Solving Mr Occam's Razor with a Drop Weight - D

[Mr.D]

Hi,

We have a saying where i come from:

Yorkshire born,
Yorkshire bred.
Strong in t' arm,
Weak in t' head.


That said I have one dear principle which is from dear old Mr Occam and his Razor.

So, here I am attempting to simulate a Fork Lift Truck at 2000kg, traveling at approx 10 cm per second and impacting a stationary Racking Upright. I have constructed a drop weight of 50kg. How high must this weight be dropped from in order to approx simulate the FLT impact?

Or, from which hieght does the 50kg weight need to be dropped in order to create an impact force of 400NM?

The first person to give me the correct answer gets to have me as a slave for a week!

Regards,

D,

 

[Doc Al]

Simulating a collision? Not easy!

So, here I am attempting to simulate a Fork Lift Truck at 2000kg, traveling at approx 10 cm per second and impacting a stationary Racking Upright. I have constructed a drop weight of 50kg. How high must this weight be dropped from in order to approx simulate the FLT impact?

I doubt you'll find my answer all that helpful: It depends on what you are trying to simulate. Most likely, you would find that giving the drop weight the same kinetic energy as the truck would be more useful. In that case, since the KE (= 1/2mV^2) of the truck equals about 10 J, the weight would have to be lifted to a height of 10/(50x9.8) = 0.02 M. But if you wanted to have the same momentum, which is mV = 200 Kg-M/s, the height would have to be 0.82 M.

Or, from which height does the 50kg weight need to be dropped in order to create an impact force of 400NM?

Do you mean 400 N? (NM = Newton-Meter is a measure of energy, not force.) Now that's tough. Because the force depends on the details of the impact, specifically the duration of the collision. Impossible to predict with the given information. For example: dropping the weight onto a mattress vs dropping it onto an iron slab--the force will be quite different.

Perhaps an engineer will be able to give you some additional insight.

 

[Mr.D]

Doc Al,

Thanks Doc, 820mm gives me a start.

Let explain what I'm trying to do. I have invented a device to prevent/mitigate Fork Lift Truck impacts to warehouse pallet racking. The typical FLT weighs 2000kg. The impact protection device is a multilateral polymer based sleeve that attaches to the front and lateral sides of the racking column.

Momentum of the truck is important as is the stationary steel upright. I was informed that the industrial standard that impact protection devices should withstand is 400NM??

 

[KingNothing]

I have a question. It's kind of on a tangent. If Newton Meter is a unit of energy, shouldn't it be a unit of work, because work=force*distance?

Also, the 'industrial standard that protection devices withstand" would be measured in Newtons.

 

[Doc Al]

units of energy/work

If Newton Meter is a unit of energy, shouldn't it be a unit of work, because work=force*distance?

Sure. Any unit of energy is also a unit of work. 1 NM = 1 Joule.