A lift of mass 2000 kg is supported by thick steel ropes. If maximum upward acceleration of the lift be 1.2 `m//s^(2)`, and the breaking stress for the ropes be `2.8 xx 10^(8)`N/`m^(2)`, what should be the minimum diameter of the rope?

Explanation

Here, m = 2000 kg, a = 1.2`m//s^(2)`
Breaking stress = 2.8`xx 10^(8) N//m^(2)`
Let the diameter of the rope be D.
when the lift moves upwards, the tension in the rope is
T = m(g + a) = 2000(9.8 + 1.2) = 22000N.
Now, breaking stress = `("force")/("area") = (T)/(pi D^(2)//4 ) = (4T)/(pi D^(2))`
or, `2.8xx 10^(8) = (4 xx 22000 xx 7)/(22 xx D^(2))`
or, `D^(2) = (4 xx 22000 xx 7)/(22 xx 2.8 xx 10^(8)) = 10^(-4)` or, D = 1 cm