A bouy is attached to three tugboats by three ropes. The tugboats are engaged in a tug-0f-war. One tugboat pulls west on the buoy with a force `vec(F)_(1)` of magnitude `1000N`. The second tugboat pulls south on the buoy with a force `vec(F)_(2)` of magnitude `2000N`. The third tugboat pulls northeast (that is, half way between north and east), with a force `vec(F)_(3)` of magnitude `2000N`
a. Express each force in unit vector from `(hat(i),hat(j))`.
b. Calculate the magnitude of the resultant force.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Express each force in unit vector form

1. **Identify the forces and their directions**:
- Tugboat 1 pulls west with a force \( F_1 = 1000 \, \text{N} \).
- Tugboat 2 pulls south with a force \( F_2 = 2000 \, \text{N} \).
- Tugboat 3 pulls northeast with a force \( F_3 = 2000 \, \text{N} \).

2. **Convert each force into unit vector form**:
- For \( F_1 \) (west):
\[ \vec{F_1} = -1000 \hat{i} \, \text{N} \]
- For \( F_2 \) (south):
\[ \vec{F_2} = -2000 \hat{j} \, \text{N} \]
- For \( F_3 \) (northeast):
- The angle for northeast is \( 45^\circ \).
- The components can be calculated using:
\[ \vec{F_3} = 2000 \cos(45^\circ) \hat{i} + 2000 \sin(45^\circ) \hat{j} \]
Since \( \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \vec{F_3} = 2000 \left(\frac{1}{\sqrt{2}}\right) \hat{i} + 2000 \left(\frac{1}{\sqrt{2}}\right) \hat{j} = 1000\sqrt{2} \hat{i} + 1000\sqrt{2} \hat{j} \]

3. **Final expressions for each force**:
\[ \vec{F_1} = -1000 \hat{i} \, \text{N} \]
\[ \vec{F_2} = -2000 \hat{j} \, \text{N} \]
\[ \vec{F_3} = 1000\sqrt{2} \hat{i} + 1000\sqrt{2} \hat{j} \, \text{N} \]

### Part (b): Calculate the magnitude of the resultant force

1. **Sum the forces**:
\[ \vec{F}_{\text{net}} = \vec{F_1} + \vec{F_2} + \vec{F_3} \]
Substituting the expressions:
\[ \vec{F}_{\text{net}} = (-1000 \hat{i}) + (-2000 \hat{j}) + (1000\sqrt{2} \hat{i} + 1000\sqrt{2} \hat{j}) \]

2. **Combine the \( \hat{i} \) and \( \hat{j} \) components**:
- For \( \hat{i} \):
\[ F_{\text{net}, i} = -1000 + 1000\sqrt{2} \]
- For \( \hat{j} \):
\[ F_{\text{net}, j} = -2000 + 1000\sqrt{2} \]

3. **Resultant force vector**:
\[ \vec{F}_{\text{net}} = \left(-1000 + 1000\sqrt{2}\right) \hat{i} + \left(-2000 + 1000\sqrt{2}\right) \hat{j} \]

4. **Calculate the magnitude of the resultant force**:
\[ |\vec{F}_{\text{net}}| = \sqrt{(F_{\text{net}, i})^2 + (F_{\text{net}, j})^2} \]
Substituting the values:
\[ |\vec{F}_{\text{net}}| = \sqrt{\left(-1000 + 1000\sqrt{2}\right)^2 + \left(-2000 + 1000\sqrt{2}\right)^2} \]

5. **Simplifying the expression**:
- Calculate \( (-1000 + 1000\sqrt{2})^2 \) and \( (-2000 + 1000\sqrt{2})^2 \) separately, then add them and take the square root.