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A force of 25 kN applied to a piece of steel produces an extension of 2 mm. Assuming the elastic limit is not exceeded, determine (a) the force required to produce an extension of 3.5 mm, (b) the extension when the applied force is 15 kN. |
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Answer» From Hooke’s law, extension x is proportional to force F within the limit of proportionality, i.e. x α F or x = kF, where k is a constant. If a force of 25 kN produces an extension of 2 mm, then 2 = k(25), from which, constant k = \(\cfrac{2}{25}\) = 0.08 (a) When an extension x = 3.5 mm, then 3.5 = k(F), i.e. 3.5 = 0.08 F, from which, force F = \(\cfrac{3.5}{0.08}\) = 43.75 kN (b) When force F = 15 kN, then extension x = k(15) = (0.08)(15) = 1.2 mm |
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