Question

A mass 'M' is suspended by a rope from a rigid support at point 'P'. Another rope is tied at end 'Q' and pulled horizontally with a force 'F'. If the rope makes an angle $\theta$ with vertical, then the tension in the string 'PQ' is

• A. $F \sin \theta$
• B. $\frac{F}{\sin \theta}$
• C. $F \cos \theta$
• D. $\frac{F}{\cos \theta}$

Hint:

In equilibrium problems, the sum of forces in any direction must be zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The system is in equilibrium under three forces acting at point Q: the tension \( T \) in rope PQ, the horizontal pulling force \( F \), and the vertical downward force (weight \( Mg \)). We need to express tension \( T \) in terms of \( F \) and \( \theta \).

Step 2: Key Formula or Approach:
Apply equilibrium conditions at point Q:
- Sum of horizontal forces = 0
- Sum of vertical forces = 0

Step 3: Detailed Explanation:
Let \( T \) be the tension in the string PQ.
The string makes an angle \( \theta \) with the vertical.
Resolving the tension \( T \) into components:
- Horizontal component = \( T \sin \theta \) (acting towards the left)
- Vertical component = \( T \cos \theta \) (acting upwards)
Since the point Q is pulled horizontally with force \( F \) towards the right, for horizontal equilibrium:
\[ T \sin \theta = F \] Solving for \( T \):
\[ T = \frac{F}{\sin \theta} \]

Step 4: Final Answer:
The tension in the string 'PQ' is \( \frac{F}{\sin \theta} \).