# Deceleration of a falling load of a rope

1. Feb 23, 2012

### tommy060289

Hey everyone

1. The problem statement, all variables and given/known data
Im trying to calculate the force encountered by an object falling due to gravity that is caught by a rope.

Obviously F=ma and since mass is constant I need to find the rate of deceleration. I've tried to using the equation v^2 = u^2 +2ax as v = 0 and u =1.57 m/s so I need to find x to calculate the deceleration.

In order to calculate s I was going to use the equation for young's modulus which is σ/ε and calculate the strain, As E = 200 GPa and I just need to find the stress on the rope, but here in lies my problem, as I can not figure out how to do this. I know the CSA as r = 0.003 m but I don't know what to use for my force value as surely the force is what I am trying to work out in the first place, and I don't think using mass * gravity is sufficient as the force must be greater as we have a velocity

2. Relevant equations

F = m*a

v^2 = u^2 + 2ax

x = l*σ/E

σ =F/A

3. The attempt at a solution

My attempt is described above but I just can't figure out what to use for F

2. Feb 23, 2012

### PeterO

That equation you have quoted applies to constant acceleration.

imagine that instead of a rope, you had a spring. As the spring stretches under the load, the force gets stronger and stronger, meaning higher and higher acceleration, so a formula for a constant acceleration situation is hardly appropriate.

The formula could be used to find the average force - the constant force that would achieve the same event [stopping the falling mass].

You could just as easily use work energy considerations however.

Gravity has done work accelerating the mass as it falls, then the rope does work stopping the falling mass.
If the object fell 3 m, then its weight force times 3m is the amount of work done gaining energy [3mg].
If the object then stopped in 3 cm, clearly the force would have to be 100 times the weight force to do the same amount of work. That force however would not be constant, so 100mg is only the average force [100mg x 0.03 = 3mg; total work the same].

3. Feb 23, 2012

### tommy060289

But I'm still struggling going about calculating the distance in which the object is stopped. Is there anyway to calculate this. I could use young's modulus but again, the problem lies in calculating the stress?

4. Feb 23, 2012

### PeterO

Even a steel rope behaves just as really strong spring - provided you do not strain it beyond its yield point. so make use of E = 0.5kx2 and relate k back to the Young's modulus

You are not going to get a number answer, unless you know the mass, how far it fell and young's modulus of the catching rope/wire, and how thick and long it was to begin with.
Of course you could just use M for mass, h for how far it fell, g for gravity, γ for Young's modulus, etc. and derive an algebraic solution.

5. Feb 23, 2012

### tommy060289

thanks peter, ill try that!

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