"G force" for Dummies

[Wholly Cow]

I am trying to find a simple table for g force but i am having a hard time finding something that can be explained in laymans terms.
An object (with a weight of 50kg) is dropped straight down 3.81 Meters
(which is equal to 12.5 feet, or 1 story) to solid concrete resulting in a
stopping distance of 1cm. How many gs of force does the object
“experience”? Can you help me?

[Madison]

One "G force" is a layman's explanation of acceleration and is representative of the force of gravity at sea level or 9.8 m/s^2.

There are a number of factors that affect a body's acceleration, so I don't believe that you'll find a simple table that provides such information. Sorry.

You must first calculate the final velocity of your falling body as it impacts the concrete. From that, you can calculate the acceleration in m/s^2 and divide that by 9.8 to get the value in "g-forces" as it slows from the velocity to zero over .01 m.

V(final)^2=V(initial)^2+2(acceleration)(distance) is the formula you'll want to use.

[JBotAlan]

Never have I been so happy to find my old physics notes! (nerd alert )

The object travels the 3.81 m accelerating at 9.8m/s/s (gravity), so we can figure out how long it takes to fall.

distance = initial velocity * time + 0.5 * acceleration * time^2
3.81 = 0 * t + 0.5 * 9.8 * t^2
3.81 = 4.9 * t^2
3.81 / 4.9 = t^2
sqrt(3.81/4.9) = t
t = about 0.8818 s

Now, we can figure out how fast it was going at the impact.

final velocity = initial velocity + acceleration * time
Vf = 0 + 9.8 * 0.8818
Vf = about 8.64164 m/s

Now, we know that it decelerates from 8.64... m/s to 0 m/s over 1cm. Time to find the acceleration.

final velocity ^ 2 = initial velocity ^ 2 + 2 * acceleration * distance
We know that 1 cm = 1/100 m.
0 = 8.6414 ^ 2 + 2 * acceleration * 0.01
-74.6779 = 2 * acceleration * 0.01
-74.6779 = 0.02 * acceleration
acceleration = -3733.897 m/s/s
Divide that by gravity (9.8)
We get an answer of 381 Gs.

Someone please check this; I didn't take the weight into account but I think it's irrelevant here...

Wish I had my physics teacher right here now.

JBot

[eugenebrooks]

The prior post is correct. The answer can be more easily
understood from the point of view of conservation of energy.

The total change in gravitational potential is m g h, where h is
the distance of the drop, 3.81 meters.

To get the average force, f, applied during the stop equate this total
energy to force times, f, times the distance of the stop, 0.01 meters.

50 * 9.80 * 3.81 = f * 0.01

Now, f = m a, so we have

50 * 9.8 * 3.81 = 50 * a * 0.01

And finally, putting a in terms of gees, a factor times the
acceleration of gravity,

50 * 9.8 * 3.81 = 50 * gees * 9.8 * 0.01

We see that the mass cancels, and the acceleration of gravity
cancels, and the number of gees is 3.81/0.01 = 381

At this point, you should see that it is not hard to go
a little further and account for the gravitational potential
for that last centimter of crunch into the concrete, and
then get into some deep thought about the gee of gravity
and when it kicks in, so to speak.

Eugene

[Craig Roys]

Resident Physics teacher here.

The previous posts are correct: 1 g = 9.81 m/s/s
It is just a way to compare apparent weight to actual weight on earth. Typically used in cases using circular motion (jet, space shuttle, rollercoaster, etc.). To say you are experiencing 3 g's would be to say that you feel 3 times heavier that you would standing still on earth because you are experiencing 3 times earth's acceleration due to gravity (i.e. a 150lb person experiencing 3 g's feels as if they weigh 450lbs).
This apparent extra weight comes from forces (in addition to gravity) acting on an object - usually centripetal force.

One last nit-picky thing. g's are represented with a lower case g so as not to be confused with the gravitational constant: G=6.67x10^11 N*m^2/kg^2.

Hope that helps.

[Qbranch]

Yay for impulse forces!

You all have it right... its actually amazing sometimes how many times the force of gravity you realize when things hit the ground.

-q

[DonRotolo]

OK, students, now consider the case where both the object being dropped and the surface it's dropped upon do not yield. That is, reduce the distance in which it decelerates from 1 cm to zero.

What kind of G force does the object endure then? What becomes the limiting factor?

(All you engineers and physics teachers out there, shhhh...)

Don

[Guy Davidson]

I'm a student, and I'd have to guess the limiting factor is how much the object yields. I guess it will bounce back up to a certain height, that height also a factor of how much the object yields.

[sciencenerd]

Originally Posted by Don Rotolo

OK, students, now consider the case where both the object being dropped and the surface it's dropped upon do not yield. That is, reduce the distance in which it decelerates from 1 cm to zero.

What kind of G force does the object endure then? What becomes the limiting factor?

(All you engineers and physics teachers out there, shhhh...)

Don

That's an impossible situation, but if it were to happen both the object and the floor would experience an infinite force.

-Resident (confused) physics student

[Qbranch]

Ok... here we go:

Force = Change in Momentum/Change in Time

So... the less time we have to stop... the limit as time approaches zero... is infinity. Just kind of mathematically/physically restating whats above i know... but wanted to make sure it fit the formulas too.

So... i'm guessing the point of asking that question was purely academic?

-q

[DonRotolo]

Yes. There absolutely must be some physical time in which the object consumes to go from full speed t a stop, it cannot physically be zero.

Assume a drop onto a polished hard steel surface:

If we take the case of a scoop of mashed potatoes, that time is very long, relatively. It will not bounce very high as the energy dissipated is not reflected back to the object but consumed by deforming the 'ball'.

If we take the case of a hardened steel ball, the ball will elastically deform, that deformation energy will be returned mostly to the ball (it's relative to the mass) and the ball will bounce higher than a 'super ball'. What kind of force (in g) would it take to dent in a hardened steel ball? A lot. The ball experiences several thousand g

Don

[eugenebrooks]

For extra points, for the steel ball hitting the steel
plate, how can you estimate the collision time?

Eugene

[JesseK]

Ok, let's calculate something practical.

Your CMU Cam is mounted 48" off the ground for your 5' robot. Your overall robot weight 105lbs with an estimated 20% of the weight above 36" (i.e. fairly low CoG). Your robot attempts to drive up the ramps of another robot but behold, upon reaching the top you've misaligned it with half of your robot hanging off. With the slightest correction-intended jiggle of the joystick, your beautiful metalic creation falls sideways and crashes into the floor, with the CMU Cam hitting first.

Calculate how many G's you just put your CMU Cam through upon impact and propose to me whether or not it survived. I'd say the carpet gives (maybe) 1/4" of stopping distance.