Calculating stress, strain and Young's modulus?

[rockchalk1312]

After a fall, a 77 kg rock climber finds himself dangling from the end of a rope that had been 15 m long and 9.1 mm in diameter but has stretched by 2.3 cm. For the rope, calculate (a) the strain, (b) the stress, and (c) the Young's modulus.


strain = ΔL/L
stress = F/A
Young's modulus = (F/A)/(ΔL/L)


I haven't gotten any of these right and I have no idea what I'm doing wrong on them.

a) 2.3 cm = .023 m / 15 m = .0015

b) A = ∏r² = (∏)(.0091/2)² = 6.503E-5 m²
stress = ((77)(9.8))/(6.503E-5) = 1.16E7 N/m²

c) (754.6/6.503E-5) / (.023/15) = .756

None of those were right; also I thought Young's modulus was supposed to be a huge number so obviously that's not right. Any help is appreciated!

 

[Chestermiller]

The strain was correct, although they may have given the result with more digets 0.00153.

The calculation of the cross sectional area is correct.

The calculated stress is correct, but perhaps should be expressed as 11.6 MPa.

The calculated Young's modulus, which is the stress divided by the strain is incorrect. You made a mistake in arithmetic. 11.6/0.00153 = 7584 MPa = 7.58 GPa.

 

[rockchalk1312]

All of those helped me! The computer just wanted them entered in an odd way. Thank you!