Stress/strain/elongation question

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To determine the elongation of a steel cable supporting a 711kg object, the correct approach involves calculating stress using the formula stress = F/A, where F is the weight of the object (mg) and A is the cross-sectional area. The strain can then be calculated using strain = stress/Y, and elongation can be found with delta L = Lo * strain, which is valid as long as the material remains in the elastic range. For the second part, when the object is accelerated upward at 3.3m/s², the correct force calculation should be F = ma + mg, as the total force must account for both the weight and the additional force due to acceleration. This adjustment leads to a greater elongation than in the static case, aligning with expectations. The methodology for both parts is confirmed as accurate.
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First part of the question:
A high speed lifting mechanism supportsa 711kg object with a steel cable 24.9m long and 3.83cm(squared) in cross sectional area. Determine elongation of the cable.

I think I got this first part right... can anyone confirm my approach.

stress = F/A
F = mg = 711*9.8
A = .0383

strain = stress/Y

and then

elongation delta L = Lo * strain

Is this how you find elongation?

Second part of the question
By what additional amount does the cable increase in length if the object is accelerated upward at a rate of 3.3m/s2?

This was my approach to the second part. not sure if this is correct...

stress = F/A
but this time F = ma - mg (is this correct?)

everything else will be the same as the first part.

can someone please verify both these please.

Thanks in advance
 
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donjt81 said:
First part of the question:
A high speed lifting mechanism supportsa 711kg object with a steel cable 24.9m long and 3.83cm(squared) in cross sectional area. Determine elongation of the cable.
I think I got this first part right... can anyone confirm my approach.
stress = F/A
F = mg = 711*9.8
A = .0383
strain = stress/Y
and then
elongation delta L = Lo * strain
Is this how you find elongation?
As long as you are in the elastic range, yes. Your approach is correct since

S = \frac{P}{A}

you can then relate strain (e) to stress via Young's Modulus, E by Hooke's Law

S = eE

Once you have the strain, use the definition of engineering strain to find the increase in length:

e = \frac{\Delta L}{L_o}

donjt81 said:
By what additional amount does the cable increase in length if the object is accelerated upward at a rate of 3.3m/s2?
This was my approach to the second part. not sure if this is correct...
stress = F/A
but this time F = ma - mg (is this correct?)
Close, but it will be
F = ma + mg

You can reason this out. If the mass is accelerating, would you expect it to weigh more or less than it does staically? The other way to think of it is that the acceleration is in the upward direction, but the reaction force, the force that causes the increased deflection, is opposite because of Newton's 3rd law.
 
OK that makes sense. Because when i calculated it with F = ma - mg, I was getting a smaller elongation than the first part. It didnt make sense.

But if we do F = ma + mg then the elongation will be more than the first part and it makes more sense.

thanks again
 
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