Question
Use a step by step method and solve five problem on triangle law of vector addition
Answers
GPT-4o mini
The triangle law of vector addition states that if two vectors are represented as two sides of a triangle taken in order, then the resultant vector can be represented by the closing side of the triangle taken in the opposite order.
Let's solve five problems using this law, step by step.
### Problem 1
**Given two vectors:**
- Vector A: 5 units, at an angle of 0° (along the positive x-axis).
- Vector B: 7 units, at an angle of 90° (along the positive y-axis).
**Steps to solve:**
1. **Draw the vectors:** Start with vector A along the x-axis. From the head of vector A, draw vector B vertically upwards.
2. **Form the triangle:** The tail of vector A and the tail of vector B will meet at one point, forming a right triangle.
3. **Determine the resultant vector:** The resultant vector R can be found by drawing a line from the tail of the first vector (A) to the head of the second vector (B).
4. **Calculate the magnitude of R using Pythagorean theorem:**
\[
R = \sqrt{A^2 + B^2} = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74} \approx 8.6 \text{ units}
\]
5. **Calculate the direction (angle θ) using trigonometry:**
\[
\tan(\theta) = \frac{B}{A} = \frac{7}{5}
\]
\[
\theta = \tan^{-1}\left(\frac{7}{5}\right) \approx 54.5°
\]
**Result:** Magnitude of R ≈ 8.6 units at an angle of 54.5° from the x-axis.
---
### Problem 2
**Given two vectors:**
- Vector A: 10 units, at an angle of 30°.
- Vector B: 10 units, at an angle of 120°.
**Steps to solve:**
1. **Resolve the vectors into components:**
- \( A_x = 10 \cos(30°) = 10 \times \frac{\sqrt{3}}{2} \approx 8.66 \)
- \( A_y = 10 \sin(30°) = 10 \times \frac{1}{2} = 5 \)
- \( B_x = 10 \cos(120°) = 10 \times -\frac{1}{2} = -5 \)
- \( B_y = 10 \sin(120°) = 10 \times \frac{\sqrt{3}}{2} \approx 8.66 \)
2. **Add the components:**
- \( R_x = A_x + B_x = 8.66 - 5 = 3.66 \)
- \( R_y = A_y + B_y = 5 + 8.66 \approx 13.66 \)
3. **Calculate the magnitude of resultant R:**
\[
R = \sqrt{(R_x)^2 + (R_y)^2} \approx \sqrt{(3.66)^2 + (13.66)^2} \approx \sqrt{13.46 + 186.96} = \sqrt{200.42} \approx 14.14 \text{ units}
\]
4. **Calculate the angle θ:**
\[
\tan (\theta) = \frac{R_y}{R_x} = \frac{13.66}{3.66}
\]
\[
\theta = \tan^{-1}\left(\frac{13.66}{3.66}\right) \approx 74.74°
\]
**Result:** Magnitude of R ≈ 14.14 units at an angle of 74.74°.
---
### Problem 3
**Given two vectors:**
- Vector A: 8 units, at an angle of 0°.
- Vector B: 15 units, at an angle of 180° (negative x-axis).
**Steps to solve:**
1. **Locate the vectors:** Vector A points to the right (positive x), and vector B points to the left (negative x).
2. **Add the components:**
- \( A_x = 8, \quad A_y = 0 \)
- \( B_x = -15, \quad B_y = 0 \)
Therefore,
- \( R_x = 8 - 15 = -7 \)
- \( R_y = 0 + 0 = 0 \)
3. **Determine the magnitude of R:**
\[
R = \sqrt{(-7)^2 + 0^2} = \sqrt{49} = 7 \text{ units}
\]
4. **Determine the direction:**
Since R is along the negative x-axis, the angle is:
\[
\theta = 180°
\]
**Result:** Magnitude of R = 7 units at an angle of 180°.
---
### Problem 4
**Given two vectors:**
- Vector A: 12 units, at an angle of 45°.
- Vector B: 12 units, at an angle of 225°.
**Steps to solve:**
1. **Resolve the vectors into components:**
- For A:
- \( A_x = 12 \cos(45°) \approx 12 \times 0.707 \approx 8.49 \)
- \( A_y = 12 \sin(45°) \approx 12 \times 0.707 \approx 8.49 \)
- For B:
- \( B_x = 12 \cos(225°) \approx 12 \times -0.707 \approx -8.49 \)
- \( B_y = 12 \sin(225°) \approx 12 \times -0.707 \approx -8.49 \)
2. **Add the components:**
- \( R_x = A_x + B_x \approx 8.49 - 8.49 = 0 \)
- \( R_y = A_y + B_y \approx 8.49 - 8.49 = 0 \)
3. **Magnitude of R:**
\[
R = \sqrt{(0)^2 + (0)^2} = 0 \text{ units}
\]
4. **Direction:**
Since R is 0, it has no direction.
**Result:** Magnitude of R = 0 units.
---
### Problem 5
**Given two vectors:**
- Vector A: 9 units, at an angle of 60°.
- Vector B: 5 units, at an angle of 180°.
**Steps to solve:**
1. **Resolve the vectors into components:**
- For A:
- \( A_x = 9 \cos(60°) = 9 \times 0.5 = 4.5 \)
- \( A_y = 9 \sin(60°) = 9 \times \frac{\sqrt{3}}{2} \approx 7.79 \)
- For B:
- \( B_x = 5 \cos(180°) = 5 \times -1 = -5 \)
- \( B_y = 5 \sin(180°) = 0 \)
2. **Add the components:**
- \( R_x = A_x + B_x = 4.5 - 5 = -0.5 \)
- \( R_y = A_y + B_y = 7.79 + 0 = 7.79 \)
3. **Calculate the magnitude of R:**
\[
R = \sqrt{(-0.5)^2 + (7.79)^2} \approx \sqrt{0.25 + 60.75} \approx \sqrt{61} \approx 7.81 \text{ units}
\]
4. **Calculate the angle θ:**
\[
\tan(\theta) = \frac{R_y}{R_x} = \frac{7.79}{-0.5}
\]
This indicates it's in the second quadrant:
\[
\theta = 180° + \tan^{-1}\left(\frac{7.79}{-0.5}\right) \approx 180° + 99.46° = 279.46°
\]
**Result:** Magnitude of R ≈ 7.81 units at an angle of approximately 279.46°.
These problems encapsulate the processes involved in applying the triangle law of vector addition.
Let's solve five problems using this law, step by step.
### Problem 1
**Given two vectors:**
- Vector A: 5 units, at an angle of 0° (along the positive x-axis).
- Vector B: 7 units, at an angle of 90° (along the positive y-axis).
**Steps to solve:**
1. **Draw the vectors:** Start with vector A along the x-axis. From the head of vector A, draw vector B vertically upwards.
2. **Form the triangle:** The tail of vector A and the tail of vector B will meet at one point, forming a right triangle.
3. **Determine the resultant vector:** The resultant vector R can be found by drawing a line from the tail of the first vector (A) to the head of the second vector (B).
4. **Calculate the magnitude of R using Pythagorean theorem:**
\[
R = \sqrt{A^2 + B^2} = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74} \approx 8.6 \text{ units}
\]
5. **Calculate the direction (angle θ) using trigonometry:**
\[
\tan(\theta) = \frac{B}{A} = \frac{7}{5}
\]
\[
\theta = \tan^{-1}\left(\frac{7}{5}\right) \approx 54.5°
\]
**Result:** Magnitude of R ≈ 8.6 units at an angle of 54.5° from the x-axis.
---
### Problem 2
**Given two vectors:**
- Vector A: 10 units, at an angle of 30°.
- Vector B: 10 units, at an angle of 120°.
**Steps to solve:**
1. **Resolve the vectors into components:**
- \( A_x = 10 \cos(30°) = 10 \times \frac{\sqrt{3}}{2} \approx 8.66 \)
- \( A_y = 10 \sin(30°) = 10 \times \frac{1}{2} = 5 \)
- \( B_x = 10 \cos(120°) = 10 \times -\frac{1}{2} = -5 \)
- \( B_y = 10 \sin(120°) = 10 \times \frac{\sqrt{3}}{2} \approx 8.66 \)
2. **Add the components:**
- \( R_x = A_x + B_x = 8.66 - 5 = 3.66 \)
- \( R_y = A_y + B_y = 5 + 8.66 \approx 13.66 \)
3. **Calculate the magnitude of resultant R:**
\[
R = \sqrt{(R_x)^2 + (R_y)^2} \approx \sqrt{(3.66)^2 + (13.66)^2} \approx \sqrt{13.46 + 186.96} = \sqrt{200.42} \approx 14.14 \text{ units}
\]
4. **Calculate the angle θ:**
\[
\tan (\theta) = \frac{R_y}{R_x} = \frac{13.66}{3.66}
\]
\[
\theta = \tan^{-1}\left(\frac{13.66}{3.66}\right) \approx 74.74°
\]
**Result:** Magnitude of R ≈ 14.14 units at an angle of 74.74°.
---
### Problem 3
**Given two vectors:**
- Vector A: 8 units, at an angle of 0°.
- Vector B: 15 units, at an angle of 180° (negative x-axis).
**Steps to solve:**
1. **Locate the vectors:** Vector A points to the right (positive x), and vector B points to the left (negative x).
2. **Add the components:**
- \( A_x = 8, \quad A_y = 0 \)
- \( B_x = -15, \quad B_y = 0 \)
Therefore,
- \( R_x = 8 - 15 = -7 \)
- \( R_y = 0 + 0 = 0 \)
3. **Determine the magnitude of R:**
\[
R = \sqrt{(-7)^2 + 0^2} = \sqrt{49} = 7 \text{ units}
\]
4. **Determine the direction:**
Since R is along the negative x-axis, the angle is:
\[
\theta = 180°
\]
**Result:** Magnitude of R = 7 units at an angle of 180°.
---
### Problem 4
**Given two vectors:**
- Vector A: 12 units, at an angle of 45°.
- Vector B: 12 units, at an angle of 225°.
**Steps to solve:**
1. **Resolve the vectors into components:**
- For A:
- \( A_x = 12 \cos(45°) \approx 12 \times 0.707 \approx 8.49 \)
- \( A_y = 12 \sin(45°) \approx 12 \times 0.707 \approx 8.49 \)
- For B:
- \( B_x = 12 \cos(225°) \approx 12 \times -0.707 \approx -8.49 \)
- \( B_y = 12 \sin(225°) \approx 12 \times -0.707 \approx -8.49 \)
2. **Add the components:**
- \( R_x = A_x + B_x \approx 8.49 - 8.49 = 0 \)
- \( R_y = A_y + B_y \approx 8.49 - 8.49 = 0 \)
3. **Magnitude of R:**
\[
R = \sqrt{(0)^2 + (0)^2} = 0 \text{ units}
\]
4. **Direction:**
Since R is 0, it has no direction.
**Result:** Magnitude of R = 0 units.
---
### Problem 5
**Given two vectors:**
- Vector A: 9 units, at an angle of 60°.
- Vector B: 5 units, at an angle of 180°.
**Steps to solve:**
1. **Resolve the vectors into components:**
- For A:
- \( A_x = 9 \cos(60°) = 9 \times 0.5 = 4.5 \)
- \( A_y = 9 \sin(60°) = 9 \times \frac{\sqrt{3}}{2} \approx 7.79 \)
- For B:
- \( B_x = 5 \cos(180°) = 5 \times -1 = -5 \)
- \( B_y = 5 \sin(180°) = 0 \)
2. **Add the components:**
- \( R_x = A_x + B_x = 4.5 - 5 = -0.5 \)
- \( R_y = A_y + B_y = 7.79 + 0 = 7.79 \)
3. **Calculate the magnitude of R:**
\[
R = \sqrt{(-0.5)^2 + (7.79)^2} \approx \sqrt{0.25 + 60.75} \approx \sqrt{61} \approx 7.81 \text{ units}
\]
4. **Calculate the angle θ:**
\[
\tan(\theta) = \frac{R_y}{R_x} = \frac{7.79}{-0.5}
\]
This indicates it's in the second quadrant:
\[
\theta = 180° + \tan^{-1}\left(\frac{7.79}{-0.5}\right) \approx 180° + 99.46° = 279.46°
\]
**Result:** Magnitude of R ≈ 7.81 units at an angle of approximately 279.46°.
These problems encapsulate the processes involved in applying the triangle law of vector addition.