Question

A 1000-kg object hangs from the lower end of a steel rod 5.0 m long that is suspended vertically. The diameter of the rod is 0.80 cm and Young's modulus for the rod is 210,000 MN/m2. What is the elongation of the rod due to this object?
A) 0.047 cm
B) 1.2 cm
C) 0.12 cm
D) 1.8 cm
E) 0.46 cm

Answer 1

Answer:

$\Delta L=4.642\,mm$

Explanation:

Given:

mass of the hanging object, m= 1000 kg

original length of the steel rod, L=5 m

diameter of the rod, d= 0.80 cm

Young's modulus for the rod, E = $210,000 MPa$

Firstly we find the value of stress:

$\sigma = \frac{Force}{area}$

$\sigma = \frac{1000\times 9.8}{\pi\times 4^2}$

$\sigma = 194.9648 MPa$

Now strain,

$\epsilon= \frac{\sigma}{E}$

$\epsilon= \frac{194.9648 }{210000}$

$\epsilon= 0.00093$

We know,

$\epsilon= \frac{\Delta L}{L}$

$0.00093= \frac{\Delta L}{5000}$

$\Delta L=4.642\,mm$

Answer 2

Answer:E

Explanation:

Given

mass of object $m=1000 kg$

Length of steel rod $L=5 m$

diameter of rod $d=0.8 cm$

Young modulus $E=210,000 MN/m^2$

And we know $E=\frac{stress}{strain}$

$stress=\frac{Load}{Area}$

$Strain=\frac{\Delta L}{L}$

Therefore $\Delta =\frac{FL}{AE}$

$\Delta =\frac{1000\times 9.8\times 5}{\frac{\pi (0.8\times 10^{-2})^2}{4}\times 210000\times 10^6}$

$\Delta =\frac{20\times 9.8}{\pi \times 0.64\times 210\times 100}$

$\Delta =4.64\times 10^{-3} m$

$\Delta =0.464 cm$