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| jackw72 |
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 jackw72 World Chat Champion

Joined: 18 Nov 2006 Karma :     
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 Posted: 01:42 - 25 Jan 2012 Post subject: Force on impact calculation. |
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Hi,
Anyone got any idea what calculation I would use to calculate the force a human being would have whilst sprinting 100m.
I.e. how hard would a man running 8 meters per second hit another object.
I have tried the kinetic energy formula but am wondering if anybody knows what calculation would be used for this?
Thanks,
Jack  ____________________ 08/11/2007 A2 Passed.
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| at106 |
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 at106 Brolly Dolly
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| UnknownStuntm... |
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 UnknownStuntm... World Chat Champion

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| Stevey |
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 Stevey Two Stroke Sniffer
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| D O G |
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 D O G World Chat Champion

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| jackw72 |
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 jackw72 World Chat Champion

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| smegballs |
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 smegballs World Chat Champion
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| jackw72 |
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 jackw72 World Chat Champion

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| D O G |
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 D O G World Chat Champion

Joined: 18 Dec 2006 Karma :     
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 Posted: 12:32 - 25 Jan 2012 Post subject: |
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He cannot stop dead - that would require infinite force.
If you want to know the kinetic energy a 90kg man moving at 8.6m/s has, that is something different.
This site tells you that it is 3.3kJ. |
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| jackw72 |
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 jackw72 World Chat Champion

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| D O G |
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 D O G World Chat Champion

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| smegballs |
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 smegballs World Chat Champion
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| ThoughtContro... |
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 ThoughtContro... World Chat Champion

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| jackw72 |
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 jackw72 World Chat Champion

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| D O G |
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 D O G World Chat Champion

Joined: 18 Dec 2006 Karma :     
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 Posted: 14:59 - 25 Jan 2012 Post subject: |
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| jackw72 wrote: | Ok guys let me put a scenario out but it is kinda complex.
I have just started playing American football and did a test 100m at said speed and weight. I wondered what sort of impact me running into someone stationary at that speed would be equal to.
The faster a mass moves the more force it gets (is there a formula for working out how much the mass increases as it gets faster?) |
It builds up more energy, a 'force' is different - a force is the mechanism of energy transfer from one body to another, the higher the force, the faster the energy is transferred.
| jackw72 wrote: | So lets say I am facing a man who weighs 80kg who isn't moving running towards him at 8.6 meters per second at 90k with a 3.3 kj energy.
I think working out the newtons force would be best considering there is no way to tell how far a man @ 80kg would move in this equation without going into masses of detail. |
I'm struggling to understand what you are trying to do, too. I think you are trying to give a 'real world' comparative to the energy in the impact. The easiest way to do this is to assume that all the energy you have when running is used up in the collision - by breaking bones, brusing flesh, acelerating the other guy backwards, stopping you, making sound waves, producing heat etc etc etc.
That energy is 3.3kJ, which is the equivalent of the same energy a 1,000kg car would lose if it crashed into a hypothetical brick wall at 6mph (2.56m/s).
Or, if you could convert it directly into heat energy, and transfer that with no losses to 1kg of water, you could raise its temperature by 0.8 of a degree (celcius). |
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| Frost |
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 Frost World Chat Champion

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| jackw72 |
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 jackw72 World Chat Champion

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 Posted: 16:01 - 25 Jan 2012 Post subject: |
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| D O G wrote: |
I'm struggling to understand what you are trying to do, too.  I think you are trying to give a 'real world' comparative to the energy in the impact. The easiest way to do this is to assume that all the energy you have when running is used up in the collision - by breaking bones, brusing flesh, acelerating the other guy backwards, stopping you, making sound waves, producing heat etc etc etc.
That energy is 3.3kJ, which is the equivalent of the same energy a 1,000kg car would lose if it crashed into a hypothetical brick wall at 6mph (2.56m/s).
Or, if you could convert it directly into heat energy, and transfer that with no losses to 1kg of water, you could raise its temperature by 0.8 of a degree (celcius). |
Yes that is what I am trying to do. I want to know at the point of impact how much force I am putting out onto the other person (or person weighing wall if it is easier) but in a way that a simpleton could understand.
I hit the man/wall with x amount of force, x amount of newtons etc.
Does my mass (although not physically) increase as I am running faster and by how much? 1.003% per mph?
I dunnahhhh its confusning everyone but thank you for your expertise so far I hope I am challenging you  ____________________ 08/11/2007 A2 Passed.
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| ThoughtContro... |
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 ThoughtContro... World Chat Champion

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| el_oso |
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 el_oso World Chat Champion

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| mickfulton |
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 mickfulton Nova Slayer

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 Posted: 19:29 - 29 Jan 2012 Post subject: |
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If a runner stops dead from 8.6 m/s in 0.01s then the deceleration would be 8.6/0.01= 860m/s/s
The Force required to do this is F=ma ......... 90kg * 860= X N
Hope this helps ........you need to know the time it takes to stop the runner and this tells you the required force to stop them, or the force the give to the thing stopping them.
It gets more complicated if you take in to account collisions (inelastic or otherwise), so don't bother
Force is something which changes the shape, speed or direction....so if the runner is at constant speed then effectively he is experiencing NO FORCE (i.e. balanced forces) in the horizontal plane.
Maybe try reading a basic newtonian mechanics book ____________________ Current half-cars :2004 SV650, VX535 1991
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| Paulington |
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 Paulington World Chat Champion

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 Posted: 23:17 - 07 Feb 2012 Post subject: |
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Simple physics. In all these calculations I am talking about collisions and everything happening in 'Physicsland', a magical place where no drag forces exist and in all collisions whether elastic or inelastic transfer of energy is total and 100% efficient.
Let's say the man is an average weight of 70kg and that he is running at 8ms^-1.
Ke = 0.5mv^2
Which is calculated as thus:
0.5 x [70x(8^2)] = 2240 Joules.
That means that if that man were to hit an object he would impart 2240 joules of energy into the object.
F=ma and any of the SUVAT equations are NOT THE CORRECT WAY to calculate this as has been done above as you have asked how much force a running man would have, that is the wrong terminology.
A running man has kinetic energy which he has gained from chemical energy in his muscles. The kinetic energy you have referred to is the 'force' but it is better called 'energy'.
The best way of determining the force in Newtons that the runner hitting the object has imparted is using the equations of motion. To do this you would have to use the equations of motion, more specifically 'v = u+at' and 'F=ma'.
Firstly, you would have to know the speed the runner was doing the INSTANT before he hit the wall, that is U, V is the final speed, i.e. 0ms^-1 and T is the amount of time it took the runner to go from max speed to a complete stop after hitting the wall, around 0.1-0.4 seconds.
Then, knowing 'a', you can plug that in to F=ma where a is the answer from the equation above and m is the mass of the man in question.
Or in one single equation:
F = m[(v-u)/t].
This would give you the force in Newtons that the runner applied to the object he hit with his body. ____________________ "Four wheels move the body, two wheels move the soul."
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Old Thread Alert!
The last post was made 14 years, 254 days ago. Instead of replying here, would creating a new thread be more useful? |
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