# Homework Help: Need help with Basic Kinematics question!

1. Jan 9, 2012

### gretchenm

1. The problem statement, all variables and given/known data

Here is my question: A crane is moving a stainless steel coil weighing 497.3 slugs. The crane's cable snaps and the coil falls for 3/4 seconds before hitting the ground. What is the FORCE the coil hits the ground with?

2. Relevant equations

F=ma
PE=KE??

3. The attempt at a solution

I have already converted slugs to lbf...
F = ma = 497.3 slugs= 497.3 lbf s^2/f * 32.2 f/s^2= 16013.06 lbf

Now I am feeling pretty stupid trying to figure out how to calculate the force that it hits the ground with. I am sure I have to use the .75 seconds somewhere, but for some reason can't figure it. Any help? Thanks! :)

Last edited: Jan 9, 2012
2. Jan 9, 2012

### BruceW

The force it hits the ground at depends on how easily the object deforms. Are you sure the question wasn't asking for the momentum with which it hits the ground?

You've already written down the force on the object due to gravity, and you have the time it falls for. Now it is straightforward to calculate the accumulated momentum of the object.

3. Jan 9, 2012

### cupid.callin

Well i guess you must use, F = dp/dt

but dt here wont be 3/4. 3/4 is time of falling from which you will calculate final velocity ...

4. Jan 9, 2012

### gretchenm

Hmm, I agree with your statement about the force depending on how easily the object deforms. Do we have to use the elasticity modulus of stainless steel in there somewhere, maybe? I think it is 180 x 10^9 N/m^2 if I looked it up correctly.

5. Jan 9, 2012

### gretchenm

So then final velocity would be...
v(f)=0+32.2 (.75)=24.15 ft/s...right? Then where can we go with this?

6. Jan 9, 2012

### cupid.callin

As BruceW said, the force of collision depends of type of collision (elastic or inelastic) and time of collision.

7. Jan 9, 2012

### gretchenm

So I think what I am getting from the answers that have been posted, is that this is a question that does not give enough information to correctly answer....how would I go about calculating the momentum then? Would it just be 16013.06 lbf times the final velocity of 24.15 ft/s? (386715.399 lbf ft/s*1.355817948=524315 J/s) I just used a conversion factor I looked up to get SI units. Anyone agree with this calculation? (or not..)

8. Jan 9, 2012

### cupid.callin

I did not see the values but method is fine. p=mv !!
And yes the info. given is not enough. But maybe the questions has some assumptions that author thought will be obvious to a reader, perhaps a note before starting a series of questions.

9. Jan 9, 2012

### gretchenm

Yeah, I think I am just going to stick with calculating momentum and be done with this! Thanks everyone for all of the input! :)