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Each problem below refers to a vector \(\mathbf{V}\) with magnitude \(|\mathbf{V}|\) that forms an angle \(\theta\) with the positive \(x\)-axis. In each case, give the magnitudes of the horizontal and vertical vector components of \(\mathbf{V}\), namely \(\mathbf{V}_{x}\) and \(\mathbf{V}_{y}\), respectively. \(V \mid=17.6, \theta=67.2^{\circ}\)

Short Answer

Expert verified
Horizontal component: 6.88, Vertical component: 16.20

Step by step solution

01

Understand Vector Components

The vector \(\mathbf{V}\) can be broken down into two components: the horizontal component \(\mathbf{V}_x\) and the vertical component \(\mathbf{V}_y\). These components can be calculated using trigonometric functions based on the given angle \( \theta\) and the magnitude of the vector \(|\mathbf{V}|\).
02

Calculate Horizontal Component

Use the cosine function to find the horizontal component \(\mathbf{V}_x\). The formula is given by:\[\mathbf{V}_x = |\mathbf{V}| \cdot \cos(\theta)\]Substitute the given values: \[|\mathbf{V}| = 17.6 \quad \text{and} \quad \theta = 67.2^{\circ}\]\[\mathbf{V}_x = 17.6 \cdot \cos(67.2^{\circ})\]
03

Perform Calculation for \(\mathbf{V}_x\)

Use a calculator to evaluate the cosine function:\[\cos(67.2^{\circ}) \approx 0.391\]Now calculate:\[\mathbf{V}_x = 17.6 \cdot 0.391 = 6.88\]
04

Calculate Vertical Component

Use the sine function to find the vertical component \(\mathbf{V}_y\). The formula is:\[\mathbf{V}_y = |\mathbf{V}| \cdot \sin(\theta)\]Substitute the given values:\[|\mathbf{V}| = 17.6 \quad \text{and} \quad \theta = 67.2^{\circ}\]\[\mathbf{V}_y = 17.6 \cdot \sin(67.2^{\circ})\]
05

Perform Calculation for \(\mathbf{V}_y\)

Use a calculator to evaluate the sine function:\[\sin(67.2^{\circ}) \approx 0.921\]Now calculate:\[\mathbf{V}_y = 17.6 \cdot 0.921 = 16.20\]
06

Review and Compile Solutions

The horizontal component \(\mathbf{V}_x\) is approximately 6.88, and the vertical component \(\mathbf{V}_y\) is approximately 16.20. Ensure that these values make sense based on the initial vector orientation and magnitude.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are crucial for breaking down vectors into their components when vectors are oriented at an angle. These functions include sine, cosine, and tangent, which relate the angles and sides of right triangles. When dealing with vector decomposition, the cosine function helps find the horizontal component, while the sine function is used for the vertical component.

In this context, the angle \( \theta \) created by vector \( \mathbf{V} \) with the horizontal axis is key. By applying these functions:
  • Cosine \( (\cos) \): This function measures the ratio of the adjacent side (horizontal component) over the hypotenuse (vector magnitude).
  • Sine \( (\sin) \): This function finds the ratio of the opposite side (vertical component) over the hypotenuse.
Utilizing these trigonometric functions allows for accurate calculations of vector components.
Horizontal Component
The horizontal component \( \mathbf{V}_x \) of a vector is the portion of the vector that runs parallel to the \( x \)-axis. To find this component, we use the cosine function because it relates the angle with the closest side of the triangle formed by the vector. The formula used is:
  • \( \mathbf{V}_x = |\mathbf{V}| \cdot \cos(\theta) \)
Here:
  • \(|\mathbf{V}|\) is the magnitude of the vector.
  • \(\theta\) is the angle the vector makes with the horizontal axis.
By substituting the given values \( |\mathbf{V}| = 17.6 \) and \( \theta = 67.2^{\circ} \), you can determine that \( \mathbf{V}_x \approx 6.88 \). This value represents how much of the vector's strength is directed along the horizontal direction.
Vertical Component
The vertical component \( \mathbf{V}_y \) of a vector signifies the vector's projection in the direction perpendicular to the horizontal, essentially along the \( y \)-axis. The sine function is employed here, as it correlates the angle to the opposite leg of the reference triangle. The calculation is straightforward:
  • \( \mathbf{V}_y = |\mathbf{V}| \cdot \sin(\theta) \)
For the given vector with a magnitude \( |\mathbf{V}| = 17.6 \) and angle \( \theta = 67.2^{\circ} \), compute \( \mathbf{V}_y \approx 16.20 \). This component indicates how much of the vector is acting in the upward or downward direction in a plane. Understanding these components provides insights into the vector's overall behavior in the specified orientation.
Magnitude of a Vector
The magnitude of a vector is a measure of its size or length, irrespective of its direction. It represents the total strength or intensity of the vector. In vector decomposition, the magnitude is crucial because it serves as a base to calculate the vector's horizontal and vertical components using trigonometric functions.

The magnitude often comes from calculations based on the vector's two-dimensional or three-dimensional coordinates. In problems where the vector is defined by components in the plane:
  • The magnitude \( |\mathbf{V}| \) combines both horizontal and vertical effects. Think of it as the hypotenuse in a right triangle where the vector components are the other sides.
In the given problem, \( |\mathbf{V}| = 17.6 \) provides the overall influence, which is split into effectively measurable parts using trigonometric relationships. Computing and understanding the magnitude is key in physics and engineering, where precise force measurements are pivotal.

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Most popular questions from this chapter

Distance and Bearing Problems 17 through 22 involve directions in the form of bearing, which we defined in this section. Remember that bearing is always measured from a north-south line. A man wandering in the desert walks \(2.3\) miles in the direction \(\mathrm{S} \mathrm{} 31^{\circ} \mathrm{W}\). He then turns \(90^{\circ}\) and walks \(3.5\) miles in the direction \(\mathrm{N} \mathrm{} 59^{\circ} \mathrm{W}\). At that time, how far is he from his starting point, and what is his bearing from his starting point?

Draw \(135^{\circ}\) in standard position, locate a convenient point on the terminal side, and then find \(\sin 135^{\circ}, \cos 135^{\circ}\), and \(\tan 135^{\circ}\).

Refer to right triangle \(A B C\) with \(C=90^{\circ}\). Begin each problem by drawing a picture of the triangle with both the given and asked for information labeled appropriately. Also, write your answers for angles in decimal degrees.If \(c=5.678 \mathrm{ft}\) and \(a=4.567 \mathrm{ft}\), find \(A\)

Use a calculator to find a value of \(\theta\) between \(0^{\circ}\) and \(90^{\circ}\) that satisfies each statement below. Write your answer in degrees and minutes rounded to the nearest minute. $$ \csc \theta=7.0683 $$

To further justify the Cofunction Theorem, use your calculator to find a value for each pair of trigonometric functions below. In each case, the trigonometric functions are cofunctions of one another, and the angles are complementary angles. Round your answers to four places past the decimal point. $$ \sec 56.7^{\circ}, \csc 33.3^{\circ} $$

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