[Find by vector method the horizontal force and the force inclined at an angle of `60^@` to the vertical whose resultant is a vertical force P.]

Options

A. `P, 2P`
B. `P, Psqrt(3)`
C. `2P, Psqrt(3)`
D. None of these

Answer

To solve the problem of finding the horizontal force (F1) and the force inclined at an angle of 60 degrees to the vertical (F2) such that their resultant is a vertical force P, we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Diagram**:
- Draw a vertical line representing the vertical force P.
- Draw a horizontal force F1 to the right.
- Draw the force F2 inclined at 60 degrees to the vertical. This means that F2 makes a 30-degree angle with the horizontal.

2. **Resolve the Forces**:
- The vertical component of F2 can be expressed as:
\[ F2 \cos(60^\circ) \]
- Since the resultant is a vertical force P, we can set up the equation:
\[ P = F2 \cos(60^\circ) \]

3. **Substitute the Value of Cosine**:
- We know that \(\cos(60^\circ) = \frac{1}{2}\). Thus, we can rewrite the equation:
\[ P = F2 \cdot \frac{1}{2} \]

4. **Solve for F2**:
- Rearranging the equation gives:
\[ F2 = 2P \]

5. **Find the Horizontal Force F1**:
- The horizontal component of the force F2 can be expressed as:
\[ F2 \sin(60^\circ) \]
- Since F1 is the horizontal force, we can write:
\[ F1 = F2 \sin(60^\circ) \]

6. **Substitute the Value of Sine**:
- We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). Thus, we can substitute:
\[ F1 = (2P) \cdot \frac{\sqrt{3}}{2} \]

7. **Simplify the Expression for F1**:
- This simplifies to:
\[ F1 = P \sqrt{3} \]

### Final Answers:
- The horizontal force \(F1\) is:
\[ F1 = P \sqrt{3} \]
- The force inclined at 60 degrees to the vertical \(F2\) is:
\[ F2 = 2P \]