OK, ill make it little bit difficult.
Lets say a concrete block holds an element that has a weight of 100 kN. The block cracks, is destroyed if the weight is 200kN. So from what hight should the element fall to destroy the concrete block? I want to compere the stress done by the element to the block if it just stands there and if it is in motion.
''Force is what makes it hurt, which is determined''
So why the apple standing still on my head doesnt hurt, but falling from tree does. Apple produces force F=mg, does it get bigger if it falls? think not. How to compere stress done to my head between these two apple conditions?
Because the apple sitting on your head isn't experiencing a decelleration. The decelleration of an apple falling onto your head (and being slowed down by your head) is what produces the force which you feel.
Thank you xxChrissxx now it makes sence. So to know how big will be the force i need to know how much the impact would take time? And the time of the process is dependent from the elasticity of my head? Am i thinking right? Now it makes sence to put a pillow on my head.
Well strictly it can only be done empirically, as collisions are generally very complex and no two collions are ever precisely thae same. It depends on how the bricks are orientated, how they strike the ground. What the two materials are made of, what happends during the impact event. etc etc.
However with some simple assumptions and some information on the impact, you'll have to make an educated stab at guessing the impact time you can work out the drop height to attain a force necessary to break something.
Suppose you are standing on planet Earth now the apple is falling down on your head freely. Then its force should be F = mg OR GMm/(r - h)2
where, m = mass of apple
M = mass of earth
r = distance of apple from earth
h = your height
when this apple will fall on the surface of head it would create pressure.
P = F / A
as much pressure it would create that much pain you would get.
You have already taken revenge of this pain according to NEWTONS LAW by creating same but opposite force. F = - mg
so the apple can be smashed
The force upon collision isn't the same as when it is standing still. When something picks up speed and collides into something, there will be a transfer of energy resulting in a large momentary jump in the force, which can be described by the impulse - the impulse is the change in momentum, or for that matter, the distribution of momentum across the interacting objects - it can also describe the "distribution" of force through time.
I've left this thread for two days, now I come back to see that my reason is wrong? are you guys saying that it's not "IMPULSE" that's causing the hurt?
Should I remind you people that the definition of impulse is "the change in momentum through time" which means the change in velocity. The apply hurts him because it's decelerating, i.e. it's because the impulse his head is applying on the apple!
Impulse also explains why if a fast car hits a wall the damage is more than a slow car hitting the wall, it's because the impulse the wall applies on the latter is less!
Dont panic impluse is the same thing. Impulse is acutally more correct as it takes into account the distrubution over time.
impulse = f * t
The method we used assumes an average force, but is a more simple explinations as it doesnt require calculus. Its still the force that does the hurting, impulse describes how the force acts better.
Eg. When we say that its force, we acutally mean impulse. But for someone who is unfamiliar with the concepts this caould be confusing. Becuase what we are waying is tha tthe force is acting over the impact time.
Apple 1 KG, stops from 10m/s in 1 S.
Force = 10N
The impluse applied is 10(n) for 1(s) = 10Ns
Using the rate of change of mumentum and time, we will still get a 10Ns impulse, but your method will describe the curve of how the force acts. The area unde rthe force-time curve will be the same in both incidences. So for example it would start and end at zero, curve and peak at 20N