Reducing the Impact Load of a Falling Object

Don't let it drop. You cannot change the final force at rest, so your only choice is to not allow it to fall quickly. Now how you cushion this descent will change depending on how the mass is supported.

Reduce the mass. Reduce the height. Distribute the load over a wider area. Ask the teacher.

Load distribution changes will change the pressure and thus may minimize any damage. But the force will remain the same.

ok

Wait a minute, here's a snotty homework answer.

Take the one thousand kilograms of steel that has a volume of about 1.27E5 cc. Press roll this steel into a single one half millimeter thick square sheet of steel. The square sheet of steel will now be about sixteen meters on each side of the square. Dropping this sheet of steel 2 meters (square side down) in a normal atmosphere will require a displacement of over 500 cubic meters of air displacement. Building a 2 meter deep packing square face container like box to hold this sheet of steel will both permit suspending the steel before dropping and restrict the airflow as the sheet descends. This will thus add a dampening factor (k) to the kinematic equation, thus limiting the velocity of the sheet at impact at the bottom of the container.

I will leave the final numerical calculation of this differential equation as an exercise for the student. Don't forget, there will be a midterm exam next Friday.

Yep. That's how I saw it.

_{The little twit!!!}

Don't forget that if you mirror finish the top surface you will help global warming and get extra credit.

I was going to suggest a very springy structure, but I prefer your answer.

Oh, remember no homework credit will be given using this method without including the calculation for the total area between the sheet metal and the sides of the box.

i dont know why everybody concluded its homework question????????????

the real problem is fan assembly that weights 1000 kg and rotating at 100 rpm at height of 10 m may fell down and creates risk so the solution is to put kind of support at 2 m from the fan. so can this support withstand this impact load?

is there any wrong in simplifying real situation to basic theory of mechanics?

This bears no resemblance to your first post. Now we have, " creates risk so the solution is to put kind of support at 2 m from the fan."

Why not just fix the fan?

Wait...you have a fan that weighs 1000kg and you are worried it might fall? There are so many variables with this problem we cant begin to come up with an answer. As LynLynch says, fix the fan so it cannot fall because at even short lengths, fan blades spinning at 100rpm and 1000kg can pack quite a punch, if the blades are made of a tough material they could do a serious amount of damage.

If this is not some kind of homework assignment, you need to get somebody with some expertise to evaluate the situation in person so they can engineer safety measures for you.

Drew

Redfred-- GA for this particular thread given the nature of the OP. But I think Aaron Robinson's answer is a good approach to the generalized problem.

I myself have approached a similar problem involving potential impact loading on a piece of delicate machinery using a similar impulse=momentum approach. Doing this made obvious what we all know instinctively; that the more the flexibility (which increases the int "dt" of the impact) in the contacting bodies the less the destructive forces.

Which gives some supporting theory to the old WWII Seabee saying that "Waves could break anvils with rubber mallets"

One other remembrance from my abovementioned encounter with the delicate machinery problem it worth passing along. I obtained a reasonable approximation of spring constants and fundamental vibration modes for the situation at hand (square root k/m) by testing the structure in three dimensions with a spring scale and a dial indicator strategically positioned. For large structures with fundamental vibration modes of a few hertz this method will also give a rough demonstration of the dampening in the system by watching the decay rate of the dial indicator deflection after a quickly applied load. (Remember that stuff about critical dampening in single degree of freedom vibration systems?)

A selection of simple spring type force gauges for a range of forces over several orders of magnitude along with a dial indicator and a selection of simple mounting hardware is an excellent tool set for the mechanical engineer's kit. One can often in a few minutes assess the character of a problem where otherwise hours would be spent studying drawings and doing calculations followed by more hours defending one's calculations and assumptions.

Ed Weldon

Thanks for the GA. Yes, Aaron Robinson is on the path to solve this problem. It's the path I started down in my attempt to impress the OP that he was in more trouble than he thought. (Or at least the impression I and others had of him/her.) I debated pointing out that from the energy transfer equation, this 1000kg mass will be moving at 6 m/s when the mystery platform would have to catch this. When he/she announced leaving us, I found adding that calculation result pointless.

Now many well known systems today do stop this much falling material successfully. The scenario that immediately comes to my mind that is closest to this stated problem is the landing of an aircraft. But these systems are designed by well trained engineers and craftsmen that know what they are doing. This person clearly did not.

I still think that this is a homework problem, and that the OP added the details because he/she was caught and hoped to draw out more useful information. I still like my sheet metal answer though. That was fun.

Oh Papa... so now it is a 1000 kg fan sitting 10 meters up and you want to minimize the damage when it falls down. Is it so??? So why did you not say so at the beginning?? Well, attach a bloomin' parachute to the fan, so instead of falling it will just gently float down. It is obvious that the parachute will have to be properly dimensioned and spread in readiness.

even you if you did many thing to maintain your machines they still break

Is this fan in a cooling tower?

As the previous posts say, you simply can not let a rotating equipment with that mass to fall even by 1/2 meter. It is not the impact, it is the consequentials.

You have to ensure it does not fall. carry out a proper design and with sufficient factor of safety and it will not fall. After all there are bridges couple of centuries old, but not falling due to

- Proper - rather may be considered over-factor of safety, but in these cases worth it.

- Periodic maintenance and checks.

Do these and you don't have to worry about it falling.

Do not try to support the fallen, avoid it from being fallen .

Catch it before it hits the ground. This will reduce the impact considerably. Besides, you say that you want to find (know) the impact load. Now if you reduce the falling force of the mass, how will you ever know the true impact load??? So what is the purpose of this whole thing???

its my fault i have posted in this forum

i am done

Goodbye

Your problem isn't posting in this forum. Your problem is the minimal information you gave with your posts. You ask people who know something about physics a question with serious consequences and you get upset by our answers. Perhaps you are better off in a less technical forum.

Anon

The problem here is the OP is highly annoyed because...

He wants the fan to fall and the others are forcing him to avoid it.

The force with which he wants the fan to fall, make it really look like a homework in sheep's skin.

GA Aaron. You've given us a useful approach to numerical analysis of this type of problem. This is a good analytical approach to this type of problem, often forgotten by many mechanical engineers and worth keeping alive in one's bag of engineering tricks.

Ed Weldon

"assumes that nether object has a significant amount of plastic deformation."

in reality this is not true because the structure either will have significant plastic deforemation or will fall together with object.

Catch it like cricketers!

Let's assume that there is a good reason for asking the question.

In which case, judicious use of nylon rope seems appropriate.

Provide additional hub keyed to the vertical fan shaft above blades. Make arrangements for stopper and fixer structure, which can stop or block the falling fan assembly. Such a safety system will not interfere with the rotation of the fan and as well work as safety limit protection.

I hope your intention is to avoid falling risks of the fan by extra safety provision.

Thank you for your comments and getting the reason behind my question

can you more clarify your idea

if we put additional hub assembled to the shaft it will rotate with the whole fan

so how be can we fix the additional hub.

Thank you guest for the reply. The hub can be an outer shell or even a flange keyed to the vertical fan shaft above the blade. As you have stated the hub or flange will be rotating along with the fan.

But the stopper unit is placed just below the hub or flange. This stopper is just a hollow outer shell, which is just few mm bigger than the fan shaft as well few mm smaller dia than the flange or hub. This stopper will not be touching the fan shaft as well the flange or hub, since it is to be supported by external linkage to wall fixer.

The fan shaft will be having a free rotation inside stopper shell. The hub or flange during fall will be arrested by the fixered shell and work as an additional safety to avoid the impact of the falling fan.

*It is so simple , cheap and ensures a secondary safety..

Hope I made my expression clear. if you need any drawings ,I can draw a sketch, scan and post. best wishes.

i will be very thankfull if you can post a sketch for the arrangement

but before that i would like you to see picture of the fan hub assembled to the gearbox shaft

GUEST,

Please find the rough diagrams for your referenxe. A fan sheild outer guard is also suggested as per optiom A[shown in dotted lines]

Thanks a lot dear

Use a bed of Poliestirene Foam ISOPOR spheres. This is used in the Bremen (Ge) Free Fall Drop Tower for reduce the impact of the 2,000 kg Chamber that falls in vacuum at a height of 100 m.

See < http://en.wikipedia.org/wiki/Fallturm_Bremen >.

It has a 123-meter-high drop tube (actual drop distance is 110 m), in which for 4.74 seconds (with release of the drop capsule), or for over 9 seconds (with the use of a catapult, installed in 2004) weightlessness can be produced. The entire tower, formed out of a reinforced concrete shank, is 146 meters high.