Newton's Law: Mass at Rest vs Falling from 100m

[be unique]

Experts tell me that:
"At average gravity on Earth (conventionally, g = 9.80665 m/s2), a kilogram mass exerts a force of about 9.8 Newtons. An average-sized apple exerts about one Newton of force, which we measure as the apple's weight..
After one second, you're falling 9.8 m/s..
.. falling from 50m high is the equivalent of getting hit by a car going 112 km/h, 31m/sec."
-
This means that a mass at rest = the mass falling 5 meters. Falling 50meters gives 3 times the force on impact.
I want to replicate a force of mass 1kg at 44m/sec , as in a fall from 100 meters , by placing mass at rest.
Is this a) 5kg at gravity
b) 800 kg at gravity ?
or c) drop 4.5kg for 5 meters ?

 

[berkeman]

This means that a mass at rest = the mass falling 5 meters. Falling 50meters gives 3 times the force on impact.
I want to replicate a force of mass 1kg at 44m/sec

That makes no sense. You are mixing up concepts (and your units would not be consistent anyway).

Impacts are characterized by "impulse", not force.

 

[Janus]

Experts tell me that:
"At average gravity on Earth (conventionally, g = 9.80665 m/s2), a kilogram mass exerts a force of about 9.8 Newtons. An average-sized apple exerts about one Newton of force, which we measure as the apple's weight..
After one second, you're falling 9.8 m/s..
.. falling from 50m high is the equivalent of getting hit by a car going 112 km/h, 31m/sec."
-
This means that a mass at rest = the mass falling 5 meters. Falling 50meters gives 3 times the force on impact.
I want to replicate a force of mass 1kg at 44m/sec , as in a fall from 100 meters , by placing mass at rest.
Is this a) 5kg at gravity
b) 800 kg at gravity ?
or c) drop 4.5kg for 5 meters ?

The force of impact depends on the distance over which the deceleration of the impact takes place.
The part about being hit by car being the same as falling from 50m just means that you would be moving at 31 m/sec when you hit the ground after falling 50 meters. How much force this equates to is a different matter. If the ground were covered with a thick foam pad, it could decelerate you over a few feet and the force you would experience would be much less than if you hit bare ground.
To produce the same force with an object at rest as an object that impacts after falling 100m, you would have to know the nature of the impact.

 

[scottdave]

The impulse of a moving mass coming to a stop in a short time period is different than a mass just pushing down due to weight.
Do you know how to find the momentum of a 1 kg at 44 m/s ?
What is the momentum of the 5kg (not moving)? How about the 800 kg? What about the mass which has fallen 5 meters?

 

[be unique]

Exactly . Can i produce the effect of 1kg. 44km/sec by a mass at rest? What mass would it be?

 

[Mister T]

Exactly . Can i produce the effect of 1kg. 44km/sec by a mass at rest? What mass would it be?

You mean, can you exert a force of 9.8 N by having a 1.0 kg object, moving at a speed of 44 km/s, come to a stop? Sure. You just to make it accelerate at 9.8 m/s².

It would take a time $\Delta t=\frac{\Delta v}{a}=\frac{44 000 \ \mathrm{m/s}}{9.8 \ \mathrm{m/s^2}}\approx4500 \ \mathrm{s}$. About $1\frac{1}{4}$ hour.

 

[be unique]

Sorry I made a typo ( 44km/sec) the original q is " I want to replicate a force of mass 1kg at 44m/sec , as in a fall from 100 meters , by placing mass at rest."
That is , a mass of weight x is exerting force vertically down by gravity without velocity , at rest . Then what formula gives this mass to produce 44 Newtons?

 

[be unique]

The idea is similar to this:
"Indentation hardness measures the resistance of a sample to material deformation due to a constant compression load from a sharp object; they are primarily used in engineering and metallurgy fields. The tests work on the basic premise of measuring the critical dimensions of an indentation left by a specifically dimensioned and loaded indenter."
So I'm looking for a load at rest .

 

[Ibix]

So I'm looking for a load at rest .

Stand on you bathroom scale. The needle will point to some number.

Now imagine jumping onto the scale (don't actually do it - you may damage the scale even if you are very light). The needle will swing from zero to some very high value, then swing back to a low value then back and forth until it finally settles at the same value it did when you stepped on it carefully (assuming you didn't bust the spring).

Your question is: what mass can I place carefully and gently on the scale so that the needle shoots up really high then swings back and forth before settling at a lower value? Given my description (and the fact that people build enormous drop-test rigs with considerable cost and safety implications), what do you think the answer is?

 

[be unique]

Why do you say that 40 Newtons is large? What is this "shooting really high"? My q is about why does 1kg at rest have the same Newtons as dropping 5 meters? A length of timber / steel can break under load , both static and with momentum. So I'm asking about static load of the value of 44 Newtons .

 

[Ibix]

Why do you say that 40 Newtons is large?

I didn't. I said the overshoot on a force meter (such as a bathroom scale) when you drop a mass on it could be very large.

My q is about why does 1kg at rest have the same Newtons as dropping 5 meters?

And the basic problem is that this question makes no sense. More explanation below.

A length of timber / steel can break under load , both static and with momentum.

They do, but for very different reasons and in very different ways. An obvious example is a bullet. Shoot a piece of wood and you'll get a fairly neat bullet hole. Place a bullet against a piece of wood and apply increasing pressure and the wood will eventually crack. You will never get a bullet hole by applying a static load because wood reacts differently to being hit by a small mass at high speed compared to a large mass at low speed. Force meters attached to the wood will show completely different things in the two cases.

The calculation in your first post is, very simply, determining the velocity at which you would hit the ground if dropped from 50m. That's 31m/s or thereabouts. Then it's saying that it doesn't matter whether you hit something big and heavy and stationary (like the ground) while traveling at 31m/s or if something big and heavy moving at 31m/s (like a car) hits you. It's not saying any more than that.

You seem to be reading something about impact testing into it. That's not what it's saying and it doesn't work that way. People build gigantic impact testing machines because the only way to test how a material reacts to high velocity impacts is to hit it at high velocity.

 

[Mister T]

Sorry I made a typo ( 44km/sec) the original q is " I want to replicate a force of mass 1kg at 44m/sec , as in a fall from 100 meters , by placing mass at rest."

What is "a force of mass 1 kg"? Do you mean 9.8 N?

That is , a mass of weight x is exerting force vertically down by gravity without velocity , at rest .

$w=mg$ with or without velocity.

Then what formula gives this mass to produce 44 Newtons?

Now, you are asking a different question, it seems to me, because in what you call your "original q" which I've quoted above, there is no mention of 44 Newtons. There is a mention 44 m/s, is this a typo?

If not then the answer is given in Post #6. You just replace the 44 000 m/s with 44 m/s and you get a time of about 4.5 seconds. This is the time you must spend making the object stop, which is equal to the time it spent falling because you want the force to be the same as the force that was exerted on it as it fell, if I understand you correctly.

 

[CWatters]

...I want to replicate a force of mass 1kg at 44m/sec , as in a fall from 100 meters , by placing mass at rest."

Can't be done.

You haven't provided enough information to know what force the falling mass exerts on the ground. It depends on numerous factors like the hardness of the ground etc. If it makes an impact dent or crater you can measure the depth and estimate the stopping distance and impact force that way but it won't be very accurate because it's not likely to be a constant force.

Lets say the ground is hard and it stops in 1cm = 0.01m If we were to assume a constant deceleration we can use..

V² = U² +2as

The final velocity V = 0 so

a = -U²/2s

U = 44m/s so

a = 44²/(2*0.01)
= -96,800 m/s/s

Using F = ma that puts the force at around 96,800 Newtons. However it only takes about 44/96800 = 0.45 mS to stop so that's how long the force is exerted for.

I'm not sure what "replicate" means in this context. Sure you can apply a force of 96,800 N but the results of applying that force constantly rather than for 0.00045 seconds might be different. What matters to you?

It might be better to think in terms of energy. 1kg at 44m/s represents 968 Joules of energy.