displacement. Draw the vector addition diagram and show that it is in qualitative agreement with your numerical solution.

Step-by-Step Solution

Answer

The fourth displacement has a magnitude of approximately 143.39 m and is directed 40.6° north of west.

Step 1: Define the Coordinate System

To solve the problem using vector components, first, establish a coordinate system. Let north be the positive y-direction, east be the positive x-direction, south be the negative y-direction, and west be the negative x-direction.

Step 2: Break Down Each Displacement into Components

- **Vector 1:** 180 m west is purely horizontal, so its components areen: - - **Vector 2:** 210 m at \( 45 ^{\circ} \) east of south: - Components: - **Vector 3:** 280 m at \( 30 ^{\circ} \) east of north: - Components:

Step 3: Calculate Components for Each Vector

- **Vector 1 Components:** - \( x_1 = -180 \), \( y_1 = 0 \)- **Vector 2 Components:** - \( x_2 = 210 \times \sin(45^{\circ}) = 148.49 \) - \( y_2 = -210 \times \cos(45^{\circ}) = -148.49 \)- **Vector 3 Components:** - \( x_3 = 280 \times \sin(30^{\circ}) = 140 \) - \( y_3 = 280 \times \cos(30^{\circ}) = 242.49 \)

Step 4: Calculate Net Displacement Components

Sum the x-components:\[ x_{total} = x_1 + x_2 + x_3 = -180 + 148.49 + 140 = 108.49 \]Sum the y-components:\[ y_{total} = y_1 + y_2 + y_3 = 0 - 148.49 + 242.49 = 94 \]

Step 5: Determine Magnitude and Direction of the Fourth Displacement

Since the final position is the starting point, the resultant displacement must be zero.x-components of fourth vector \( x_4 = - x_{total} = -108.49 \)y-components of fourth vector \( y_4 = - y_{total} = -94 \)Magnitude is:\[ \sqrt{x_4^2 + y_4^2} = \sqrt{108.49^2 + 94^2} \approx 143.39 \]Direction (angle \( \theta \)) from the negative x-axis:\[ \theta = \tan^{-1}\left(\frac{y_4}{x_4}\right) = \tan^{-1}\left(\frac{-94}{-108.49}\right) \approx 40.6^{\circ} \text{ north of west} \]

Step 6: Plot the Vector Addition Diagram

Create a diagram showing vectors 1, 2, and 3, properly added head-to-tail. Vector 4 should be the one that closes the loop, allowing the total displacement to equal zero. This visual should corroborate the numerical solution found.

Key Concepts

Coordinate System

Understanding the coordinate system is crucial when dealing with vector problems in physics. It allows us to break down complex paths into simpler, manageable parts. In this exercise, we use a 2D coordinate system where:

• North is defined as the positive y-direction.
• East is the positive x-direction.
• South and West are negative y- and x-directions, respectively.

By assigning directions as positive or negative, we can effectively translate real-world movements into coordinates. This not only helps in visualizing the movement but also simplifies calculations by making use of basic arithmetic and vector mathematics.

Vector Components

Vectors are entities that have both magnitude and direction. Breaking them into components simplifies the process of adding them, especially when they are not aligned along the axes. Each vector can be split into:

• An x-component, which aligns with the horizontal (east-west) axis.
• A y-component, which aligns with the vertical (north-south) axis.

For instance, a vector of 210 meters at a 45-degree angle east of south has components:

• An x-component calculated using \( 210 \times \sin(45^\circ) \), which gives approximately 148.49 meters east.
• A y-component calculated using \( -210 \times \cos(45^\circ) \), resulting in approximately -148.49 meters south.

Understanding this breakdown is essential for vector addition, making it simpler and more accurate.

Displacement Vectors

Displacement vectors are specifically used to describe change in position. Unlike distance, which is a scalar, displacement considers direction and is represented by vectors. In the exercise, the spelunker's path consists of multiple vectors. To find out how far off a specific point she is, we add these vectors:

• Vector 1 solely influences the x-direction.
• Vectors 2 and 3, with angles involved, affect both x and y-directions.

By adding the components of each vector, we determine the overall displacement. Here, since she returns to her start, the sum of these vectors equals zero. The challenge is finding the unknown vector that completes the return journey.

Trigonometry in Physics

Trigonometry plays a significant role in solving physics problems involving vectors. Through the use of sine, cosine, and tangent functions, we can derive or infer unknown elements of a vector. Take angles into account along with the magnitude to break vectors into their components.

• For any vector, the sine of the angle helps find the component perpendicular to the reference axis (x-component if angle given with y-axis).
• The cosine of the angle finds the component parallel to the reference axis (y-component if angle given with x-axis).
• The tangent function is especially useful to find the angle of direction when components are known, using \( \tan^{-1}(\frac{y}{x}) \) as demonstrated in determining the angle of the fourth vector in our exercise.

These functions allow precise calculations of complex movements, providing a structured approach to tackle vector addition.