A body of weight W is suspended by a string It is pulled aside with a horizontal force of 2W and held at rest. The tension in it is

Options

A. 2W
B. 3W
C. `Wsqrt3`
D. `Wsqrt5`

Answer

To solve the problem, we need to analyze the forces acting on the body suspended by the string and use the principles of equilibrium.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the body acting downwards is \( W \).
- A horizontal force of \( 2W \) is applied to the body.
- Let \( T \) be the tension in the string.

2. **Draw a Free Body Diagram**:
- The weight \( W \) acts downward.
- The tension \( T \) acts along the string, making an angle \( \theta \) with the vertical.
- The horizontal force \( 2W \) acts to the right.

3. **Resolve the Tension into Components**:
- The tension \( T \) can be resolved into two components:
- Vertical component: \( T \cos \theta \) (balances the weight)
- Horizontal component: \( T \sin \theta \) (balances the horizontal force)

4. **Set Up the Equations for Equilibrium**:
- Since the system is at rest (in equilibrium), we can write two equations:
- For vertical forces:
\[ T \cos \theta = W \quad \text{(1)} \]
- For horizontal forces:
\[ T \sin \theta = 2W \quad \text{(2)} \]

5. **Eliminate \( \theta \)**:
- To eliminate \( \theta \), we can square both equations and add them:
\[ (T \cos \theta)^2 + (T \sin \theta)^2 = W^2 + (2W)^2 \]
- This simplifies to:
\[ T^2 (\cos^2 \theta + \sin^2 \theta) = W^2 + 4W^2 \]
- Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ T^2 = 5W^2 \]

6. **Solve for Tension \( T \)**:
- Taking the square root of both sides gives:
\[ T = \sqrt{5} W \]

### Final Answer:
The tension in the string is \( T = \sqrt{5} W \).