Find the horizontal force and a force inclined at an angle of `60^(@)` with the vertical so that the resultant is a vertical force of P kg wt.

Options

A. P, 2P
B. P, `Psqrt(3)`
C. `P sqrt(3),2P`
D. none

Answer

To solve the problem of finding the horizontal force \( F_1 \) and the force \( F_2 \) inclined at an angle of \( 60^\circ \) with the vertical such that the resultant force is a vertical force of \( P \) kg wt, we can follow these steps:

### Step 1: Define the Forces
Let:
- \( F_1 \) be the horizontal force.
- \( F_2 \) be the force inclined at \( 60^\circ \) with the vertical.

### Step 2: Resolve \( F_2 \) into Components
Since \( F_2 \) is inclined at \( 60^\circ \) with the vertical, we can resolve it into horizontal and vertical components:
- The vertical component of \( F_2 \) is \( F_2 \cos(60^\circ) \).
- The horizontal component of \( F_2 \) is \( F_2 \sin(60^\circ) \).

### Step 3: Write the Equations for Resultant Forces
The resultant force in the vertical direction must equal \( P \) kg wt. Therefore, we can write:
\[ F_2 \cos(60^\circ) = P \]
And since the horizontal forces must balance out (as there is no resultant horizontal force), we have:
\[ F_1 = F_2 \sin(60^\circ) \]

### Step 4: Substitute the Values of Cosine and Sine
Using the values:
- \( \cos(60^\circ) = \frac{1}{2} \)
- \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)

We can substitute these into our equations:
1. From the vertical force equation:
\[ F_2 \cdot \frac{1}{2} = P \implies F_2 = 2P \]
2. From the horizontal force equation:
\[ F_1 = F_2 \cdot \frac{\sqrt{3}}{2} = 2P \cdot \frac{\sqrt{3}}{2} = \sqrt{3}P \]

### Step 5: Final Answers
Thus, the horizontal force \( F_1 \) and the inclined force \( F_2 \) are:
- \( F_1 = \sqrt{3}P \)
- \( F_2 = 2P \)

### Summary of Results
- Horizontal Force \( F_1 = \sqrt{3}P \)
- Force Inclined \( F_2 = 2P \)