final velocity calculation 5

[11xminurti3]

Hi, I am trying to work out impact velocity, im tryng to use v^2=u+2*as

I know what my distance (s) is
I know what my inital velocity (u) is

But i dont know my acceleration
And I dont know my time

I was trying to use f=mg (i know what the mass is) then use a=f/m - but obviously end up with 9.81 as there just the same equation re arranged

any ideas, been racking my brain on this one for a few hours

cheers

 

[hydtools]

Pick two impact duration times: 0.010sec, 0.005sec

m*v/t = 32kg*7m/s/0.01sec =

m*v/t = 32kg*7m/s/0.005 =

Ted

 

[DRWeig]

11xminurti3,

Forget the math for a minute and try this thought on. Impact is a two-way street. The moving object applies a force to the static one, and the static one applies an equal and opposite reaction force to the falling object.

If I drop an egg from 6 feet onto my new Tempurpedic mattress, it doesn't break. The soft, conforming, yet firm mattress gives the egg a good distance to decelerate, somewhere between 5 and 10 mm from my eyeball estimate of the dent it made. The force was very low.

If I drop an egg from 6 feet onto my garage floor, I have a mess to clean up. The concrete made the poor little egg decelerate to zero in almost-zero distance. The force was high.

Those are two cases in which the speed and energy of the object (egg) are the same. In the high-dollar mattress case, the resulting impact force was not enough to even break the yolk inside the egg, much less its shell. In the garage floor case, the resulting impact force was plenty to smash the egg into about a 1-meter diameter slippery mess.

There are other examples you could try with your fine bed and a garage floor to demonstrate the reaction force that the bombarded surface provides.

How high will a tennis ball bounce off of a superb mattress versus a slab of oil-stained concrete?

How much will you be spending on a new Nikon camera when you drop it on the concrete versus the top of a magnificently-engineered piece of sleeping equipment?

How hard will I be thrown into the steering wheel if I'm going 100 MPH and stop in a few feet or so when I hit a huge tree? That was a component of the impact force that moved me. If I crash into a rubber band instead, and it stretches half a mile before it brings me to a stop, I'll have virtually no force at all applied to my body. Both driving incidents were in the same car at the same speed.

In each case, you need to know the deceleration distance to determine the force. I hope I didn't come off as too sarcastic, but you have been given the same information in every prior post.

By the way, I'm not a Tempurpedic salesman. However, after my wife wore me down enough, I now own one. It's the best thing that ever happened to my back and my sleep quality.

Best to you,

Goober Dave

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[hydtools]

OK. Use work = energy F*d = 0.5*m*v^2

F = KE/d d is the distance the rock or egg travels after it impacts the surface Increasing d decreases F, the force due to impact. d is NOT the distance the rock or egg falls to impact.

Ted

 

[moon161]

Peak forces during a collision can be large, especially with rigid objects, but it actually takes work to yield or break something. If you are dealing with a structure that can in any way be treated as a spring, say a diving board, you can relate kinetic energy to structural displacement.

Say I drop a bowling ball 5 m onto a coil spring from a car, what is the peak force developed by the spring?

Bowling ball = 5kg
h= 5m
KE= mgh = 250J

Spring
k = 25000 N/m
F = kx

Spring energy storage:
E = kx[sup]2[/sup]/2
x = (2E/k)[sup]1/2[/sup]
x = (500/25000)[sup]1/2[/sup]
x = .14m

F = .14m x 25000 N/m = 3500 N

Knowledge of the spring design, (wire dia, coil dia, steel yeild stress) will tell you if the spring will yield at the peak deflection or not.

Find your physics book and work the kinematics back and forth. I would dig for my copy of Shigley & Mischke, Machine design as well. You need it to be transparent to you.

 

[GregLocock]

http://www.eng-tips.com/faqs.cfm?fid=1245

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[Walterke]

v^2 = 2*g*h = 7m/s

please tell your boss you're not up for this task and need someone else to do it. Before anyone gets hurt.


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[hydtools]

Walterke, good catch. We all missed it.

Ted

 

[11xminurti3]

Walterke, please explane what is wrong that equation.

I appreciate the advice, the rock is falling on a solid surface, the significance of the answer I am after is very minor, a best geuss would do to be honest but I did want something a little better than that

 

[rb1957]

"Force at impact = KE/distance = 318N" ... what on earth distance did you use to derive this (i suspect the 2.5m height) ?

i ask you, did the impact take place over this distance ? (of course not)

have you looked at the othe threads about impacts ?

have even lokked at wiki for impacts ?

have you done anything more than post here ?

sorry (but not really) ... we've tried to guide you but you haven't been able to follow.

i 2nd walter's suggestion, or hit the books, research this yourself, and then you'll be able to follow our leads.

 

[Walterke]

Nevermind 11xminurti3. The calculation is correct. Just the way you're writing it down is a bit unorthodox imho.

However, if it takes you 'a few hours' to figure out the acceleration of a falling object I really don't think you should be the one doing these calculations.

What kind of educational background do you have?

P.S.: I really don't mean to be rude, it's just the the next part of your calculation will be a LOT more complicated then this is, and if you don't have the proper knowledge on the subject, I'm worried about the results. As an engineer myself I make enough errors as is. Just because you know (or know where to find) the formulas doesn't mean you know how to solve this problem, as you may interpret them incorrectly.

Perhaps explain why you need this a bit more, and we might be able to help you.

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[IRstuff]

" if it takes you 'a few hours' to figure out the acceleration of a falling object..."

Not just that, but the fact that the fundamental problem is covered in high school physics. The only question should have been, "what deceleration time or distance do I use?"

TTFN
faq731-376

 

[hydtools]

Walterke, your first comment is correct. v^2 = velocity squared which is not 7m/s
v^2 = 49m^2/sec^2

Ted

 

[EnergyMix]

I think it might be safe to assume that since 11xminurti3 has a 32kg rock that it probaly is NOT going to fall on an mattress. Even if it did, it would probably squash the mattress like the proverbial egg. That's a pretty hefty rock, guys. It's also really not falling very far.

To put it bluntly, it will go "splat." The question is not going to be "how high will this bounce," but "how big a hole is this going to make?" Of course, depending on the material the rock is comprised of*, the rock might break into pieces upon impact (similar to an egg hitting a cement floor.)

For a ROUGH estimate, assume a displacement on the order of 0.01 mm (assuming it's a hard surface.) If you're hitting sand or soft soil and it really will dig a hole, you might want to do a second calculation with a larger displacement.

*correct grammar would insist that I say "of which the rock was comprised." Bah, humbug.

 

[moon161]

If you only want the barest indication of what will happen when it hits, I will tell you this. I would not get my toes under the rock, with our without steel toe boots.

 

[miningman]

I dont know if this is a positive contribution to this thread, but with my background in underground mining, I do know a thing or two about falling rocks..... some alot more than 32kg. Plus the original question is very similiar indeed to a query that was put to me recently by a structural engineer who was getting heavily involved in an area he should have stayed away from.

Is this 32kg rock dropping straight from the back of a tunnel , or is it rolling down the side of a mountain?? Obviously all theoretical calculations are highly dependent on the answer to this question. Then I have to point out that the geometry of all falling rocks will vary between something approximating a perfect sphere, to something that approximates a perfect cube. Depending on what I was designing, I would want to know what the estimated surface area of the contact zone is. Even all this is very theoretical and I would have to design for a point load contact over a very small area.

Penetration of an underengineered system can easily be imagined, but perhaps the areal extent of the penetration would not be critical. I dont know if the OP is out of his area of expertise, but many posters here have, with the best of intentions, made a number of assumptions that might not be valid. Just IMHO of course.