Chapter 9: Problem 12
Find the \(x\) - and \(y\) -components of the given vectors by use of the trigonometric functions. The magnitude is shown first, followed by the direction as an angle in standard position. $$9750 \mathrm{N}, \theta=243.0^{\circ}$$
Short Answer
Expert verified
The x-component is -4423.5 N and the y-component is -8687.25 N.
Step by step solution
01
Identify Given Information
The vector has a magnitude of 9750 N and a direction of \(\theta = 243.0^{\circ}\). We need to find the \(x\)- and \(y\)-components of this vector.
02
Decompose Vector into Components
To find the components, we will use trigonometric functions. The cosine function gives the \(x\)-component and the sine function gives the \(y\)-component for a vector with a given angle \(\theta\).
03
Calculate the X-Component
The \(x\)-component is calculated by the formula \(x = \text{magnitude} \times \cos(\theta)\). Here, \(x = 9750 \times \cos(243.0^{\circ})\). First, calculate \(\cos(243.0^{\circ})\).
04
Value of Cosine
Since \(\theta = 243.0^{\circ}\) is in the third quadrant, where cosine values are negative, we find \(\cos(243.0^{\circ}) = \cos(243.0 - 180^{\circ}) = \cos(63.0^{\circ})\) and convert using the fact that cosine is negative in the third quadrant.
05
Compute Negative Cosine
Using a calculator, find \(\cos(63.0^{\circ}) \approx 0.4540\). Therefore, \(\cos(243.0^{\circ}) = -0.4540\).
06
Compute X-Component Value
Substitute the value back to find \(x = 9750 \times -0.4540 = -4423.5\, \text{N}\).
07
Calculate the Y-Component
The \(y\)-component is found using \(y = \text{magnitude} \times \sin(\theta)\). Use \(y = 9750 \times \sin(243.0^{\circ})\). First, calculate \(\sin(243.0^{\circ})\).
08
Value of Sine
Since \(243.0^{\circ}\) is in the third quadrant where sine values are negative, \(\sin(243.0^{\circ}) = \sin(243.0 - 180^{\circ}) = \sin(63.0^{\circ})\) and sine is negative in the third quadrant.
09
Compute Negative Sine
Using a calculator, find \(\sin(63.0^{\circ}) \approx 0.8910\). Therefore, \(\sin(243.0^{\circ}) = -0.8910\).
10
Compute Y-Component Value
Substitute the value back to find \(y = 9750 \times -0.8910 = -8687.25\, \text{N}\).
11
State the Components
The \(x\)-component of the vector is \(-4423.5\, \text{N}\) and the \(y\)-component is \(-8687.25\, \text{N}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Functions
When analyzing vectors, trigonometric functions are invaluable tools. They help us break a vector into its horizontal (x-component) and vertical (y-component) parts. This is particularly useful in physics and engineering applications where vectors often need to be decomposed to understand their effects in different directions.
The key trigonometric functions involved are:
The key trigonometric functions involved are:
- Cosine: This function relates the x-component of a vector to its magnitude and the angle (θ) from the standard position. The formula is given by: \( x = ext{magnitude} \times \cos(\theta)\).
- Sine: This function is associated with the y-component. The formula here is: \( y = ext{magnitude} \times \sin(\theta)\).
- Cosine is negative in the third quadrant.
- Sine is negative in the third quadrant as well.
Magnitude and Direction
Magnitude and direction form the essential attributes of a vector. The magnitude represents the size or length of the vector, and it is always a positive quantity.
Direction is often illustrated as the angle made by a vector with a reference line, usually measured from the positive x-axis in a counter-clockwise direction, known as standard position.
In our example, the magnitude of the vector is 9750 N, indicating how strong this force is. The direction is an angle, 243.0°, describing where this force is applied relative to a standard direction.
To visualize magnitude and direction:
Direction is often illustrated as the angle made by a vector with a reference line, usually measured from the positive x-axis in a counter-clockwise direction, known as standard position.
In our example, the magnitude of the vector is 9750 N, indicating how strong this force is. The direction is an angle, 243.0°, describing where this force is applied relative to a standard direction.
To visualize magnitude and direction:
- Magnitude: This can be thought of as the length or the size of the arrow that represents the vector on a graph.
- Direction: Often, this is like the aim of the arrow, indicating which way the arrowhead is pointing.
Angle in Standard Position
The angle in standard position is a way to systematically describe the rotation of a vector from a baseline—typically the positive
x-axis. It is always measured counterclockwise and helps in determining the trigonometric function values needed for vector decomposition.
This angle is significant because:
This angle is significant because:
- It gives a uniform reference point to measure all angles.
- In the unit circle, it helps to identify angle-related properties such as quadrant location.
- It assists in determining signs of the sine and cosine functions.
- Cosine is negative because, in this quadrant, x-values of unit circle coordinates are negative.
- Sine is also negative, as y-values here are negative too.