Inertia physics question

[Bill Haywood]

If I drop a 1 pound weight ten feet, how much inertia does it gain?

Or to put it another way, how many pounds of weight will it be equivalent to?

[gull]

F = M * A

M is constant; A is constant; therefore F is constant.

[Bill Haywood]

[ QUOTE ]
F = M * A

M is constant; A is constant; therefore F is constant.

[/ QUOTE ]

?

[Silent A]

[ QUOTE ]
If I drop a 1 pound weight ten feet, how much inertia does it gain?

Or to put it another way, how many pounds of weight will it be equivalent to?

[/ QUOTE ]

What are you asking? I can think of two interpretations:

1) the gain in inertia due to changes in relativistic mass as you lift/drop the weight (doesn't really make much sense here).

2) if you dropped a 1 pound weight on your foot from 10 feet up, how much would it hurt (in terms of an equivalent weight being placed directly on your feet - but not dropped).

If it's (1) the difference is freakishly trivial.

If it's (2) then it all depends on how easily your foot and the weight deform when they collide. Dropping a 1 pound loaf of bread and a 1 pound iron ball will have very different effects on your foot.

If you mean something else, please clarify.

[gull]

[ QUOTE ]
[ QUOTE ]
F = M * A

M is constant; A is constant; therefore F is constant.

[/ QUOTE ]

?

[/ QUOTE ]

Weight is a force.

[Fels krone]

The inertia will not change when you drop the object. Inertia is resistance to motion and is dependent on mass only (and shape for rotational considerations).

The weight is the force of gravity acting on the object. It will be the same throughout 10 ft. When the object impacts, the impulse it produces will equivalent to what weight?; Im guessing thats what you were asking. You would integrate the momentum over the time interval of the impact to find the energy transfer; then find the force of the impact.

[thylacine]

[ QUOTE ]
If I drop a 1 pound weight ten feet, how much inertia does it gain?

Or to put it another way, how many pounds of weight will it be equivalent to?

[/ QUOTE ]

You have a Catch-22 here.

If you were able to clarify your question enough that someone could figure out what you were asking then you probably wouldn't need to ask the question to start with.

[ApeAttack]

[ QUOTE ]
If I drop a 1 pound weight ten feet, how much inertia does it gain?

Or to put it another way, how many pounds of weight will it be equivalent to?

[/ QUOTE ]

As the ball falls, it gains kinetic energy as its velocity increases. As wiki puts is, "It should be emphasised that 'inertia' is a scientific principle, and thus not quantifiable."

It seems like you are talking about 'impulse'
http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html

F = ma = m(dv/dt)
Impulse = F*dt = m*dv = change in momentum

where dv is the change in velocity of an object due to a collision
dt is the amount of time it takes for the object to slow down (very short if a falling body collides with the ground).
F is the average force over dt.

The dv goes up as you allow the ball to fall farther increasing the F it can deliver (assuming dt is the same).

[m_the0ry]

[ QUOTE ]
If I drop a 1 pound weight ten feet, how much inertia does it gain?

[/ QUOTE ]

p(momentum) = m(mass) * v(velocity)

This involves tedious math. Convert pounds to kg using earth's gravitational acceleration, solve for speed in a 10 foot window and you have your answer.



[ QUOTE ]
Or to put it another way, how many pounds of weight will it be equivalent to?

[/ QUOTE ]

inertia increases kinetic energy which means the weight will exert a larger force when it hits something but the inertia itself does not effect the actual mass or weight of the object in question. Technically if the weight moves fast enough it would gain mass but that is a relativistic effect, and if you have found some way to move a 1 pound weight at near the speed of light you have more important things to do with your time than ask these questions cause I bet the military would be really interested.

[PairTheBoard]

[ QUOTE ]
If I drop a 1 pound weight ten feet, how much inertia does it gain?

Or to put it another way, how many pounds of weight will it be equivalent to?

[/ QUOTE ]

Inertia is the principle that an object does not change its velocity unless a force acts upon it. The falling Weight does not change that principle.

Under simplification assumptions, the velocity of the Weight at impact can be calculated using the acceleration due to gravity of 32ft/sec^2 and the formulas,

v = gt
d = .5gt^2

where v=velocity, g=acceleration due to gravity, t=time falling, and d=distance fell. So,

t=sqrt(20/32) seconds
v=32*sqrt(20/32) = sqrt(20*32) = about 25 ft/sec

The weight of the "Weight" is defined to be the measure of the force of gravity acting upon it. That will remain 1 pound (plus a tiny bit more due to being 10 feet closer to the center of the earth). If you took the "Weight" 50,000 miles away from earth it would weigh a lot less than 1 pound. 10 feet won't make much difference though. However the Mass of the object would stay the same in all cases (ignoring miniscule relativistic effects) and would still be measured (I believe) as 1 pound.

The Momentum of the Weight at impact is it's mass times it's velocity. Measured in Pound-Feet/sec it will be about 25. Notice though that before being dropped the Momentum of the 1 pound Weight is 0.

The Kinetic Energy at impact is (.5)mass*velocity^2, so in pound*(feet/sec)^2 will be about 625. Once again it was 0 before being dropped.

Momentum and Kinetic energy must be preserved upon impact. So what happens depends on the mass of the object being struck as well as how much Kinetic energy gets transfered into heat upon impact. Making the simplifying assumption that no Kinetic energy gets transfered into heat (perfect bounce-abilty?), if the impacted object is the same mass it will bounce away at the same velocity as the impacting object while the impacting object will become stationary. If the impacted object is a smaller mass it will bounce away while the impacting object will follow at a reduced speed according to the simultaneous solution of the momentum-kinetic energy equations. If the impacted object is a larger mass it will bounce away while the impacting object bounces backwards, again according to simultaneous solutions of the equations.

As I understand it, up until early in the 19th century, the Kinetic Energy was not well understood. People thought Momentum defined everything. Then somebody did an experiment with falling balls, watching the imprint they made in clay. I think they took two balls with identical size but different densities and thus different masses. They dropped them from different heights so that they would have the same Mass*Velocity=Momentum at impact. If Momentum defined everything the theory was that they should make the same imprint in the clay. But they didn't. The one with higher Kinetic Energy according to the dominating velocity-squared term made the bigger impression in the clay. I could be wrong about the time of the experiment though.

PairTheBoard

[Chips_]

If you were catching a 1 pound object that fell 10 feet you would need 1 pound of force pushing up to keep it at constant velocity and then you would need roughly another pound of force acting over about 1 second to stop it. I skipped explaining all the math. A force of two additional pounds upward for half a second would also stop the object. After the object stopped and came to rest you would need to reduce your upward force to 1 pound to keep the object at rest and counterbalance gravity.

The guys that are saying inertia is a principle and not quantifyable are right. Inertia is a principle (an object in motion stays in motion unless acted upon by an outside force, an object at rest stays....). As a principle it is not quantifyable or equivalent to weight.

As an object falls it builds momentum which is mass times velocity (which is increasing as it falls). That is quantifyable and the farther something drops the more momentum it builds up. To stop momentum you need to apply a weight or force acting over a certain period of time.

[Matt R.]

I agree with everything Chips said. Some of the other stuff in this thread is strange, and people appear to be using terms as they are not commonly defined in mechanics.

As Chips said, inertia is not quantifiable, it is a principle. Also, weight is not generally thought of as the force of gravity that is acting on an object. It is the force exerted by the ground to keep you in the same inertial frame as the ground; i.e., the ground pushes up on you with a force to prevent you acclerating "through" the ground. Thus, standing on a scale measures weight, and if you are accelerating radially relative to the center of the earth it changes your weight (e.g. riding an elevator).

So yeah, I would basically just like to add that I believe the more precise way to frame your question is "how much more force is required to stop an object of mass M if it is accelerated through a distance H as opposed to if it were stationary."

So, you just find the final velocity with the equation
vf^2 = 2gH
where vf is final velocity, g is acceleration due to gravity, and H is the height you drop it from (assuming zero initial velocity). This comes from kinematics.

This gives you the final velocity vf, and you plug this into the momentum equation p = m*vf (m is mass of object). Force is simply F = dp/dt, so before we could answer the question, we need to know how long it takes to decelerate the object from velocity vf to zero after it contacts the ground. Or, more precisely, velocity as a function of time after impact. Chips alluded to this when he assumed 1 second to decelerate the object, but of course it would occur much more quickly than this if it hit a solid surface. I don't know enough about deformation of solids upon a collision to answer such a question, unfortunately /images/graemlins/tongue.gif. You could make some simplifying assumptions, such as assuming the deceleration is constant, but I wouldn't even know the order of magnitude of the amount of time it takes the ground to "stop" the mass.

Force is a vector quantity, so you would add this force you come up with to the force of gravity (m*g) to come up with the force exerted by the ground to stop the object. I believe this is what you were trying to ask with your original question.

[Bill Haywood]

Okay, I'm trying to refine my question, forgive me for not using terms as precisely as a physicist.

We put a one kilo weight on a scale, it measures one kilo.

We drop a kilo from ten meters, what will the scale read?

(Theoretically speaking, where would it peak? I'm sure there are big problems with building a scale that could actually read such a situation with accuracy.)

Here's my agenda, if that helps clarify my question.

Floor 60 of building W has 50 floors above it, which weigh Y tons. If those top fifty floors free fall straight down 10 feet, they will exert a force equivalent to a much larger weight. The effective weight would Y tons times ?

I'm wanting a figure like: "a rock weighs the equivalent of 12 x as much (or whatever it would be) when dropped ten feet."

I realize a precise number would be deadly complicated and require endless assumptions. I just want a rough sense of how much force changes with velocity gathered over 10 feet of free fall. 3x as much? 100x as much?

[PairTheBoard]

[ QUOTE ]
Okay, I'm trying to refine my question, forgive me for not using terms as precisely as a physicist.

We put a one kilo weight on a scale, it measures one kilo.

We drop a kilo from ten meters, what will the scale read?

(Theoretically speaking, where would it peak? I'm sure there are big problems with building a scale that could actually read such a situation with accuracy.)

Here's my agenda, if that helps clarify my question.

Floor 60 of building W has 50 floors above it, which weigh Y tons. If those top fifty floors free fall straight down 10 feet, they will exert a force equivalent to a much larger weight. The effective weight would Y tons times ?

I'm wanting a figure like: "a rock weighs the equivalent of 12 x as much (or whatever it would be) when dropped ten feet."

I realize a precise number would be deadly complicated and require endless assumptions. I just want a rough sense of how much force changes with velocity gathered over 10 feet of free fall. 3x as much? 100x as much?

[/ QUOTE ]

The answer you're looking for is going to be much a much larger number than you would think by looking at the Momentum, 25 pound-feet/sec, at impact. The determining factor is going to be its Kinetic Energy which I should have shown in my previous post as

(625/2) = about 313 pound*(feet/sec)^2.

The idea is similiar to the example I gave about dropping balls into clay in such a way that they have the same momentum but different Kinetic Energy. The Kinetic Energy determines how much the Clay is deformed in the impact.

The scale the object lands on must absorb the Kinetic Energy. It does so by applying Work to the object by way of Force applied over distance according to the formula,

W = Fd

and where Force = mass*acceleration is measured in units of
pound*feet/(sec)^2

So assuming the Scale is of some Spring-Type where the Spring resists the Force of Gravity on Weights with greater and greater Force as the Spring contracts, the answer will depend on how that Resistance varies as a function of the Spring's displacement distance. If it is a spring which compacts quickly I think you will get a bigger answer than if the Spring compacts more gradually.

PairTheBoard

[Max Raker]

[ QUOTE ]


We drop a kilo from ten meters, what will the scale read?


[/ QUOTE ]

This question cannot be answered. The answer depends on how much time the scale takes to stop the mass. The less time it takes to stop the mass (once it hits the scale) the less force it will take.

[Borodog]

As people have stated, it depends on the surface you are droppin the wieght onto. If you drop it on cement, the force will be very large. If you drop it on thick foam (more time/distance to slow down and stop the weight), it will be significantly less.

[Fels krone]

I did tests like these in a controls lab; what we used was an accelerometer. Not sure exactly how it works, but it sends out voltage that is related to the acceleration it is going through. You screw it on to your test object, then you get an acceleration vs time history for the impact. I think we used 1000 samples per second, and you can find the time history of the impact. As opposed to the scale.

[Silent A]

We can estimate the maximum force though.

A typical spring scale might deflect about 1 cm for a 100 kg man. So the spring coefficient would be about 100 kg * 10 m/s2 / 0.01 m = 100,000 N/m.

The energy of a dropped ball is = mass * gravity * height = mgh

The energy stored in a spring = spring_coefficient * deflection ^2/2 = kx^2/2

If we make these two energies equal ...

kx^2/2 = mgh

x = sqrt(2mgh/k)

F = maximum force on scale = kx = sqrt(2mghk)

F = sqrt(2 * 1kg * 10m/s2 * 10m * 100,000N/m) = sqrt(20,000,000) = 4472 N = about 450 kg.

For the 1 lb from 10 feet question ...

1 lb = 0.5 kg and 10 feet = 3 m (roughly)

so F = sqrt(2*.5*10*3*100,000) = 1732 N = about 350 pounds.

Notice that part of the above equation is "k", the spring coefficient. The bigger "k" is (i.e. the more the spring resists deflecting) the greater the maximum force. The stiffer the spring (or ground surface, or the object itself) the greater the force. Which is why a 1 pound iron weight hurts when you drop it on your foot but not a 1 pound loaf of bread. Iron is much stiffer than bread (higher k) and thus the maximum force is much higher.

[Bill Haywood]

Thanks Silent A, I believe that's the estimate I'm looking for.

So to extend this to my purposes:

After the initial floor collapsed in a World Trade tower, the force of the upper block falling 3 meters and hitting the next floor down was around 350 times its weight.

Or more conservatively: it hit with a force of several hundred times its resting weight.

[Silent A]

Don't take this too far. It all depends on how much the lower floor can deflect without collapsing itself. Dynamic loading is much more difficult to analyze than simple static loading.

But it's safe to say that the lower floors would not deflect very much and there would be a considerable amplification in terms of the maximum applied force. But remember that the energy involved is also important, not just the maximum force applied.

I'm trained as a civil engineer but I've never been exposed to trying to interpret these kinds of forces or what you'd need to consider to analyze it properly.

[Bill Haywood]

For anyone interested, here is an email I wrote to a scientist and his response. He indicates that the collapsing upper towers hit the initial floor below with force "several to 10s of times" greater than their static weight.

> Dear Sir,

> I’m looking for a bit of information that would be helpful in explaining the tower collapse to resistant minds.
>
After an initial floor gives way, the upper tower drops 3 meters and smacks into the next floor down.
>
I readily accept that the 3 meters of momentum provides enough force to pulverize anything below.
>
But it would be helpful to express this as a number, as in, “the force of the tower falling 3 meters is comparable to 10x the weight of the upper floors before they fell.”
>
> In your Counter Punch article you state:
>
> “F/(m*g) = 1 + dv/(g*dt) = 1 + 0.5/(9.81*0.01) = 6.1,
>
> a load of six times the weight of the upper block”
>
>I just want to be sure. Is that the figure I’m looking for? The force of the upper block falling 3 meters was comparable to 6x its weight?
>
> I know from trying to ask other people that this requires gross simplifications that go against the nature of scientists. But I’m just trying to make a polemical point that is a fair approximation of reality.

Reply from Manuel Garcia of Lawrence Livermore Laboratory:

Yes. To be precise, yes given the assumptions of the example. The assumptions of the example are reasonable simplifications of real life (e.g., time duration of "contact" at 0.01 seconds). For different numbers (say 0.005 seconds) the estimated magnitude of the dynamic force will be different. In real life, with 3D complexity and over the course of the collapse, there will be a range of contact times. Calculating to minute detail would require massive computation. However, from the example we can see that for reasonable estimates of contact times (e.g., a tennis ball on a racket is only a few thousands of a second) we arrive at a dynamic force that is several to 10s of times the static weight of the upper "static load" of the building.

I arrived at 0.01 seconds as a contact time (a characteristic number) for the WTC Towers floors, based on considerations of the time it would take to engage all of their inertia into bodily motion; this is given by the travel times across and through the floor slabs/frameworks by elastic waves (the same as "earthquake" waves). The large size and thickness of the WTC floors adds time (bigger distances) but the stiffness of metal and concrete (as opposed to sound waves in air and water, and even waves in the looser packing of the uppermost layers of the Earth) make for higher speed waves. It all boiled down to 0.01 seconds, say within a factor of 3 either way.

Good luck. I find that some people are genuinely interested to get over their gaps in physics knowledge, while others are just cemented into their fantasies. Don't waste your time on the latter.

http://www.websurdity.com/2007/02/28/unc...-an-inside-job/ (http://www.websurdity.com/2007/02/28/uncomfortable-questions-was-the-death-star-attack-an-inside-job/)

Manuel Garcia article: http://www.counterpunch.org/physic11282006.html

[Silent A]

By the way, my calculations above assume a contact time of zero and so the maximum force becomes a function of the deformation of the falling weight and the scale. With something like an iron weight on a spring scale this is reasonable.

However, for something as complex as a falling floor it's not reasonable to assume near instantaneous contact time so the equations the scientist quotes may well dominate the pure elastic deformation equations I used to answer your initial question.

I also wonder why his "dv" is only 0.5 m/s. An object falling 3 m should be travelling at about 7.5 m/s when it hits.

ETA: OK, I read the pertinent part of the article. "dv" is only 0.5 m/s because the falling floor never comes to rest as the lower floor collapses first.

[PairTheBoard]

[ QUOTE ]
I also wonder why his "dv" is only 0.5 m/s. An object falling 3 m should be travelling at about 7.5 m/s when it hits.

ETA: OK, I read the pertinent part of the article. "dv" is only 0.5 m/s because the falling floor never comes to rest as the lower floor collapses first.

[/ QUOTE ]

That dv looks like the reduction in velocity caused by contact with the lower floor before it collapses. Does he show an estimate for the velocity of the upper floors after their 3 meter descent? I'm wondering if that might be less than what we've been figuring because it's not really a free fall. The failing support structures may be gradually deforming over the 3 meters thereby somewhat cushioning the descent.

Also, he mentions that the energy of the falling upper building is being absorbed by "waves" propagating throughout the lower part of the building. I wonder if that might change things. I have no idea how that kind of thing works.

PairTheBoard

[Silent A]

I know what he's talking about but it's hard to explain.

He's talking about the very instant the upper floor hits the lower floor. In this first fraction (which he estimates to be about 0.01 s) the floor looses 0.5 m/s of its speed as the lower floor deflects under the weight and absorbs some of the weight. He doesn't explain how he calculated this to be only 0.5 m/s but it's certainly possible to do - and probably too complicated for this article.

The forces are transmitted through the structure as pressure waves (basically sound waves). The reason he calculates their speed (the speed of sound in concrete) is to show that the time it takes the pressure waves to travel to the columns is less than the the time it takes for the impact to occur. If the time was greater, there would be time for the pressure waves to dissipate their energy and thus only transfer a fraction of the dynamic load to the columns. Since the time is so short, effectively all the load gets transferred and it's pretty safe to say that the columns were not designed for anything close to this kind of force.

So, in the initial impact (0.01 s) the falling floor loses 0.5 m/s of speed and an effective load 6 times the floor weight reaches the columns and they collapse. Once the first floor goes, the rest is inevitable.

[CORed]

The impact force of the dropped object is going to depend on the physical characteristics of the object and the surface you drop it onto. A one pound steel ball dropped onto a concrete floor will exert a lot more force over a much shorter period of time than a one pound teddy bear dropped onto a pillow. I think you are assuming that there is a single number that answers your question. There isn't The momentum and kinetic energy involved are easily calculated, but the impact force isn't without more information.