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 ______.

पर्याय

• `[(6"M"("g + a"))/(pi"S")]^(1/2)`

• `[(4"M"("g + a"))/(pi"S")]^(1/2)`

• `[("M"("g + a"))/(pi"S")]^(1/2)`

• `[("M"("g - a"))/(pi"S")]^(1/2)`

Answer 1

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 `underline([(4"M"("g + a"))/(pi"S")]^(1/2))`.

Explanation:

The maximum stress produced in a rope is given by

`sigma_"max" = "Force"/"Area" = "Mg"/(pi"r"^2)`

As the lift is accelerating with acceleration a, then

`sigma_"max" = ("M"("g" +- "a"))/(pi "r"^2)`

`=> "r"^2 = ("M"("g" +- "a"))/(pi"S")   ....["Given", sigma_"max" = "S"]`

`"d"^2/4 = ("M"("g" +- "a"))/(pi"S")    ...[because "r" = "d"/2]`

`=> "d" = sqrt((4"M"("g" +- "a"))/(pi"S"))`

As acceleration is maximum I.e., g' = g + a, so

d = `sqrt((4"M"("g" + "a"))/(pi"S"))`

OR

Stress × area = M (g +a)

`"S" xx pi "d"^2/4` = M (g +a)

`"d"^2 = (4"M"("g + a"))/("S"pi)`

d = `[(4"M"("g + a"))/("S"pi)]^(1//2)`