Impact force of falling object

[APcraze]

Homework Statement

I am doing a project on pressure sensitive films and finding if they indicate the correct amount of impact. I have carried out an impact test by dropping a mass of 5KG from a height of 1m. I have calculated the velocity to be 4.43m/s. The kinetic energy just before impact is equal to the gravitational potential energy at the height in which it was dropped =49J. The distance traveled after impact was 0.1m. Using the work-energy principle the impact force is found to be 490N.

The attempt at a solution
This test I used a hexagon shaped mass and the area which contacted the film was 116.63m^2.

The pressure film is indicating a pressure force of 6.5MPa=65kgf/cm^2.
The impact force that has been calculated is 490N=49.9kgf.

The calculated and indicated pressure forces can be seen to be far off. The manufacturer of the pressure film has claimed that the films give an estimated pressure and not necessarily exact.

I just want to ask if anyone sees where I am going wrong? I also want to ask if area needs to be considered in the calculation.

Thanks for your time.

Alexander.

 

[mfb]

This test I used a hexagon shaped mass and the area which contacted the film was 116.63m^2.

You might want to check that number (or the unit).

490 N is the average force. Peak force can be significantly higher if your deceleration is not uniform. How does the deceleration look like?

 

[Nidum]

The distance traveled after impact was 0.1m

How can that be ? Pressure sensitive film only distorts a small amount under load .

 

[APcraze]

How can that be ? Pressure sensitive film only distorts a small amount under load .

Sorry I forgot to mention.

The pressure sensitive film had been adhered to a portion of a 737 fuselage laying flat on the ground. The mass did penetrate the surface very lightly and bounced back up 0.1m.

 

[mfb]

The mass did penetrate the surface very lightly and bounced back up 0.1m.

That is a completely different statement. It indicates a much shorter stopping distance.

 

[haruspex]

Using the work-energy principle the impact force is found to be 490N

Further to mfb's observation about peak force, if the force is not constant then ΔE/Δs will not give you the average force either. The average force is defined as mΔv/Δt, which only gives the same answer for constant acceleration.

 

[Nidum]

Too many variables .

To calculate the impact load even approximately would require detailed information about the dynamic response characteristics of the fuselage section .

Some possible ways to get better results :

Use instrumentation fitted to the fuselage section to measure its response to the drop load .

Use a well supported solid and heavy anvil block instead of the fuselage section .

Put pressure sensitive film on a platen and mount on top of a load cell .

Make a spring block anvil with accurately known dynamic response characteristics .

Fit an accelerometer to the dropping load .

Many other possibilities but need more information about your project .

 

[mfb]

A technical detail:

Further to mfb's observation about peak force, if the force is not constant then ΔE/Δs will not give you the average force either. The average force is defined as mΔv/Δt, which only gives the same answer for constant acceleration.

There are other acceleration profiles where the answer happens to match by chance - everything where the elapsed time is the same as you would get with a constant acceleration.