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

`\Delta L=4.642\,mm`

Step-by-step 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 = (Force)/(area)`

`\sigma = (1000* 9.8)/(\pi* 4^2)`

`\sigma = 194.9648 MPa`

Now strain,

`\epsilon= (\sigma)/(E)`

`\epsilon= (194.9648 )/(210000)`

`\epsilon= 0.00093`

We know,

`\epsilon= (\Delta L)/(L)`

`0.00093= (\Delta L)/(5000)`

`\Delta L=4.642\,mm`

Answer 2

Answer:E

Step-by-step 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=(stress)/(strain)`

`stress=(Load)/(Area)`

`Strain=(\Delta L)/(L)`

Therefore
`\Delta =(FL)/(AE)`

`\Delta =(1000* 9.8* 5)/((\pi (0.8* 10^(-2))^2)/(4)* 210000* 10^6)`

`\Delta =(20* 9.8)/(\pi * 0.64* 210* 100)`

`\Delta =4.64* 10^(-3) m`

`\Delta =0.464 cm`