Question 1169043: Using trigonometry determine the resultant vector
a) 8m [N] and 12 m [W]
b) 15N [30degreesSofE] and 8 N [E]
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's determine the resultant vector for each case using trigonometry.
**a) 8 m [N] and 12 m [W]**
1. **Represent the vectors in component form:**
* Vector 1 (A): 8 m [N] lies along the positive y-axis.
$\mathbf{A} = (0, 8)$
* Vector 2 (B): 12 m [W] lies along the negative x-axis.
$\mathbf{B} = (-12, 0)$
2. **Add the component vectors to find the resultant vector (R):**
$\mathbf{R} = \mathbf{A} + \mathbf{B} = (0 + (-12), 8 + 0) = (-12, 8)$
3. **Calculate the magnitude of the resultant vector:**
$|\mathbf{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(-12)^2 + (8)^2} = \sqrt{144 + 64} = \sqrt{208}$
$|\mathbf{R}| = \sqrt{16 \times 13} = 4\sqrt{13} \approx 14.42$ m
4. **Calculate the direction of the resultant vector:**
The angle $\theta$ that the resultant vector makes with the positive x-axis can be found using the arctangent function:
$\tan(\theta) = \frac{R_y}{R_x} = \frac{8}{-12} = -\frac{2}{3}$
$\theta = \arctan(-\frac{2}{3}) \approx -33.69^\circ$
Since $R_x$ is negative and $R_y$ is positive, the resultant vector lies in the second quadrant. To find the angle relative to the positive x-axis, we add $180^\circ$:
$\theta_{actual} = -33.69^\circ + 180^\circ = 146.31^\circ$
To express the direction in a standard compass bearing format, we can find the angle relative to the North or West direction. The angle with the negative x-axis (West) is $33.69^\circ$. Therefore, the direction is approximately $33.69^\circ$ North of West [N $33.69^\circ$ W].
Alternatively, the angle with the positive y-axis (North) is $90^\circ - 33.69^\circ = 56.31^\circ$. Therefore, the direction is approximately $56.31^\circ$ West of North [W $56.31^\circ$ N].
Both [N $33.69^\circ$ W] and [W $56.31^\circ$ N] are correct ways to express the direction.
**Resultant vector for a): Approximately 14.42 m [N 33.69° W]**
**b) 15 N [30° S of E] and 8 N [E]**
1. **Represent the vectors in component form:**
* Vector 1 (A): 15 N [30° S of E]
$A_x = |\mathbf{A}| \cos(30^\circ) = 15 \times \frac{\sqrt{3}}{2} \approx 15 \times 0.866 = 12.99$ N (East)
$A_y = -|\mathbf{A}| \sin(30^\circ) = -15 \times \frac{1}{2} = -7.5$ N (South)
$\mathbf{A} = (12.99, -7.5)$
* Vector 2 (B): 8 N [E] lies along the positive x-axis.
$\mathbf{B} = (8, 0)$
2. **Add the component vectors to find the resultant vector (R):**
$\mathbf{R} = \mathbf{A} + \mathbf{B} = (12.99 + 8, -7.5 + 0) = (20.99, -7.5)$
3. **Calculate the magnitude of the resultant vector:**
$|\mathbf{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(20.99)^2 + (-7.5)^2} = \sqrt{440.5801 + 56.25} = \sqrt{496.8301} \approx 22.29$ N
4. **Calculate the direction of the resultant vector:**
The angle $\theta$ that the resultant vector makes with the positive x-axis (East) can be found using the arctangent function:
$\tan(\theta) = \frac{R_y}{R_x} = \frac{-7.5}{20.99} \approx -0.3573$
$\theta = \arctan(-0.3573) \approx -19.67^\circ$
The negative angle indicates that the resultant vector is below the positive x-axis. Therefore, the direction is approximately $19.67^\circ$ South of East [S $19.67^\circ$ E].
**Resultant vector for b): Approximately 22.29 N [S 19.67° E]**
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