# Stress/strain/elongation question

1. Nov 18, 2005

### donjt81

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.

2. Nov 18, 2005

### FredGarvin

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:

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

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.

3. Nov 18, 2005

### donjt81

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