9702/Oct Nov/13/2013/Q22

A lift is supported by two steel cables each of length 20 m.

Each of the cables consists of 100 parallel steel wires, each wire of cross-sectional area 3.2×10^–6 m². The Young modulus of steel is 2.1×10¹¹Nm^–2.

Which distance does the lift move downward when a man of mass 70 kg steps into it?
A 0.010 mm

B 0.020 mm

C 0.10 mm

D 0.20 mm

Solution:

Answer: C

Hooke’s law: F = ke

The lift is supported by 2 steel cables, each of which consists of 100 parallel steel wires. So, there is a total of 200 parallel steel wires in parallel.

For parallel spring,
Effective spring constant, k_eff = k₁ + k₂ + k₃ + … 

First, we need to find the spring constant, k for 1 steel wire.

For 1 wire,
Young modulus, E = stress / strain = (F/A) / (e/L) = FL / Ae
Hooke’s law: F = ke
Young modulus, E = (ke)L / Ae = kL / A
Spring constant, k = EA / L = (2.1 × 10¹¹) (3.2 × 10^–6) / 20 = 33600 Nm^-1 

Effective spring constant, k_eff = 200k
Hooke’s law: F = k_eff e
Mass of person = 70kg. Weight = mg = 700N           (take g = 10 ms^-2)
Extension, e = F / k_eff = 700 / (200 × 33600) = 0.00010m = 0.10 mm

Reference: PYQ - Oct/Nov 2013 Paper 13 Q22