What Is the Stretch of a Cable Under Load?

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    Cable Stretch
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The discussion revolves around calculating the stretch of a cable under load, specifically for a wrecking ball scenario. The user initially applies the formula δ=(FL/AE) but encounters an unrealistic result of 29779.68 m. They calculate the force (F) using the density of steel, volume, and gravitational acceleration, but later realize a mistake in interpreting the modulus of elasticity (E) in GPa. The user acknowledges their error after three hours of work, indicating the importance of understanding unit conversions in engineering calculations. The thread highlights the challenges of applying theoretical formulas to practical problems in physics.
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Homework Statement


A 1.25m diameter wrecking ball hangs motionless from a 1.75 cm diameter cable. The wrecking ball is solid and is constructed of steel (density = 7800 kg/m3). If the cable is 16 m long, how much does the cable stretch? For the cable use E = 175 GPa.

Homework Equations


δ=(FL/AE)

The Attempt at a Solution



δ=(FL/AE)

F=ρ*v*g
=7800*(4/3)*pi*r^3*9.8
=78171.66 N

A=PI*R*R
=(pi*(1.75*1.75)/4)*10^-4
=2.4*10^-4

E=175000 N-m^2

L=16 m

When I plug in the values I get 29779.68 m which I know cannot be correct.

Thanks for your time
 
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Look up what G in GPa signifies.
 
I've been working on this problem for 3 hours now because I failed to realize my simple mistake. Thank you a lot.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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