Distance travelled by a falling object

[skyping]

First Post..hopefully of many!

I have to work out the minimum distance a 5.6KG object can fall vertically to 'acheive a force' of 12KG when the object hits the ground.

I have no other information and really need some help!

P.S. I am not a scientist, simply a confused manager needing a bit of help in a very small organisation!

Thanks for looking.

 

[tiny-tim]

welcome to pf!

hi skyping! welcome to pf!

I have to work out the minimum distance a 5.6KG object can fall vertically to 'acheive a force' of 12KG when the object hits the ground.

sorry, but i don't understand the question …

the force depends on the distance in which the object stops …

it will have a greater force on stone than on butter

what exactly is the problem …

and why do you believe that "12KG" is the limit?
​

 

[skyping]

Thanks for the reply tiny-tim.
This problem was given to me as I'm the only person in the company with any physics qualification...an a level 20 years ago!
Basically, one of my team is trying to deduce (estimate?) What the minimium distance a 5.6kg object could be dropped from a stationary position to exert a force of 12kg on the ground. They have estimated that this value will damage the object, but I need to work out the vertical distance and I am both very rusty and unsure een if this is even possible!

Hope this is clearer and once again thanks for looking.

 

[tiny-tim]

Basically, one of my team is trying to deduce (estimate?) What the minimium distance a 5.6kg object could be dropped from a stationary position to exert a force of 12kg on the ground. They have estimated that this value will damage the object …

i think you're going to have to tell him that force isn't the issue, and ask him to give you the data his "force of 12 kg" comes from, so that you decide scientifically what the correct question is

(i suppose they have considered the option of not dropping it on the ground?)

(hmm … has anyone dropped your team on the ground recently? )

 

[berkeman]

First Post..hopefully of many!

I have to work out the minimum distance a 5.6KG object can fall vertically to 'acheive a force' of 12KG when the object hits the ground.

I have no other information and really need some help!

P.S. I am not a scientist, simply a confused manager needing a bit of help in a very small organisation!

Thanks for looking.

hi skyping! welcome to pf!


sorry, but i don't understand the question …

the force depends on the distance in which the object stops …

it will have a greater force on stone than on butter

what exactly is the problem …

and why do you believe that "12KG" is the limit?
​

Welcome to the PF.

As tiny-tim is saying, the object produces an "impulse" when hitting the ground, not a "force":

http://en.wikipedia.org/wiki/Impulse_(physics )

.

 

[technician]

First of all force is not measured in kg and it sounds to me that you mean a force equal to the weight of 12kg. If you take the force of gravity to be 10Newtons per kg (it is actually 9.8) then the weight of 12 kg is near enough 120N.
When the 5.6kg mass falls it will exeret a force on whatever stops it and, as tiny tim hints, this depends on how quickly it is stopped.
The equation to calculate the force when an object is stopped is
F = (change in momentum)/time taken.
Momentum = mass x velocity
This means you must make an educated guess of how long the falling 5.6kg will take to stop in its collision...it could be 1/10, 1/100 or 1/1000 of a second...lets take 1/10
this means that momentum/0.1 = 120N...gives momentum = 12Ns (dont worry about the units at this point.)
This means we want the 5.6kg to have a momentum of 12Ns when it collides.
momentum = mass x velocity so we want the mass to reach a velocity of 12/5.6 = 2.1m/s
To calculate the height from which to drop the 5.6kg mass must be dropped we can use an equation v^2 = 2gs. g=10 (acceleration of gravity) and s = the height
so 2.1^2 = 2 x 10 x s gives s = 0.22m or 22cm.
If you think the 5.6kg will stop in 1/100 of a second you can work through to find the new height.
I may have completely misunderstood your original problem ! but at least there are some figures and ideas for you to look at.

 

[skyping]

Wow...thanks to everybody for their input!

I really appreciate the excellent explanations given, and it certainly has given me 'food for thought'.

I did think that there was not enough information to give an exact answer, and so I was correct in that respect, even though I got the units of force wrong (as said very very rusty)!

Many thanks again from a appreciated desk manager who will both use and broadcast this excellent forum to his colleagues!