Solving Part B of e=1500*18/(1.3x10^11 x 0.000452389)

[calculator20]

Homework Statement

A vertical steel cable of diameter 24 mm and of length 18 m supports a weight of 1500 N attached to its lower end. Calculate a) the tensile stress in the cable, b) the extension of the cable, c) the elastic energy stored in the cable, assuming its elastic limit has not been reached.

YM steel is 1.3x10^11

Relevant Equations

E = FL/eA

I’m fine with part a, answer 3.3 x10^6 but I’m getting 0.00045 for part b when it should be 2.8x10^-4, what am I missing?

e = (1500x18)/(1.3x10^11 x 0.000452389)

 

[Steve4Physics]

I agree with your calculation.

However, note that 1.3x10¹¹ Pa (you forgot the units!) is a rather small value for the Young’s modulus of steel. There are a range of values depending on the type of steel, but 2.1x10¹¹Pa is typical value. Using this value gives the extension as 2.8x10⁻⁴m (you forgot the units!), which is the required answer.

So the mistake might be having the wrong value of Young's modulus.

 

[calculator20]

I agree with your calculation.

However, note that 1.3x10¹¹ Pa (you forgot the units!) is a rather small value for the Young’s modulus of steel. There are a range of values depending on the type of steel, but 2.1x10¹¹Pa is typical value. Using this value gives the extension as 2.8x10⁻⁴m (you forgot the units!), which is the required answer.

So the mistake might be having the wrong value of Young's modulus.

Thank you for your help. I used the quoted value from the textbook but I’ve just spotted that value for copper below it is identical so it could be a mistake in the book. Thanks again