Falling Lever - Impact Force

Sep 2018
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So as I lay here contemplating back surgery and trying to avoid sneezing the remains of a kidney through assorted orifices, I contemplate a problem - well, the primary "problem" is how long ago College was, but the problem I'm here to explore today is the force generated by a falling lever.

Now, I could simply have dropped this on my head, and the math is pretty trivial ("perform it in your head as a parlor trick"-simple)

So let's imagine that we have a ladder, and this ladder has one end in contact with the surface that someone is standing upon. The other end is 14.6m in the air. The ladder weighs 63.5kg overall.

The ladder becomes unsettled and falls through its arc, and the woman reflexively reaches overhead and catches the end around 14m from the pivot. Let's say she stands 1.8m tall, and the easily compressible forearms put the initial point of impact at around 2.05m

So assume our heroine (moron?) foolishly forgot her cape and boots that morning. What kind of impact force are we looking at?

I'm going to try attaching an image here - but please bear in mind that I'm much better at math than art... Although illustrations make these things easier for me, I'm not terribly good at 'em...

I'm happy to simplify this - let's agree to ignore cross section, wind resistance, energy lost in material flexibility, ... If you want to bounce some massless balls on frictionless planes - knock yourself out. This is a magnitude problem for me. ;)

TIA for any insights!

[P.S. If it's unclear - it's a practical problem, not a homework problem (Smirk) ]
 

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Sep 2018
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Laying here still contemplating...

The only force here is gravity * mass. The ladder's center of gravity is going to roughly be its midpoint.

So the force acting is 63.5 * [9.8] = 622.3
The Torque is then Force * Distance to center of gravity from the pivot, right?

T ~ 4542.75 Nm

At this point, I lose focus and forgot why I cared about Torque. I'm sure it'll come back.

But I need a moment of Inertia in there, 'I', and I = (1/3) * m * l^2
I = 1/3 * 63.5kg * (14.6^2) ~ 309 kg*m^2 which seems like a standard unit...

So I'll leave this note here in case it tickles someone else, and revisit it after a nap otherwise...

[edit]

And T=Iα right
So then α ~ 14.717 rad/s^2

That seemed an important memory to me a moment ago...
 
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ChipB

PHF Helper
Jun 2010
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Morristown, NJ USA
You've figured out how to determine the ladder's acceleration at the moment that it is parallel to the ground (i.e., the moment arm is half of 14.6m). But since it tips from vertical, as it rotates the moment arm that gravity is acting on changes from 0m to 7.3m. Hence the rotational acceleration increases from 0 rad/s^2 to some value a little less than 14.717 rad/s^2 as it falls (a little less because the woman catches it before it hits the ground). Once you have a formula for acceleration as a function of time you would have to perform an integration to determine the rotational velocity at the instant the woman catches it. That's a pretty difficult calculation.

There is a better way to solve this, using energy principles. As the ladder falls its initial gravitational potential energy (PE) is converted to kinetic energy (KE). Then as the woman slows the ladder's fall she does work on the ladder, and KE goes to zero. From conservation of energy we know that the work she does is equal to the change in PE of the ladder.

$ W = \Delta PE = mg \Delta h $.

Here delta h is the change in the height of the center of mass from beginning (h1=7.3m) to when the ladder stops moving. You haven't told us what the distance is that the woman requires to stop the ladder's movement. It's not instantaneous, so let's assume that she arrests its movement at height 1.5m. Then h2= 1/2(1.5 m x 14.6/14) = 0.72m. So delta h for the center of mass of the ladder is h2-h1 = 0.72m - 7.3m = -6.58m.

The work she performs in stopping the ladder is calculated from W = Fd, where F is the force she applies to the end of the ladder and d is the distance over which she applies that force. Using the values I suggested above, d= 1.5 - 2.05 = -0.55m.

So now we have: $W = \Delta PE$
$ F \times \ -0.55m = 65 kg \ \times \ 9.8\ m/s^2 \ \times \ -6.58m$

Solve for F. Does the result seem reasonable? Should we consider using a different values for 'h' and 'd'?
 
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Sep 2018
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Hi Chip,

Thanks for taking up the mantle!

Just a short observation before an appointment with a surgeon - if it were at 1.5m, "she'd" be in worse shape than "she" is. ;)

Because this is applied real-world, and we'd hope there's only single member stupid enough to populate our sample, some of this has to be spit-balled. I would say we couldn't possibly be below 1.6, 1.63m.

Let's split the difference and say 1.615 (yeah, that's some silly significant digits for a spitball. 1.62 so I don't feel silly then, preserving the 2.05); d = -0.43

h2 then is 1/2(1.62 * 14.6/14) = 0.84m, delta-h then is 0.84 - 7.3 = -6.46m (your spitball is now working out close enough to my spitball me-thinks)

So F=(65*9.8*−6.46)/-0.43 ~=9570 and that's expressed in Newtons

Now, proving that I'm a heathen and simultaneously localizing myself, we admit that I tend to only think of Newtons when they follow "fig", and feel more comfortable scaling to pound-force. (yeah yeah. I'll leave the Slugs at home). Roughly about 2150lbf.

Bearing in mind it's a thought exercise as a distraction, and alternating betwixt the calculator on my phone and doing the simpler math in my head - did we (thanks to your assistance) just derive the roughly-scaled magnitude of the answer?

Again, I do appreciate the pointer, and hope I didn't completely miss the boat.

[The remainder of the story is that doing some work (in the conversational sense, as well as the literal), I had a ladder become unbalanced while "walking it" around a commercial chiller. [Wait, is that OSHA knocking on the door now?], and there was a display case in the path of the ladder's fall. I elected to catch the ladder instead of doing a lot of "silly" math, and arrested its fall before it took out that display case. Had I taken the opportunity to perform the calculations necessary to determine the magnitude, I'd have let it go and let insurance do what insurance does. Now I've ruptured three discs, herniated two more, caused some kidney damage, and have had "some" discomfort. At times like those, I tend to retreat into my head where I can distract myself. ;) Sometimes the discomfort distracts from the distraction, and I lose the train of thought and have to start over, so if my "work" here (in the artistic as well as academic form this time] seems disjointed, I hope you'll forgive.)

I'd use this as an opportunity to say "stay in school, kids!" - but that's probably what got me in trouble in the first place. ;)