Question

A force of 10 kN applied to a component produces an extension of 0.1 mm. Determine (a) the force needed to produce an extension of 0.12 mm, and (b) the extension when the applied force is 6 kN, assuming in each case that the limit of proportionality is not exceeded.

Answer 1

From Hooke’s law, extension x is proportional to force F within the limit of proportionality, i.e. x ∝ F or x = kF, where k is a constant. If a force of 10 kN produces an extension of 0.1 mm, then

0.1 = k(10), from which, constant k = 0.1/10 = 0.01

(a) When an extension x = 0.12 mm, then 0.12 = k(F ), 

i.e. 0.12 = 0.01F, from which, 

force F = 0.12/0.01 = 12 kN

(b) When force F = 6 kN, then 

extension x = k(6) = (0.01)(6) = 0.06 mm