Finding the change in length of humerous

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SUMMARY

The discussion focuses on calculating the change in length of the humerus bone under tension, specifically using Young's modulus. The tension applied is 190N, with the initial length of the humerus set at 30cm and a uniform radius of 2cm. Young's modulus for bone is given as 16x109 Nm-2. The correct formula to determine the change in length (ΔL) is ΔL = (F*L)/(Y*A), where A is the cross-sectional area of the bone.

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Hi I'm stuck on a particular practice question in mechanics and I'm hoping I can get some help

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
 
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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.
 
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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