Question
find the magnitude of the vector sum -->a+-->b.
vector a is 0.111m long in a 90.0 degrees direction.
vector b is 0.234m long in a 300 degrees direction.
vector a is 0.111m long in a 90.0 degrees direction.
vector b is 0.234m long in a 300 degrees direction.
Answers
GPT-4o
To find the magnitude of the vector sum \(\vec{a} + \vec{b}\), follow these steps:
1. **Convert each vector into its components** (using sine and cosine functions).
2. **Add the components** of the vectors.
3. **Calculate the magnitude** of the resultant vector.
### Step 1: Vector Components
For vector \(\vec{a}\):
- Magnitude \(a = 0.111 \, \text{m}\)
- Direction \(90.0^\circ\)
The components of vector \(\vec{a}\) are:
\[ \vec{a}_x = a \cos(90^\circ) = 0.111 \cos(90^\circ) = 0 \, \text{m} \]
\[ \vec{a}_y = a \sin(90^\circ) = 0.111 \sin(90^\circ) = 0.111 \, \text{m} \]
For vector \(\vec{b}\):
- Magnitude \(b = 0.234 \, \text{m}\)
- Direction \(300^\circ\)
The components of vector \(\vec{b}\) are:
\[ \vec{b}_x = b \cos(300^\circ) = 0.234 \cos(300^\circ) = 0.234 \cos(-60^\circ) = 0.234 \times 0.5 = 0.117 \, \text{m} \]
\[ \vec{b}_y = b \sin(300^\circ) = 0.234 \sin(300^\circ) = 0.234 \sin(-60^\circ) = 0.234 \times (-\frac{\sqrt{3}}{2}) = -0.2025 \, \text{m} \]
### Step 2: Add the Components
Now, add the components of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{R}_x = \vec{a}_x + \vec{b}_x = 0 + 0.117 = 0.117 \, \text{m} \]
\[ \vec{R}_y = \vec{a}_y + \vec{b}_y = 0.111 + (-0.2025) = -0.0915 \, \text{m} \]
### Step 3: Magnitude of the Resultant Vector
The magnitude of the vector sum \(\vec{R}\) is given by:
\[ |\vec{R}| = \sqrt{(\vec{R}_x)^2 + (\vec{R}_y)^2}
= \sqrt{(0.117)^2 + (-0.0915)^2}
= \sqrt{0.013689 + 0.00837225}
= \sqrt{0.02206125}
= 0.1485 \, \text{m} \]
So, the magnitude of the vector sum \(\vec{a} + \vec{b}\) is \(0.1485 \, \text{m}\).
1. **Convert each vector into its components** (using sine and cosine functions).
2. **Add the components** of the vectors.
3. **Calculate the magnitude** of the resultant vector.
### Step 1: Vector Components
For vector \(\vec{a}\):
- Magnitude \(a = 0.111 \, \text{m}\)
- Direction \(90.0^\circ\)
The components of vector \(\vec{a}\) are:
\[ \vec{a}_x = a \cos(90^\circ) = 0.111 \cos(90^\circ) = 0 \, \text{m} \]
\[ \vec{a}_y = a \sin(90^\circ) = 0.111 \sin(90^\circ) = 0.111 \, \text{m} \]
For vector \(\vec{b}\):
- Magnitude \(b = 0.234 \, \text{m}\)
- Direction \(300^\circ\)
The components of vector \(\vec{b}\) are:
\[ \vec{b}_x = b \cos(300^\circ) = 0.234 \cos(300^\circ) = 0.234 \cos(-60^\circ) = 0.234 \times 0.5 = 0.117 \, \text{m} \]
\[ \vec{b}_y = b \sin(300^\circ) = 0.234 \sin(300^\circ) = 0.234 \sin(-60^\circ) = 0.234 \times (-\frac{\sqrt{3}}{2}) = -0.2025 \, \text{m} \]
### Step 2: Add the Components
Now, add the components of \(\vec{a}\) and \(\vec{b}\):
\[ \vec{R}_x = \vec{a}_x + \vec{b}_x = 0 + 0.117 = 0.117 \, \text{m} \]
\[ \vec{R}_y = \vec{a}_y + \vec{b}_y = 0.111 + (-0.2025) = -0.0915 \, \text{m} \]
### Step 3: Magnitude of the Resultant Vector
The magnitude of the vector sum \(\vec{R}\) is given by:
\[ |\vec{R}| = \sqrt{(\vec{R}_x)^2 + (\vec{R}_y)^2}
= \sqrt{(0.117)^2 + (-0.0915)^2}
= \sqrt{0.013689 + 0.00837225}
= \sqrt{0.02206125}
= 0.1485 \, \text{m} \]
So, the magnitude of the vector sum \(\vec{a} + \vec{b}\) is \(0.1485 \, \text{m}\).