Impact Force Physics

Question

I am currently trying to understand load cells in the measurement of impact forces from falling objects and just had a thought experiment that I do not quite understand how to calculate.

What is the difference in impact force when a soft material is placed on the surface vs below another hard surface.If I were to use something like a load cell, how would I compare the relative impact forces for:

Test A: Soft material placed above the load cell- which deflects and absorbs some energy when the falling object hits it

Test B: Soft material placed below the load cell - would the material deflect the same from underneath or is the load cell absorbing some energy in this scenario?

I guess another way of asking this question is - Is the deflection of the soft material different when placed on the surface vs below another material?

I was thinking about how a trampoline might behave if it were covered in concrete. The only thing that has changed in such a scenario is there is more static deflection initially due to the mass of the concrete, however the spring constant has not changed, so I am quite confused at how to relate this back to the impact force. Would this feel the same as jumping on a concrete floor on the ground, a trampoline or would it be somewhere in between a concrete floor and a trampoline?

Answer 1

Basic Hooke law says that trampoline surface displacement will be : $$x=\frac Fk$$ where $F$ is force affecting trampoline. Trampoline will experience a pair of forces - one due to concrete weight and another due to falling object momentum transfer to trampoline, so equation becomes : $$x=\frac 1k \left(m_c g + \frac {\Delta p} {\Delta t} \right) $$

Where $m_c$ is concrete mass. Check the scheme - A) is plain trampoline B) trampoline + concrete :

What's missing? Actually trampoline and spring in general stiffness coefficient is not constant, but is function of impacting force : $k=k(F) $. It's very easy to understand why it is so - if we will expand spring over some $x_{max} $, point of no return, then spring will experience permanent expansion - it will not return into equilibrium position anymore or even we may break it at all, thus Hooke law will not hold anymore, because it holds only for relatively small displacements. I've got a nice chart where theoretical Hooke law deflection (red line) is compared with a real one (gray dotted line), check this out :

So this means that spring and trampoline included can only be expanded until maximum displacement $x_{max} $, after which it will not generate reaction force. This gives final displacement formula : $$x=\frac 1k \left(m_c g + \frac {\Delta p} {\Delta t} \right) \delta_{x<x_{max} } + x_{max} \left(1-\delta_{x<x_{max} } \right) $$ Where $\delta_{x<x_{max} }$ is Kroneker delta function, shorthand form, namely $\delta_{x<x_{max} }\equiv\delta_{x<x_{max}, \text{true} }$

This formula gives insight, that covering trampoline with concrete will make reaching maximum trampoline displacement faster. So back to answer - it shoud feel something in between falling on trampoline alone vs falling on concrete alone. Because you may break trampoline, may reach limiting trampoline stiffness, which will induce shock wave to you and etc.

Answer 2

I thought I might pose a more specific test question to help explain the above (sorry this was difficult to do from the comments section).

If I were to use a force measuring device, would I get a different force for an object colliding with:

-a very thin piece (20mm) of concrete above it vs -a very thick piece (500mm) of concrete above it

Intuitively, it feels like they would both feel just as hard if I fell on them, would have similar impulse times (<5ms) and therefore generate the same impact force. But would I measure a different impact force from below or is there no energy absorption/loss through the thicker piece of concrete? Or is the force that would be measured below the concrete different to the force experienced by the object at impact?