A `10kg` body is suspended by a rope is pulled by means of a horizontal force to make `60^(@)` by rope to vertical.The horizontal forces is

Options

A. `10 kgwt`
B. `30 kgwt`
C. `10sqrt3 kgwt`
D. `30sqrt3 kgwt`

Explanation

To solve the problem of a 10 kg body suspended by a rope that makes an angle of 60 degrees with the vertical when pulled by a horizontal force, we can follow these steps:

### Step 1: Understand the Forces Acting on the Body
The body has two main forces acting on it:
1. The weight of the body (W = mg), acting downwards.
2. The tension (T) in the rope, which has two components:
- A vertical component (T cos(θ)) that balances the weight of the body.
- A horizontal component (T sin(θ)) that is balanced by the horizontal force (F).

### Step 2: Calculate the Weight of the Body
The weight of the body can be calculated using the formula:
\[ W = mg \]
Where:
- \( m = 10 \, \text{kg} \)
- \( g \approx 9.81 \, \text{m/s}^2 \)

Calculating the weight:
\[ W = 10 \times 9.81 = 98.1 \, \text{N} \]

### Step 3: Set Up the Equations
From the forces acting on the body, we can set up the following equations based on the components of tension:
1. Vertical component:
\[ T \cos(60^\circ) = mg \]
2. Horizontal component:
\[ T \sin(60^\circ) = F \]

### Step 4: Substitute the Known Values
Using \( \cos(60^\circ) = 0.5 \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
1. From the vertical component:
\[ T \cdot 0.5 = 98.1 \]
\[ T = \frac{98.1}{0.5} = 196.2 \, \text{N} \]

2. From the horizontal component:
\[ F = T \cdot \sin(60^\circ) = 196.2 \cdot \frac{\sqrt{3}}{2} \]

### Step 5: Calculate the Horizontal Force
Now substituting the value of T into the horizontal force equation:
\[ F = 196.2 \cdot \frac{\sqrt{3}}{2} \]
\[ F = 98.1 \sqrt{3} \]

### Step 6: Final Calculation
To find the numerical value of \( F \):
Using \( \sqrt{3} \approx 1.732 \):
\[ F \approx 98.1 \cdot 1.732 \approx 170.0 \, \text{N} \]

### Final Answer
The horizontal force \( F \) required to maintain the body at an angle of 60 degrees with the vertical is approximately **170.0 N**.

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