Question

A solid steel column is 4.0 m long and 0.20 m in diameter. Young's modulus for this steel is 2.0 × 1011 N/m2. By what distance does the column shrink when a 5000-kg truck is supported by it?

Answer 1

Answer:

3.12 x 10^-5 m

Explanation:

Length of steel column, L = 4 m

diameter, d = 0.2 m

radius = half of diameter = 0.1 m

Young's modulus, Y = 2 x 10^11 N/m^2

Mass of truck, m = 5000 kg

Force, F = mass of truck x acceleration due to gravity

F = 5000 x 9.8 = 49000 N

Area of crossection of cable,

A =  $\pi r^{2}=3.14\times\0.1\times0.1=0.0314m^{2}$

Let ΔL be the shrink in length of cable, then by the formula of Young's modulus

$Y=\frac{F\times L}{A\times\Delta L}$

$\Delta L = \frac{F\times L}{A\times Y}$

$\Delta L = \frac{49000\times 4}{0.0314\times 2\times10^{11}}$

ΔL = 3.12 x 10^-5 m

Thus, the shrink in the length of cable is 3.12 x 10^-5 m.