A 10 kg body is suspended by a rope is pulled by means of a horizontal force to make 60° by rope to vertical. The horizontal force is

Options

A. 10 kgwt
B. 30 kgwt
C. `10sqrt(3)` kgwt
D. `30sqrt(3)` kgwt

Answer

To solve the problem, we will analyze the forces acting on the body suspended by the rope and apply the principles of equilibrium.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- The weight of the body (W) acting downwards: \( W = mg \) where \( m = 10 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
- The tension (T) in the rope acting at an angle of 60° to the vertical.
- A horizontal force (F) acting to pull the body.

2. **Calculate the Weight of the Body**:
\[ W = mg = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

3. **Resolve the Tension into Components**:
- The tension can be resolved into two components:
- Vertical component: \( T_v = T \cos(60°) \)
- Horizontal component: \( T_h = T \sin(60°) \)

Knowing that \( \cos(60°) = \frac{1}{2} \) and \( \sin(60°) = \frac{\sqrt{3}}{2} \):
\[ T_v = T \cdot \frac{1}{2} = \frac{T}{2} \]
\[ T_h = T \cdot \frac{\sqrt{3}}{2} \]

4. **Apply the Equilibrium Condition in the Vertical Direction**:
Since the body is in equilibrium, the vertical component of tension must balance the weight of the body:
\[ T_v = W \]
\[ \frac{T}{2} = 98 \, \text{N} \]

5. **Solve for Tension (T)**:
\[ T = 98 \, \text{N} \times 2 = 196 \, \text{N} \]

6. **Apply the Equilibrium Condition in the Horizontal Direction**:
The horizontal force (F) must balance the horizontal component of the tension:
\[ F = T_h \]
\[ F = T \cdot \frac{\sqrt{3}}{2} \]

7. **Substitute the Value of T**:
\[ F = 196 \, \text{N} \cdot \frac{\sqrt{3}}{2} \]
\[ F = 98\sqrt{3} \, \text{N} \]

### Final Answer:
The horizontal force \( F \) required to maintain the angle of 60° with the vertical is:
\[ F = 98\sqrt{3} \, \text{N} \]