A mass 'M' is suspended by a rope from a rigid support. It is pulled horizontally with a force F. If the rope makes an angle ' `theta`' with vertical in equilibrium, then the tention in the string is

Options

A. `F sin theta`
B. `F/(sin theta)`
C. `F cos theta`
D. `F/(cos theta)`

Answer

To solve the problem, we need to analyze the forces acting on the mass \( M \) suspended by the rope and how they relate to the tension in the rope when it is pulled horizontally with a force \( F \).

### Step-by-Step Solution:

1. **Draw the Free Body Diagram (FBD)**:
- Start by drawing the mass \( M \) suspended from a rigid support by a rope.
- Indicate the angle \( \theta \) that the rope makes with the vertical.
- Mark the weight of the mass acting downward as \( W = Mg \) (where \( g \) is the acceleration due to gravity).
- Indicate the horizontal force \( F \) acting on the mass.

2. **Identify Forces**:
- The tension \( T \) in the rope acts along the rope at an angle \( \theta \) from the vertical.
- The weight \( W \) acts downward.
- The horizontal force \( F \) acts to the right.

3. **Resolve the Tension into Components**:
- The tension \( T \) can be resolved into two components:
- Vertical component: \( T \cos(\theta) \) (acting upward)
- Horizontal component: \( T \sin(\theta) \) (acting horizontally)

4. **Set Up Equilibrium Conditions**:
- Since the mass is in equilibrium, the sum of the forces in both the vertical and horizontal directions must be zero.
- For vertical forces:
\[ T \cos(\theta) = Mg \quad \text{(1)} \]
- For horizontal forces:
\[ T \sin(\theta) = F \quad \text{(2)} \]

5. **Solve for Tension \( T \)**:
- From equation (2), we can express \( T \) in terms of \( F \) and \( \theta \):
\[ T = \frac{F}{\sin(\theta)} \quad \text{(3)} \]
- From equation (1), we can express \( T \) in terms of \( Mg \) and \( \theta \):
\[ T = \frac{Mg}{\cos(\theta)} \quad \text{(4)} \]

6. **Equate the Two Expressions for T**:
- Set equations (3) and (4) equal to each other:
\[ \frac{F}{\sin(\theta)} = \frac{Mg}{\cos(\theta)} \]

7. **Rearranging to Find Tension**:
- Cross-multiply to isolate \( T \):
\[ T = \frac{F \cos(\theta)}{\sin(\theta)} = F \cot(\theta) \]

8. **Final Expression for Tension**:
- Thus, the tension \( T \) in the string is given by:
\[ T = \frac{F}{\sin(\theta)} \]