Question

A lift is tied with thick iron ropes having mass \( M \). The maximum acceleration of the lift is \( a \) m/s² and maximum safe stress is \( s \) N/m². The minimum diameter of the rope is

• A. \( \frac{2M(g + a)}{\pi s} \)
• B. \( \frac{2M(g - a)}{\pi s} \)
• C. \( \frac{4M(g + a)}{\pi s} \)
• D. \( \frac{4M(g - a)}{\pi s} \)

Hint:

When calculating the diameter of a rope based on stress, use the formula \( s = \frac{F}{A} \) and solve for \( A \), which depends on the cross-sectional area of the rope.

Answer 1

The Correct Option is C

Step 1: Understanding the force on the rope.
The maximum force acting on the rope due to the lift is the sum of the gravitational force \( Mg \) and the additional force due to acceleration \( Ma \). Thus, the total force is \( F = M(g + a) \).
Step 2: Using the stress formula.
The stress \( s \) is given by: \[ s = \frac{F}{A} = \frac{M(g + a)}{A} \] where \( A \) is the cross-sectional area of the rope. The area is related to the diameter \( D \) by \( A = \frac{\pi D^2}{4} \).
Step 3: Solving for the diameter.
Substitute the expression for \( A \) into the stress formula and solve for \( D \): \[ s = \frac{4M(g + a)}{\pi D^2} \Rightarrow D = \sqrt{\frac{4M(g + a)}{\pi s}} \] Step 4: Conclusion.
The correct answer is (C), \( \frac{4M(g + a)}{\pi s} \).