A lift of mass 2000 kg is supported by thick steel ropes. If maximum upward acceleration of the lift be 1.2 ms^–2, and the breaking stress for the ropes be 2.8 × 10⁸ Nm^–2, what should be the minimum diameter of rope? 

Here, m = 2000 kg, a = 1.2 m/s²; breaking stress = 2.8 × 10⁸ N/m².

As the lift moves upwards, the tension in the rope is:

T = m(g + a) = 2000(9.8 + 1.2) = 22,000 N.

Now, breaking stress = Force / Area = T / (πD² / 4) = 4T / (πD²)

or 2.8 × 10⁸ = (4 × 22,000 × 7) / (22 × D²)

or D² = (4 × 22,000 × 7) / (22 × 2.8 × 10⁸) = 10⁻⁴

∴ D = 10⁻² m = 1 cm.