A body placed in free space, is simultaneously acted upon by three forces `vec(F_(1)),vec(F_(2)) " and " vec(F_(3))`. The body is in equilibrium and the forces `vec(F_(1)) " and "vec(F_(2))` are known ot be 36 N due north and 27 N due east respectively. Which of the following best describes the force `vec(F_(3))`?

Options

A. 36N due south
B. 53N due `60^(@)` south of east
C. 45N due `53^(@)` south of west
D. 45N due `37^(@)` north of west.

Answer

To solve the problem, we need to find the force \( \vec{F_3} \) acting on a body that is in equilibrium, given two other forces \( \vec{F_1} \) and \( \vec{F_2} \).

### Step-by-Step Solution:

1. **Identify the Forces**:
- \( \vec{F_1} = 36 \, \text{N} \) (due North)
- \( \vec{F_2} = 27 \, \text{N} \) (due East)

2. **Equilibrium Condition**:
- For the body to be in equilibrium, the vector sum of all forces acting on it must be zero:
\[ \vec{F_1} + \vec{F_2} + \vec{F_3} = 0 \]
- Rearranging gives:
\[ \vec{F_3} = -(\vec{F_1} + \vec{F_2}) \]

3. **Calculate the Resultant of \( \vec{F_1} \) and \( \vec{F_2} \)**:
- The resultant force \( \vec{R} \) can be calculated using the Pythagorean theorem since \( \vec{F_1} \) and \( \vec{F_2} \) are perpendicular to each other:
\[ R = \sqrt{F_1^2 + F_2^2} = \sqrt{(36 \, \text{N})^2 + (27 \, \text{N})^2} \]
- Calculating the squares:
\[ R = \sqrt{1296 + 729} = \sqrt{2025} = 45 \, \text{N} \]

4. **Direction of \( \vec{F_3} \)**:
- The direction of \( \vec{F_3} \) is opposite to that of the resultant \( \vec{R} \).
- To find the angle \( \theta \) that \( \vec{R} \) makes with the east direction, we use:
\[ \tan(\theta) = \frac{F_1}{F_2} = \frac{36}{27} = \frac{4}{3} \]
- Therefore, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ \).

5. **Determine the Direction of \( \vec{F_3} \)**:
- Since \( \vec{F_1} \) is directed north and \( \vec{F_2} \) is directed east, the resultant \( \vec{R} \) will be directed northeast at an angle of \( 53^\circ \) from the east.
- Thus, \( \vec{F_3} \) will be directed opposite to this, which is \( 53^\circ \) south of west.

6. **Final Result**:
- Therefore, the force \( \vec{F_3} \) can be described as:
\[ \vec{F_3} = 45 \, \text{N} \text{ due } 53^\circ \text{ south of west} \]

### Conclusion:
The force \( \vec{F_3} \) is \( 45 \, \text{N} \) directed \( 53^\circ \) south of west.

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