Is this a homework problem? If it is then refer to your physics text book on Force calculation. To give you a hint, the basic equation is F=ma, where m is the mass of the object and a is the acceleration that it is under.

This is where I am having the problem. I know F=m*a but where does the Height comes in. I know v^2=2gh but this is where I get stuck how do I combined the two to solve the force at impact. I"m missing a step somewhere

Thanks for the respond,

Richard

Well, what is the definition of acceleration? Here's an extra point for your class, remember to include the normal force of the ball.

In addition to my previous post and a clue to the acceleration term, was the ball thrown at the ground or just released at the 2 meter height?

the ball was just release v=0 from a height of 2 meters

On Earth, G = 9.8 m/s² regardless of height object is dropped from. Momentum is a function of velocity squared: 1/2 mv². Impact force is a function of elasticities of each material and rate of change (derivative) of momentum.

Study!

thanks bill, M=0.5*m*v^2, where v=(2gh)^0.5 and m=W/g. Lets say that the ball is dropped from rest weight 2.2 lbs from a height of 2 meters on to a steel plate. I need to know the force on impact. Can anyone give me an answer?

Thanks,

Richard

Look up "impact forces" in a reference book -- you'll be surprised at what you'll learn.

In textbooks, it's assumed that the impacts are either perfectly elastic or perfectly inelastic, whereas that may not be the real case. There certainly can be cases where a steel ball falls on a slab of modeling clay, and the ball won't deform at all and the clay does nothing but deform. Modulus of elasticity must be taken into account as well as its curve -- there's a gray area in there where materials deform in elastic regions and after a certain point they begin to deform plastically, and in between it's a combination of the two.

So the answer to your question is, "no," I can't give you an answer with the information you have given me.

This sounds like: Scissor Lift Design Calculations

Hmmm . . . Suspiciously very similar!!!

What is the similarities, detectives...... Ummm, i mean engineers, lol.

Can't a guy share some engineering ideas around here, before the guru's speak, lolllll?

The similarities are:

1. You have come here twice.

2. Both times you have asked questions that could only be homework.

3. You supply no supporting data.

4. You share no engineering ideas, just homework questions.

5. Do some research, then come and ask intelligent questions.

If the question was not intelligent, how come I have not gotton an answers yet, only jebbersish i already know, Mr. Guru. Is not asking questions on forum a part of the research process??

A question can promote engineering ideas. If you don't know by now that questions is part of the learning process and also a huge part of the research process then my friend you are no Guru, just a Grump.

Carry on. What are your engineering ideas?

" Can anyone give me an answer", does not seem like much of an interchange of "engineering" ideas.

Good Luck.

I beg your pardon! You haven't gotten only gibberish!! You have gotten several good pieces of information, and with the information you supplied, no one could answer your question with a particular number.

o.k, I take that remark back since I have gotten several good pieces information

There is one point not clear and all results depend on it. What kind of ball?

At the contact both bodies are behaving in different manners.

The ball comes to the contact with an energy so that the one who said that force is M*a was wrong in this case! The force will depend on the ELASTICITY of the contact and both bodies take part to it. In fact the maximal force depends on the highest elasticity so that to only say that if falls on a steel plate it is not enough.

Your Prof did not a good job if he did not explain how a shock occurs. Make a choice is it a steel ball is it a rubber ball is it a baseball one ? Every one will have an other contact froce with the steel plate.

The ball (sphere) is made of steel, with E=29000000psi with an approximate diameter of 2". The plate is made of steel as well about 36" x 48". I am trying to figure out the dynamic force at impact. I know that the structure respond to the impact elastically and that the stress produced is proportional to the load but I guess i am trying to figure out the impact factor. Total Force= Load * Impact factor, where load is the 2.2 lbs, and the impact factor is based on the deflection.

A 2" steel ball weighs about 1.19 pounds, not 2.2 pounds.

That was just an approximation, I measure the ball to be 2 7/16" dia. I am using it to performing an experiment on cushioning effects of dynamic loading using 1 kgf (approx 2.2 lbs)

I wonder if the impact factor formula: [F(total)= W*(1+(1+ (v^2)/(g*static_deflection))^0.5] would give me adequate theoretical results.

Richard

You want the right answer? OK, here it is: You can't tell. If it lands on your noggin, you get one force; if it lands on your Aunt Sally's beer belly, you get another. Watch my lips - you must know the acceleration at impact!

Check out this site.

Ba/ael, you are the best. That is awsome.. This is the conclusion I came up with.

The ball hits the steel plate with a certain amount of momentum. The amount of force depends on how much the steel plate absorbed the momentum, the shorter the time period of absorption the higher the force. That being said if I use newton second law---> F=ma where a=v/t, then F=m(v/t) if v=245 in/s just before impact and t=0.01s (assumed impact time), then the force is approx 139 lbs at point of impact.

Richard

The mass of the body is known. The velocity at impact can be calculated knowing the height of drop. So the momentum at impact is known. Your assumed impact time has not been given prior to this, so I don't know how you arrived at 0.01 seconds. The distance the ball penetrates into the plate is very small and is related to E (Young's Modulus) for both ball and plate. For steel, this is approximately 200,000MPa (29,000,000 psi).

Here is another site which may shed further light on the subject.

I don't know if you put this question as a joke, but if you think even so simple looking question is really simple if you wanted full scale study about it you are way off my friend and the data you give are by far less than required

For example ball diameter.To give you a clue a 3,9 m diameter ball falling from 2 m (remember center of gravity hight is considered as hight) will not actually fall from too high eh?

Force ball gives and gets at impact is not of course fixed its a curve on time whose math integration gives the change in momentun that ball suffers from impact m*δV=F*δt. But none of that is known because it depends on the actual lowest point ball's center of gravity goes depending on ball and ground deformation depending on material characteristics and time of colission also depends on them. Multi differential equation here eh?

And we haven't touched aerodynamics yet that have bigger impact as diameter grows right?

How about if ball is not solid but let's say air pressure filled? Even pressure and gas type has impact on the curve. Finite element analysis here. And what point of the curve you want? max? average is fine?

The problem is even though there is no harm on stressing your mind on crap I don't really find much use on extensive analysis of problems like this unless you're retired and try to keep your brain fit or a wannabe Isaac Newton

Why not model it in any of the FEA programs? That should tell you what you want.