- 24 Jan '10 23:16 / 1 edit

Hmm, let me re-phrase.*Originally posted by joe shmo***100 lb**

An 100 pound object is dropped from a height of 10cm. Does the object exert more force on the ground when it hits the ground versus when the object is just resting on the ground?

I may have the terms mixed up but you get the idea....is there an increase? how Much? - 25 Jan '10 00:45

Hmm.*Originally posted by uzless***Hmm, let me re-phrase.**

An 100 pound object is dropped from a height of 10cm. Does the object exert more force on the ground when it hits the ground versus when the object is just resting on the ground?

I may have the terms mixed up but you get the idea....is there an increase? how Much?

The forces will be equal and opposite on the two, so I will analyze the 100lb object.

The force due to gravity will very, very slightly increase because the distance between the two masses has decreased, but that's not what you're getting at.

If we assume the force of gravity remains constant, then without another force involved the object should continue accelerating indefinitely. But it decelerates when it hits the surface.

This is because of an electric repulsion force between the atoms that increases with distance much faster than gravity decreases with distance. At some distance and velocity (touching and 0) the two forces will hit an equilibrium and acceleration will cease.

In short, YES, the object exerts more force on the ground when it hits. That force comes from electric repulsion of the electrons in the atoms.

How much? Too lazy. Next person? - 25 Jan '10 02:41 / 1 edit

Ok yes, there is an increase in Force. The increase is due to the impulse*Originally posted by uzless***Hmm, let me re-phrase.**

An 100 pound object is dropped from a height of 10cm. Does the object exert more force on the ground when it hits the ground versus when the object is just resting on the ground?

I may have the terms mixed up but you get the idea....is there an increase? how Much?

J(impulse) = Delta(mv) = Integral(F(t)•dt)

So in order to solve for the maximum force we have to have the relationship F(t).

However we could determine F avg. given the time interval for which the momentum changes. - 25 Jan '10 08:23

Do this experiment:*Originally posted by uzless***Hmm, let me re-phrase.**

An 100 pound object is dropped from a height of 10cm. Does the object exert more force on the ground when it hits the ground versus when the object is just resting on the ground?

I may have the terms mixed up but you get the idea....is there an increase? how Much?

Stand on your bathroom scale. How many kg's, or pounds, or stones or whatever, does it register?

Now, take a little jump. When you start your jump, then the scale increses. When you're in the air, then the scale says zero. When you land then the scale says more again. And when you are still again, then you can read your own wdight again.

If you plot the wieght your scale says in a function of time, then you weill see that the average weight, during this jump cycle, has an average of your actual weight. The weight more equals out exactly the weight less. - 25 Jan '10 08:40

Yep*Originally posted by joe shmo***Ok yes, there is an increase in Force. The increase is due to the impulse**

J(impulse) = Delta(mv) = Integral(F(t)•dt)

So in order to solve for the maximum force we have to have the relationship F(t).

However we could determine F avg. given the time interval for which the momentum changes.

Schoolboy physics tells us that Ft = mv-mu

ie The Impulse = the momentum change.

In practice this depends on the materiels involved, are you landing on concrete or a sprung mattress? The more 'give' on landing the larger t and the smaller F. - 25 Jan '10 19:45 / 1 editlol, ok, what i"m looking for is a number. Fabian, yes the scale moves heavier and lighter while jumping than when a person is just standing.

I'm trying to figure out if when the object lands on the ground is it landing with 100 pounds of downward force or has that downward force increased and, if so, by how much? I'm looking for a number.

Best i could do on my own with some interwebz research was to find that jumping jacks exert 3 times your body weight in force on the joints. So, if you weighed 100 pounds, you'd be hitting the floor with 300 pounds of force.

Anyone care to attempt a calc? You guys are better at this than I. - 25 Jan '10 19:47 / 1 edit

assume zero give.*Originally posted by wolfgang59***Yep**

Schoolboy physics tells us that Ft = mv-mu

ie The Impulse = the momentum change.

In practice this depends on the materiels involved, are you landing on concrete or a sprung mattress? The more 'give' on landing the larger t and the smaller F.

image you are sitting in the basement and your one hundred pound sister starts jumping up and down on the floor above you. It makes a hell of a lot more noise than if she is just walking around....yet her weight is still the same. How come? - 25 Jan '10 20:06
*Originally posted by uzless***assume zero give.**

image you are sitting in the basement and your one hundred pound sister starts jumping up and down on the floor above you. It makes a hell of a lot more noise than if she is just walking around....yet her weight is still the same. How come?**assume zero give.**

I believe this is an impossibility.

Lets do a little math and change your measurements into SI units, lets say our jumping guy has a mass of 100kg, and assume g=10.

After his little jump of 10cm (S) what is his velocity (v) on landing?

S=1/2 at^2 ; t=sqrt(2S/a)

v=at

so t = sqrt(0.02)

v = sqrt(2)

his momentum is thus 100 * sqrt(2)

Now we use Ft = mv-mu (this t is the time it takes to decelerate to zero)

(u=0 ie at rest)

sso F=mv/t

If he comes to rest in 1 second the force is 100 * sqrt(2) Newtons

If he comes to rest in 0.01 seconds the force is 10,000 * sqrt(2) Newtons

big difference - it all depends on what he lands on!

(Incidentally the decelerating force is in addition to his weight)

You can see that for an instantaneous deceleration the force would have to be infinite. - 26 Jan '10 11:48
*Originally posted by uzless*

An 100 pound object is dropped from a height of 10cm. Does the object exert more force on the ground when it hits the ground versus when the object is just resting on the ground?

I may have the terms mixed up but you get the idea....is there an increase? how Much? - 27 Jan '10 04:17

Hmm, Ok wolfgang...so if it takes the kid 1 second to come to a rest then the answer is 100 *sqrt(2) Newtons.*Originally posted by wolfgang59***[b]assume zero give.**

I believe this is an impossibility.

Lets do a little math and change your measurements into SI units, lets say our jumping guy has a mass of 100kg, and assume g=10.

After his little jump of 10cm (S) what is his velocity (v) on landing?

S=1/2 at^2 ; t=sqrt(2S/a)

v=at

so t = sqrt(0.02)

v = sqrt(2)

his momentum i ...[text shortened]... ht)

You can see that for an instantaneous deceleration the force would have to be infinite.[/b]

Man, what a thoroughly unsatisfying answer.

I was hoping for a simple 300 pounds or something. Kind of like how if you are in an airplane and the pilot banks hard right you'd feel 2 or 3 G's so you could easily just say that in the turn a 100 pound person would feel like they were 300 pounds in a 3G turn. - 27 Jan '10 05:55

100 * 2^(1/2) ~ 140 N*Originally posted by uzless***Hmm, Ok wolfgang...so if it takes the kid 1 second to come to a rest then the answer is 100 *sqrt(2) Newtons.**

Man, what a thoroughly unsatisfying answer.

I was hoping for a simple 300 pounds or something. Kind of like how if you are in an airplane and the pilot banks hard right you'd feel 2 or 3 G's so you could easily just say that in the turn a 100 pound person would feel like they were 300 pounds in a 3G turn.

140 N ~ 30 lbs - 27 Jan '10 06:00 / 2 edits

Put a G-meter on your head, and another G-meter on your shoe. Then make a jump from (say) two meters down to a concrete floor. What G-meter do you think measures the largest G?*Originally posted by uzless***I was hoping for a simple 300 pounds or something. Kind of like how if you are in an airplane and the pilot banks hard right you'd feel 2 or 3 G's so you could easily just say that in the turn a 100 pound person would feel like they were 300 pounds in a 3G turn.**

Can you guess what the head-G-meter would tell? And the shoe-G-meter?