Problem 72 - Fundamentals of Physics

Question

A fire ant, searching for hot sauce in a picnic area, goes through three displacements along level ground: \(\vec{d}_{1}\) for \(0.40 \mathrm{~m}\) southwest (that is, at \(45^{\circ}\) from directly south and from directly west), \(\vec{d}_{2}\) for \(0.50 \mathrm{~m}\) due east, \(\vec{d}_{3}\) for \(0.60 \mathrm{~m}\) at \(60^{\circ}\) north of east. Let the positive \(x\) direction be east and the positive \(y\) direction be north. What are (a) the \(x\) component and (b) the \(y\) component of \(\vec{d}_{1} ?\) Next, what are (c) the \(x\) component and (d) the \(y\) component of \(\vec{d}_{2}\) ? Also, what are (e) the \(x\) component and (f) the \(y\) component of \(\vec{d}_{3}\) ?

Step-by-Step Solution

Answer

\(d_{1x} = -0.283 \, \text{m}, \ d_{1y} = -0.283 \, \text{m}; \ d_{2x} = 0.50 \, \text{m}, \ d_{2y} = 0 \, \text{m};\ d_{3x} = 0.30 \, \text{m}, \ d_{3y} = 0.52 \, \text{m}.\)

Step 1: Understanding the given Information

We have three displacements to analyze. Displacement \( \vec{d}_1 \) is 0.40 m southwest at a 45° angle from directly south and directly west. Displacement \( \vec{d}_2 \) is 0.50 m due east, and displacement \( \vec{d}_3 \) is 0.60 m at 60° north of east.

Step 2: Finding displacement \( \vec{d}_1 \) components

Since \( \vec{d}_1 \) is at a 45° angle southwest, its x and y components can be found using trigonometry. The x component, \( d_{1x} \), is \(-0.40 \cos(45°) \), and the y component, \( d_{1y} \), is \(-0.40 \sin(45°) \). Calculate these using the identity that \( \cos(45°) = \sin(45°) = \frac{\sqrt{2}}{2} \).

Step 3: Calculating \( d_{1x} \) and \( d_{1y} \)

Substitute the values: \( d_{1x} = -0.40 \times \frac{\sqrt{2}}{2} = -0.40 \times 0.707 \approx -0.283 \, \text{m} \) and \( \ d_{1y} = -0.40 \times \frac{\sqrt{2}}{2} = -0.283 \, \text{m}. \) So, both components of \( \vec{d}_1 \) are \(-0.283 \, \text{m}. \)

Step 4: Finding displacement \( \vec{d}_2 \) components

Displacement \( \vec{d}_2 \) is due east, which means it has no y component. Thus, \( d_{2x} = 0.50 \, \text{m} \) and \( d_{2y} = 0 \, \text{m}. \)

Step 5: Finding displacement \( \vec{d}_3 \) components

The displacement \( \vec{d}_3 \) is at 60° north of east, hence we use trigonometry: the x component \( d_{3x} = 0.60 \cos(60°) \), and the y component \( d_{3y} = 0.60 \sin(60°) \). Knowing \( \cos(60°) = 0.5 \) and \( \sin(60°) = \frac{\sqrt{3}}{2} \approx 0.866 \).

Step 6: Calculating \( d_{3x} \) and \( d_{3y} \)

Substitute \( \cos(60°) \) and \( \sin(60°) \) into the equations: \( \ d_{3x} = 0.60 \times 0.5 = 0.30 \, \text{m} \) and \( \ d_{3y} = 0.60 \times 0.866 = 0.52 \, \text{m}. \)

Key Concepts

Trigonometry in Physics

Trigonometry is a critical tool in physics, especially when working with vectors. Vectors have both magnitude and direction, making them ideal for representing physical quantities like displacement. By using trigonometric functions such as sine, cosine, and tangent, we can break down vectors into their components along the x and y axes. This process is known as vector decomposition.

Let's consider the example of a displacement vector \( \vec{d}_1 \) that is 0.40 m in the southwest direction. Southwest implies that the angle of travel is 45° between the south and west directions. By using trigonometric functions, specifically cosine and sine, we can find that both the x (horizontal) and y (vertical) components of this vector are 0.40 m multiplied by \( \cos(45°) \) and \( \sin(45°) \) respectively.

These components help us to further understand the displacement in each direction. Hence, for vector \( \vec{d}_1 \):

• The x-component is \( -0.40 \cos(45°) \approx -0.283 \text{ m} \)
• The y-component is \( -0.40 \sin(45°) \approx -0.283 \text{ m} \)

Displacement Vectors

Displacement vectors play a central role in describing motion. They tell us not just how far an object has moved, but also the direction of movement. This is crucial in physics because the direction affects how we need to think about forces, motion, and energy.

In our example, we have three displacement vectors:

• \( \vec{d}_1 \): 0.40 m southwest
• \( \vec{d}_2 \): 0.50 m due east
• \( \vec{d}_3 \): 0.60 m at 60° north of east

By breaking these vectors into components, we can easily add them or perform other calculations needed to solve physics problems.

For instance,

• Vector \( \vec{d}_2 \) has only an x-component, \( d_{2x} = 0.50 \text{ m} \), because it points due east.
• Vector \( \vec{d}_3 \) requires cosine and sine functions to find its components: \( d_{3x} = 0.60 \cos(60°) = 0.30 \text{ m} \) and \( d_{3y} = 0.60 \sin(60°) = 0.52 \text{ m} \).

Understanding and analyzing displacement vectors is invaluable, as it allows us to interpret movement across various directions and forces at play.

Coordinate System in Physics

A suitable coordinate system is vital for solving physics problems effectively. It helps us define directions and reference points for measuring physical quantities. In this exercise, a 2D Cartesian coordinate system is used, where the x-axis points east and the y-axis points north.

By assigning positive values to these directions and stating so clearly, we can then determine the signs for vector components. For example:

• When a vector points southwest, like \( \vec{d}_1 \), its components are negative in both the x and y directions because southwest combines negative x (west) and negative y (south) directions.
• In contrast, a vector pointing north of east, like \( \vec{d}_3 \), has a positive x-component (east) and a positive y-component (north).

These consistent rules for sign conventions allow physicists and students to solve problems systematically and avoid confusion. Understanding and correctly using coordinate systems is foundational for solving vector and other physics problems.