Is Impact Force Equal to G-Force in Newtons?

[ClapClap]

Hi, I am dropping an object of mass m at different heights on another body. From the accelerometer I have I can get the g-force and then convert it into Newtons. What I am wondering, is the force converted into Newtons equal to the impact force the mass exerts on the body?

 

[Svein]

What I am wondering, is the force converted into Newtons equal to the impact force the mass exerts on the body?

Possibly. Use the momentum equations.

 

[ClapClap]

How can I use them?
This is the graph of the Force in Netwons my sensor recorded. I m struggling to identify where the physical impact between the two bodies begins and whwat force the maximum indentation corresponds to

 

[nasu]

Hi, I am dropping an object of mass m at different heights on another body. From the accelerometer I have I can get the g-force and then convert it into Newtons. What I am wondering, is the force converted into Newtons equal to the impact force the mass exerts on the body?

How do you "convert" it into Newtons?

 

[ClapClap]

F (N) = m x F (g) x g

F (N) = force in Newtons
m = mass dropped
F (g) = G-force
g = 9.81

 

[nasu]

So the accelerometer measures the acceleration of the falling body, right? Is the accelerometer attached to the falling body? Or to the other body?

 

[ClapClap]

Yes, exactly. It is attached to the falling body

 

[nasu]

So the first portion (the ramp before the first peak) corresponds to the fall time? I would think that the first peak may be the first collision between the two objects and then probably there are several "bouncings". Unless your second body have some structure, with a less rigid layer on top and a stiff core.

 

[ClapClap]

Yes, I would assume that the first part corresponds to the body falling.
Does the collision occur at at measurement point 27? I would think that the peak of the graph corresponds to the maximum indentation after which the body raises again.
Yes, there was bouncing involved.

 

[nasu]

But why would the acceleration decrease during falling? Was air resistance significant?
Yes, if the process is elastic the maximum acceleration will correspond to maximum displacement.
It's an interesting experiment.

 

[ClapClap]

That's what I can't understand. Then all of a sudden the acceleration peaks again at almost 2g.
Is the f-force in Newton equal to the impact force?
I am doing this for dissertation and I am so confused with the results!

 

[nasu]

If you multiply the mass of the body by its acceleration you get the force on the body. I am not sure if the acceleration measured by the accelerometer includes the affect of gravity or just of the other forces. What are the units on the vertical axis? Acceleration in units of g?

 

[ClapClap]

The graph shows the force in Newtons so the units of the vertical axis is Newtons

 

[ClapClap]

So say I am dropping the same mass on bodies of different materials. The impact force on the body impacted would always be the same right?

 

[nasu]

No, not at all.

 

[ClapClap]

Im confused then. If the force is mass times acceleration and the mass is always the same, and so its the acceleration of gravity, then impact force should be constant too?

 

[nasu]

The impact force is not given by the acceleration of gravity but by the acceleration due to the elastic (and other) forces acting during collision.
This force (which is a function of time, not a constant) depends on the material properties of the two bodies.

 

[ClapClap]

I see.
Anyway, getting back to the original problem. So the g-force converted in Newtons is simply the force exerted by the mass on the impacted body. Now, by looking at the results, how can I define at what force the impact occurs? Is the peak the maximum indentation?
By defining this and relating it to the impact duration I was going to create an acceleration vs time curve. Integrating this would give a velocity vs time curve and integrating it once again a displacement vs time. This way I'd be able to identify the maximum indentation of the mass in the body. Does this make sense?

 

[nasu]

But you already have an acceleration versus time graph. Why would you go through all this transformation then?
What you call "g-force" is not a force but an acceleration. The accelerometer measures acceleration. Assuming that the accelerometer is properly calibrated, of course.
You don't "convert it to Newtons", you use it to calculate the force producing this acceleration.

The impact starts when the force between the two bodies is non-zero, I would say. The integration works in principle. Some practical problems may occur. There were other threads about this problem. It may be useful to look them up.

 

[ClapClap]

Yes, I see what you mean I don't know how I have never realized that.
Is calculating the force producing this acceleration of any use then?

I'm sorry I don't think I understand what you mean when you say the impact begins when the force between the bodies in non-zero.