How heavy is it, really?
August 5, 2012 10:35 PM   Subscribe

Settle an argument. A 150 person falls and grabs something as they do. Assuming they have kung-fu grip, how much weight does the thing they grab need to be able to withstand without tearing, breaking, whatever. I say the weight of the person. They say it's their weight and g-forces combined. Who is right and if they are, how do you calculate those g-forces? And if they are twisting as they fall, does torque figure into it?
posted by CollectiveMind to Science & Nature (24 answers total)
Just a clarification: Has this person fallen from a height, or just tripped while standing normally?
posted by Conrad Cornelius o'Donald o'Dell at 10:39 PM on August 5, 2012

Assuming the thing they grab is free standing it totally depends on how the thing they grab is shaped- something with feet vs something with all the mass over a smaller are relative to the direction of the fall. Vector forces and all that.

Imagine if a person grabbed you and you had your feet planted together versus braced far apart in the direction of the fall. Completely different outcome.
posted by fshgrl at 10:40 PM on August 5, 2012 [2 favorites]

F=MA. So, it needs to be able to support the Force of the person's Mass Accelerating (in the negative direction) from current velocity to 0. In short, you're wrong.
posted by barely legal at 11:04 PM on August 5, 2012 [15 favorites]

To stop the faller, the grabbed thing must absorb all the faller's kinetic energy.
The kinetic energy is mass times speed squared, so the faster the faller is going the more energy the grabbed thing must absorb. The faller's speed accelerates due to gravity.
Hence you are on the losing side of your argument.
posted by anadem at 11:05 PM on August 5, 2012 [6 favorites]

as barely legal ...
posted by anadem at 11:05 PM on August 5, 2012

It depends on how fast they're falling, whether they're able to swing from the thing they're holding or otherwise redirect their fall, whether they're using their arm muscles or it's just a yank at the end as their arm extends fully.

Force is inversely proportional to the time during which the person is slowed. If that's all at once, it'll be a much higher force than if that person has swung or otherwise slowed their fall over the course of extension of their arm(s).

The force would need to be, at minimum, the weight of the person. At most, it could be several hundreds of pounds.
posted by WasabiFlux at 11:09 PM on August 5, 2012 [1 favorite]

At a minimum, the thing being grabbed would have to support both the weight of the mass being supported (not the whole person if their feet are still on the ground, for example) and the force applied, which could be described as an additional "amount of weight" in simple terms.

An extreme example would be a wooden board that place a bullet on. It will sit there where you put it. The board would support its weight without issue. Now shoot the bullet at it and it will go through. The energy of the force made the bullet sort of appear as crazy heavy on a very small area of the board and it pushed straight through.

When someone falls, they carry a force with them that is a combination of lots of factors that would be difficult to calculate as a rule of thumb. That force could 'appear' to the thing they grabbed as more than just the weight of the person falling. That being said, if they fell in such a way as to not be bearing the majority of their weight on the object, the total force being 'felt' by the object could be less than the persons weight. So to not break, it just has to able to withstand the force applied, not necessarily the weight of the person applying it.

Another way to look at it is if someone hung themselves 100 ft off a bridge by their feet from an unstretchable rope that was only rated to hold their weight. Now imagine that person jumps off the bridge with the same rope tied to their feet. The force, or 'weight', on the rope would likely be high enough to snap it.

Your falling person would have a similar effect on the object they grab. It would have to be able to absorb both the total unsupported weight and any additional force that was added by the fall, for however brief a period.
posted by qwip at 11:09 PM on August 5, 2012 [1 favorite]

Take two glass dishes and two bricks. Set one brick on top of one of the dishes. Drop the other brick off a roof onto the second dish. Does it seem that both bricks exerted the same amount of force? Do the broken glass shards help you to decide?
posted by tylerkaraszewski at 11:18 PM on August 5, 2012 [9 favorites]

I'm assuming that your somewhat informal description means that you have a person standing on a trap door, gripping the end of a rope at the precise moment the trap door instantaneously disappears.

What ensues is usually a damped harmonic oscillator with the person as a mass. It is my non-verified expectation that in such a setup no part of the spring would ever have to support a force greater than the person's weight, but I have to work instead of writing out the calculations now.

If the person attains any initial velocity before grabbing the rope, this translates into the same setup, but with different boundary conditions. In this case, the maximum forces will depend on the initial momentum (e.g. how fast the person was falling when making the connection) and lead to an increased maximum force, as per everyone else's examples.
posted by themel at 11:23 PM on August 5, 2012

Like other people have said, if the falling person has an initial velocity you are wrong - supporting just the weight is not enough. The maximum force however strongly depends on undetermined factors, such as the way the falling person is attached to the stationary point.

Here's my attempt at an explanation:

Assume falling person has velocity v at the time they grab the rope. This gives them momentum p = m*v. The total force applied to the person is F-B=F-mg, where F is the force applied by the rope and B is the weight. To stop the person in time Δt, the average force <F> must be enough for the total force to cancell al the momentum, i.e.

m*v = (<F>-mg)*Δt => <F> = mg + mv/Δt

Already this isn't saying much: we don't know anything about Δt, which depends on factors such as the manner of attachment (think bungee cord => happy fun extreme sports vs simple rope => dislocated shoulder ), but the average force must clearly be greater than gravity if there is any initial velocity. In fact, averages being what they are, the instantaneous maximum force [ which is what you must figure out to see if your attachment point will snap ] will probably be considerably greater.

Note also that, after the falling person is (instantaneously) stopped, some energy has probably been transferred to the attachment point as dynamic energy (think stretched spring) - if the falling kung-fu master doesn't let go at that exact time, they're going to bounce up again, acting as a damped harmonic oscillator as themel points out.
posted by Dr Dracator at 12:49 AM on August 6, 2012 [1 favorite]

Pedantic point: "weight" already takes g-forces into account - an object's weight is a function of its mass and the strength of gravity wherever it is measured (subject to some subtleties to do with relativity). So in a sense you're both right, and in another sense you're both wrong.
posted by A Thousand Baited Hooks at 1:56 AM on August 6, 2012 [2 favorites]

Try jumping on a scale.
posted by telstar at 2:39 AM on August 6, 2012 [1 favorite]

So this is called 'impulse loading', and is, as everybody else is saying, higher than just regular 'static loading'.
posted by Comrade_robot at 4:47 AM on August 6, 2012

Best answer: for perspective and to throw out some numbers, a 100-lb person falling 10 feet and stopping over a distance of 1 foot (like if the thing they catch onto has one foot of stretch) exerts a force of 1100 pounds. so, the answer is potentially MUCH more than the weight of the person. (i don't have the equation for this because i'm an aerialist, not a science gal. but i know this answer because i'm an aerialist and we do sometimes drop and catch things and need to know how much weight our rigging has to support.)
posted by nevers at 5:38 AM on August 6, 2012

in fact, i think the "kung-fu grip" part of your question tells you the answer. someone can hold their bodyweight by one hand with just an average-joe grip, at least briefly. you need a kung-fu grip to hold 1100 pounds. and a kung-fu shoulder to keep it from dislocating over a force of 1100 pounds.
posted by nevers at 5:44 AM on August 6, 2012

Any amount of falling (read: movement) and you need more then the weight of the faller. So even falling from standing if more than their weight.

Falling a significant distance multiplies the above several times. Up to terminal velocity, 120mph-ish.
posted by French Fry at 6:51 AM on August 6, 2012

Response by poster: Conrad, just from normal standing.
posted by CollectiveMind at 7:24 AM on August 6, 2012

Response by poster: Fshgirl, let's say they were grabbing a rope swing vs a curtain
posted by CollectiveMind at 7:25 AM on August 6, 2012

Response by poster: Barely legal, thanks. Not the first time.
posted by CollectiveMind at 7:27 AM on August 6, 2012

Response by poster: Anadem, so that means if they grab the thing high when they are just starting to fall, it won't have to absorb as much energy as if they grab the thing just inches from the ground, right?
posted by CollectiveMind at 7:28 AM on August 6, 2012

Response by poster: nevers, yes and people dislocate shoulders and thumbs with average joe grip too. I meant their grip would stop their body from their hand even though the rest of their body would keep moving, unhinging things as the force moved away from that grip.
posted by CollectiveMind at 7:37 AM on August 6, 2012

Response by poster: French Fry, it isn't a significant distance. It's just like a knee giving out while standing and collapsing and grabbing something on the way down.
posted by CollectiveMind at 7:41 AM on August 6, 2012

If the railing is arm's distance below their shoulder, and they're catching themselves, then they're not going fast, so add some small fraction to their weight to see how much they'd exert with no feet on the ground. If their feet or knees are still on the ground, then it's about half of their weight as load, so your opponent is right about force, but it's still probably 150 or really close to what you need.

Also, consider how many people can pull up even their body weight. Not many.

As a completely unscientific, practical guess, you need 120% of 150 pounds, just because you don't want to have to repair them when someone tries to use it to stand on to hang a picture.
posted by cmiller at 9:10 AM on August 6, 2012

What about gymnast in a high bar routine? What force is being exerted on the bar to keep from breaking?
posted by brent at 8:22 PM on August 6, 2012

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