# Chapter 15 - Fluids and Elasticity - Exercises and Problems: 28

A hanging mass of 55 kg will stretch the wire by 1%

#### Work Step by Step

We can use Young's modulus to solve this question: $Y = \frac{F/A}{\Delta~L/L}$ For aluminum, $Y = 69\times 10^9~N/m^2$ If the wire stretches by 1%, then $\Delta L / L = 0.01$ We can find the force required to stretch the wire by 1% as: $Y = \frac{F/A}{\Delta~L/L}$ $F = Y~A~(\frac{\Delta L}{L})$ $F = (69\times 10^9~N/m^2)(\pi)(5.0\times 10^{-4}~m)^2~(0.01)$ $F = 541.9~N$ If this is the weight of the hanging mass, we can find the mass as: $M = \frac{weight}{g}$ $M = \frac{541.9~N}{9.80~m/s^2}$ $M = 55~kg$ A hanging mass of 55 kg will stretch the wire by 1%.

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