# Calculate force exerced to stop a falling barbell by straps [closed]

## Problem

Consider a weightlifting bar with total mass $$2m$$, that is positioned in a height $$h$$ over two so called safety straps as in this video. Then the bar should fall straight down onto the safety straps.

Now I want to calculate (in dependence of $$h$$ and $$2m$$) the maximum force exerted by the bar on the straps during the impact and thus (to make it more illustrative), what mass $$2M$$ of another barbell, which would rest just statically on the straps would produce an equivalent load on the straps as the dynamic load of the falling mass $$2m$$ will do?

The answer should be an error margin of max. 20 percent.

## What I tried so far

### A simple model

It is clear that the force depends essentially from the braking distance of the bar, which is determined of how far different parts of the system (for example the straps) stretch. From the video one can see, that almost all of the non zero braking distance comes from the fact that the bar bends (and only to a minor part from the stretch of the straps). I assumed Hook's law for the bar bending and from symmetry I considered only one of the straps and one side of the bar (taking essentially the half mass $$2m$$).

So I just modeled the problem as a body of mass $$m$$, that falls the distance $$h$$, and is then catched by a hanging spring with spring constant $$D$$. If the spring is then stretched at maximum by $$s$$, conservation of energy tells me that

$$mgh = \frac{1}{2}D s^2 - mgs$$

which you can solve for $$s$$ and get:

$$s = \frac{mg \pm \sqrt{(mg)^2 + 2Dmgh}}{D}$$

This yields to a maximum force $$F = Ds$$ as

$$F = mg + \sqrt{(mg)^2 + 2Dmgh}$$

The "equivalent mass" would then be $$M = \frac{F}{g}$$.

So if I would have $$D$$, it would give me a simple model of what I want.

To get $$D$$, I solve the energy conservation by $$D$$ and get:

$$D = \frac{2mg(h+s)}{s^2}$$

### Getting parameters from the video

Now I tried to get the values for $$h$$ and $$s$$ from the video by knowing the diameter of an olympic plate ($$45\,\mathrm{cm}$$) as indicated in the following picture: The mass $$2m$$ was known from the audio stream of the video (7:55) as $$2 m = 370\,\mathrm{kg}$$. So I get

$$D \approx 110 \, \frac{\mathrm{kN}}{\mathrm{m}}$$

or a force of

$$F \approx 25,8 \,\mathrm{kN}$$

This finally leads to the following plot: ### Measuring the acceleration directly via video analysis

To confirm the model above, I tried to measure the acceleration and thus the force directly using the video analysis tool tracker:

I also used the $$45\,\mathrm{cm}$$ diameter of the plates as length reference. Since I didn't know the frame rate of the slow motion part of the video, I just took the real time part with a framerate (as read from the metadata) of $$30 \,\mathrm{fps}$$.

To check the consistency I did a linear fit for the free falling part of the motion and got an acceleration about $$10.35 \,\frac{\mathrm{m}}{\mathrm{s}^2}$$ which is pretty close to $$9.8 \,\frac{\mathrm{m}}{\mathrm{s}^2}$$ I would expect for free fall. Then I measured the acceleration during the impact like in this screenshot: The result was

$$a = 76.93\,\frac{\mathrm{m}}{\mathrm{s}^2}$$

The force I am looking force should be the sum of the (absolute value of) the net force $$ma$$ and the gravitational force $$mg$$:

$$F = m(a+g) \approx 16\,\mathrm{kN}$$

This is inconsistent by a factor $$1.6$$ with the results from my model, but I don't see the error.

Maybe the maximal slope is a bit higher, because of the discrete time steps, but I cannot imagine that this is by factor $$1.6$$.

## Alternative Approach

As another source I have found a paper called Mechanical properties of weightlifting bars which analyzed the elastic properties of some common weightlifting bars. But I was not sure, of how to adapt the information of the paper properly to my setup.

• I assumed Hook's law for the bar bending Unlikely, especially for high deformations. fast10.vsb.cz/lausova/lesson05-17.pdf – Gert Jan 25 '20 at 19:39
• I don't have an answer, but wow, it's absolutely incredible that those thin straps didn't break. What are these things made of? – knzhou Jan 25 '20 at 22:03
• @sammygerbil: It's not a assignment question. The source of the question was just that I have seen the video and wanted to know the forces acting on the straps. I didn't know that feedback type questions are not welcome here. The other way around I know that it is not welcome to ask questions where you don't show your own effort so solve the problem. So I tried to show what I was able to do by myself. But as I learned now, if you show too much effort, it turns into a "feedback" question which seems also not to be welcome :-( – Sarah Jan 28 '20 at 11:15
• @sammygerbil: I think I see (partly) your problem. What I showed is really what I was able to do by myself so far, I am not sure if I could do the analysis better. However I am really interested in the results and in how a good analysis would look like. So do you have a suggestion of how to change the questions to get a better focus? Maybe I could just ask for the force as in the problem statement in the beginning and say that I want a error margin of x Percent (for example 20 %), then show what I tried and delete the other questions? – Sarah Jan 28 '20 at 11:33
• I don't know Sarah. The usual advice here is to post the different issues in separate questions but I dislike that idea because it involves so much duplication or cross-linking. What you really need is a discussion whereas this forum works best for (and expects) questions with a single answer with little or no discussion. The question is about to be closed (4/5 votes). – sammy gerbil Jan 28 '20 at 12:03