A force `F _(1),` of magnitude `2.0N` and directed due east is exerted on an object. A second force exerted on the object is `F _(2)=2. 0 N,` due north. Wha tis the magnitude and direction of a third force, `F _(3),` which must be exerted on the object so that the resultant force is zero ?

Options

A. `1.4N, 45^(@)` north of east
B. `1.4 N, 45 ^(@)` south of west
C. `2.8 N, 45 ^(@)` north of east
D. `2.8 N , 45 ^(@)` south of west

Answer

To solve the problem, we need to find the third force \( F_3 \) that will make the resultant force on the object zero when two forces \( F_1 \) and \( F_2 \) are already acting on it.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The first force \( F_1 \) has a magnitude of \( 2.0 \, \text{N} \) and is directed due east.
- The second force \( F_2 \) has a magnitude of \( 2.0 \, \text{N} \) and is directed due north.

2. **Determine the Resultant of \( F_1 \) and \( F_2 \)**:
- Since \( F_1 \) and \( F_2 \) are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant force \( R \).
\[ R = \sqrt{F_1^2 + F_2^2} = \sqrt{(2.0 \, \text{N})^2 + (2.0 \, \text{N})^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \, \text{N} \approx 2.83 \, \text{N} \]

3. **Determine the Direction of the Resultant Force**:
- The angle \( \theta \) of the resultant force with respect to the east direction can be found using:
\[ \tan \theta = \frac{F_2}{F_1} = \frac{2.0 \, \text{N}}{2.0 \, \text{N}} = 1 \]
- Therefore, \( \theta = 45^\circ \) north of east.

4. **Find the Third Force \( F_3 \)**:
- To make the resultant force zero, the third force \( F_3 \) must be equal in magnitude and opposite in direction to the resultant \( R \).
- Thus, \( F_3 \) must have a magnitude of \( 2\sqrt{2} \, \text{N} \) and will be directed \( 45^\circ \) south of west (or equivalently \( 225^\circ \) from the east direction).

5. **Final Result**:
- The magnitude of the third force \( F_3 \) is \( 2.83 \, \text{N} \) and its direction is \( 45^\circ \) south of west.

### Summary:
- Magnitude of \( F_3 \): \( 2.83 \, \text{N} \)
- Direction of \( F_3 \): \( 45^\circ \) south of west