Problem 27
Question
Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j. $$ |\mathbf{v}|=40, \quad \theta=30^{\circ} $$
Step-by-Step Solution
Verified Answer
The vector in terms of i and j is \( 20\sqrt{3} \mathbf{i} + 20\mathbf{j} \).
1Step 1: Identify vector components
The vector components can be found using trigonometric functions: cosine for the horizontal component and sine for the vertical component. The formulas used are \( v_x = |\mathbf{v}| \cdot \cos(\theta) \) for the horizontal component and \( v_y = |\mathbf{v}| \cdot \sin(\theta) \) for the vertical component.
2Step 2: Calculate the horizontal component
Using the formula for the horizontal component, we have: \[ v_x = 40 \cdot \cos(30^{\circ}) \]Cosine of 30 degrees is \( \frac{\sqrt{3}}{2} \), so: \[ v_x = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} \]
3Step 3: Calculate the vertical component
Using the formula for the vertical component, we have: \[ v_y = 40 \cdot \sin(30^{\circ}) \]Sine of 30 degrees is \( \frac{1}{2} \), so: \[ v_y = 40 \cdot \frac{1}{2} = 20 \]
4Step 4: Write the vector in terms of i and j
The vector \( \mathbf{v} \) can be written as a sum of its components using unit vectors \( \mathbf{i} \) and \( \mathbf{j} \). Thus, we have:\[ \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} = 20\sqrt{3} \mathbf{i} + 20\mathbf{j} \]
Key Concepts
Trigonometric FunctionsUnit VectorsCosine and SineHorizontal and Vertical Components
Trigonometric Functions
Trigonometric functions are essential tools in mathematics that help us relate angles to side lengths in right triangles. They include sine, cosine, and tangent. These functions are particularly important in physics and engineering because they allow us to break down a vector into components.
The main trigonometric functions used for vector components are:
The main trigonometric functions used for vector components are:
- Sine (\( \sin \theta \)): Used to calculate the vertical component of a vector, related to the opposite side in a right triangle.
- Cosine (\( \cos \theta \)): Used for determining the horizontal component, associated with the adjacent side.
Unit Vectors
Unit vectors play a key role in expressing direction without altering magnitude. They are essential in denoting the direction of the vector components. By definition, a unit vector has a magnitude of one.
In two dimensions, the principal unit vectors are:
In two dimensions, the principal unit vectors are:
- \( \mathbf{i} \): Represents the direction of the x-axis (horizontal direction).
- \( \mathbf{j} \): Indicates the direction of the y-axis (vertical direction).
Cosine and Sine
Cosine and sine are trigonometric functions that help calculate the components of a vector.
Cosine (\( \cos \theta \):
When you have a vector with a given angle θ with the horizontal axis, you use the cosine function to find the horizontal component. This is done with the formula: \( v_x = |\mathbf{v}| \cdot \cos(\theta) \).
For example, with a vector magnitude of 40 and an angle of 30 degrees, this becomes:\( v_x = 40 \cdot \cos(30^{\circ}) = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} \).
Sine (\( \sin \theta \):
Similarly, the sine function is used to determine the vertical component of a vector through the relationship: \( v_y = |\mathbf{v}| \cdot \sin(\theta) \).
Continuing our example:\( v_y = 40 \cdot \sin(30^{\circ}) = 40 \cdot \frac{1}{2} = 20 \).
Grasping these functions' application is vital for resolving vectors in physics and engineering.
Cosine (\( \cos \theta \):
When you have a vector with a given angle θ with the horizontal axis, you use the cosine function to find the horizontal component. This is done with the formula: \( v_x = |\mathbf{v}| \cdot \cos(\theta) \).
For example, with a vector magnitude of 40 and an angle of 30 degrees, this becomes:\( v_x = 40 \cdot \cos(30^{\circ}) = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} \).
Sine (\( \sin \theta \):
Similarly, the sine function is used to determine the vertical component of a vector through the relationship: \( v_y = |\mathbf{v}| \cdot \sin(\theta) \).
Continuing our example:\( v_y = 40 \cdot \sin(30^{\circ}) = 40 \cdot \frac{1}{2} = 20 \).
Grasping these functions' application is vital for resolving vectors in physics and engineering.
Horizontal and Vertical Components
Vectors can be split into horizontal and vertical components to simplify analysis and solve practical problems. This is typically done using trigonometric functions based on the vector's angle with respect to an axis.
- Horizontal Component (\( v_x \): The horizontal component of a vector describes its movement along the x-axis. It is found using the cosine function as in \( v_x = |\mathbf{v}| \cdot \cos(\theta) \).
- Vertical Component (\( v_y \): This component represents the vector's influence along the y-axis and is calculated using the sine function: \( v_y = |\mathbf{v}| \cdot \sin(\theta) \).
Other exercises in this chapter
Problem 26
A point is graphed in polar form. Find its rectangular coordinates. $$ (6,2 \pi / 3) $$
View solution Problem 26
15–36 Sketch the graph of the polar equation. $$r=\cos \theta-1$$
View solution Problem 27
23-28 (a) Calculate proj, \(\mathbf{u} .\) (b) Resolve \(\mathbf{u}\) into \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) where \(\mathbf{u}_{1}\) is parallel to \(\
View solution Problem 27
\(25-48=\) Write the complex number in polar form with argument \(\theta\) between 0 and 2\(\pi .\) $$ \sqrt{2}-\sqrt{2} i $$
View solution