Question

5 x 10 N/m2 What distance will it be compressed when a statue that has a mass of 201,000 kg is placed on A 6.90 m long marble column has a cross sectional area of 1.93 m 2 has a Young's top of it? modulus of 50 x Number Units

Answer 1

$\text{Solution :}\\ *Length \ of\ the\ rod \ \ L = 6.90m\\ *Cross\ sectional \ area \ of \ the \ rod \ \ A =1.93m^2\\ *Young's \ modulus \ \ \ Y = 50.5\times 10^9N/m^2\\ *Mass \ placed\ \ m = 201000\ kg\\ \ \\ \ \\ \text{Young's modulus is expressed by ; }\ \ \ \\ Y = \dfrac{F\ L}{A\ \Delta L}\\ Here; \ Weight \ works \ as \ the \ force\ \ F = mg\\ Then;\\ \Delta L= \dfrac{m\ g\ L}{A\ Y} \\ \text{Plugging in the given values, we get;}\\ \Delta L = \dfrac{201000kg\times 9.8m/s^2\times6.90m}{1.93m^2\times (50.5\times 10^9N/m^2)}\\ \Delta L = 0.0001394\ m\\ \Delta L = 0.139 \ mm$