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Finding the change in length of humerous

  • Thread starter vf_one
  • Start date
14
0
Hi I'm stuck on a particular practice question in mechanics and I'm hoping I can get some help

1. Homework Statement

The tension in the humerous (bone of upper arm) is 190N as a result of swinging a ball tied onto a string. What is the resulting change in the length of the humerous? (Assume that length of humerous is initially 30cm and that it has uniform radius of 2cm and that Young's modulus for bone tension is 16x109Nm-2)

2. The attempt at a solution

I don't know how to solve this question so I looked at the units and then tried to solve it from there. Have no idea if it is right or not.

a) F/Young's modulus= 190/16x109Nm-2 = 1.1875x10-8m2

b) Initial volume= 0.3 x pi x 0.022 = 3.76...x10-4m2

b-a= resulting volume

Resulting volume/ cross sectional area (pi x 0.022) = 0.2999999m or 0.30m (final length)

Is this the right way to work it out or am I completely wrong?

Thanks
 

rl.bhat

Homework Helper
4,433
5
Young's modulus = Y = Stress / strain =(F/A)/(ΔL/L)

(ΔL/L) = (F/A)/Y

ΔL = (F*L)/Y*A

Substitute the values and find ΔL.
 
14
0
Oh thank you! I've never done a question on young's modulus before so this one was a good practice.
 

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