Could use some physics help

[vep]

I'm a rock climber and recently I've become curious about the gear we use and the "consequences" of us falling.
While climbing you use already set bolts in which you clip quickdraws. You then clip the rope into the quickdraw. Every single piece of gear is tested to 24 kN.
What I've been wondering is how much force does a falling body produce on a piece of gear exactly? And seeing as I'm really not knowledgeable in physics or math in general, I thought I'd ask for some help. And Yahoo questions thing requires me to sign up which I can't be annoyed with.

The scenarios to which I would like to know the answers are as follows:
(in the scenarios, we are presuming it's a piece of iron falling, not a human body, if it matters) (Also, when saying we have a 10 meter fall, we are presuming it is 5 meters above the quickdraw/bolt which then turns into 10 meters after falling 5 additional meters under the bolt)

1) What is the maximum force a body of 75kg can produce after reaching maximum velocity?
2) What is the force a body of 75kg would produce after a 1 meter fall?
3) What is the force a body of 75kg would produce after a 5 meter fall?
4) What is the force a body of 75kg would produce after a 10 meter fall?

If anything is unclear, I would be glad to try and clarify.
Thank you in advance.

Edit: we are trying to calculate the impact on the bolts and quickdraws, not the rope, so we can ignore shock absorption the rope would create, if that helps.

[vep]

You kinda need to take into consideration the elasticity of the cord you are using. You cannot calculate the exerted force without knowing the stopping distance.

Yes, I agree. But for the example, let's presume there is no rope and thus not factoring it. Let's just presume you drop a piece of steel from specified heights and find out how much force it would produce. I'm not interested in being 100% precise, just want to know what force an object of 75kg would produce.

Edit: we are trying to calculate the impact on the bolts and quickdraws, not the rope, if that helps.

[Deleted]

You kinda need to take into consideration the elasticity of the cord you are using. You cannot calculate the exerted force without knowing the stopping distance.

This. But...
For a 1 meter fall with a 0.030626m stopping distance, the force is 23999 so that's your maximum
For a 5 meter fall with a 0.15313m stopping distance, the force is 23999 so that's your maximum.
For a 10 meter fall with a 0.30626m stopping distance, " " " " " " "

etc etc.

- - - Updated - - -

Yes, I agree. But for the example, let's presume there is no rope and thus not factoring it. Let's just presume you drop a piece of steel from specified heights and find out how much force it would produce. I'm not interested in being 100% precise, just want to know what force an object of 75kg would produce.

Edit: we are trying to calculate the impact on the bolts and quickdraws, not the rope, if that helps.

Doesn't matter, If it hits the floor, you need to find out how much give the floor would have. If there is no give, and it "stops dead" then the force is infinite

[vep]

This. But...
For a 1 meter fall with a 0.030626m stopping distance, the force is 23999 so that's your maximum
For a 5 meter fall with a 0.15313m stopping distance, the force is 23999 so that's your maximum.
For a 10 meter fall with a 0.30626m stopping distance, " " " " " " "

etc etc.

- - - Updated - - -



Doesn't matter, If it hits the floor, you need to find out how much give the floor would have. If there is no give, and it "stops dead" then the force is infinite

I'm not entirely sure I understand. You're saying that a falling body of 75kg would produce 23999 newtons? Which in turn relates to around 23 kN? That sounds a bit odd to me. But as we didn't calculate the rope's shock absorption, does it really make that much of a difference? I mean, I left it out because the variables are big depending on how much rope is pulled out, right? A longer piece of rope should stretch more thus absorb more force?

For the second quote: i most definitely don't understand. How can it produce infinite force?

- - - Updated - - -

And the kinetic energy of that falling body:

Ke=0,5*75kg*((14m/s)^2)

Ke=7350 J

So I'm presuming J stands for joules? If so, how does that translate to newtons? Google was of no help :/
Also, I thought the formula to use would have been F=ma... Boy was I wrong.

[Revi]

I'm not entirely sure I understand. You're saying that a falling body of 75kg would produce 23999 newtons? Which in turn relates to around 23 kN? That sounds a bit odd to me. But as we didn't calculate the rope's shock absorption, does it really make that much of a difference? I mean, I left it out because the variables are big depending on how much rope is pulled out, right? A longer piece of rope should stretch more thus absorb more force?

For the second quote: i most definitely don't understand. How can it produce infinite force?

At those stopping distances it does, making those your minimum stopping distance before gear breaks. Luckily, with climbing rope the stopping distance is much greater so the force exerted on your gear from those falls would be much, much lower than 23999.

[Deleted]

I'm not entirely sure I understand. You're saying that a falling body of 75kg would produce 23999 newtons? Which in turn relates to around 23 kN? That sounds a bit odd to me. But as we didn't calculate the rope's shock absorption, does it really make that much of a difference? I mean, I left it out because the variables are big depending on how much rope is pulled out, right? A longer piece of rope should stretch more thus absorb more force?

For the second quote: i most definitely don't understand. How can it produce infinite force?

Yes, if it had such a small stopping distance. The force at impact depends on the stopping distance. That's just how it works. Whether that stopping distance depends on the rope's elasticity, or the object you hit having some 'give' when you hit it. The shorter the stopping distance, the greater the force. Which is why if you fall 50 meters onto concrete that has very little give in it, you go splat, but if you fall onto a sponge that has a lot of give in it, you are fine.

If the stopping distance is zero, then the force is infinite. This isn't possible though, since there will always be a stopping distance.

[zoefschildpad]

The Force of an impact is related to how long it takes to come to a full stop. There is no such thing as an instantly full stop without infinite force, something will give way. In this case the elasticity of the rope is important because the more elastic the rope, the longer it takes to stop and the less force is on your equipment and your body.

That is also why bungee jumping with an unelastic rope is going to tear you apart, but doing it with a bungee cord is fine.

[Deleted]

[/QUOTE]So I'm presuming J stands for joules? If so, how does that translate to newtons? Google was of no help :/
Also, I thought the formula to use would have been F=ma... Boy was I wrong.[/QUOTE]

Joule = W*s
W = Newton/s
(Newton/s)*s = Newton
Joule is measured in newton, am i right?

[vep]

Thanks for all the help, guys I've come to realize it is a more daunting task than I had imagined.

Wikipedia had a really good article about fall factor in climbing.

We've learned about fall factor, I just didn't go into details in it and had no clue it was actually that in-depth. What we were taught is basically that fall factor 2 is bad.

[Deleted]

So I'm presuming J stands for joules? If so, how does that translate to newtons? Google was of no help :/
Also, I thought the formula to use would have been F=ma... Boy was I wrong.

You're not wrong actually! Force = mass x acceleration is what you are using, but because you're working out the force of something stopping, it's decceleration, not acceleration. Acceleration is meters per second per second, or the rate at which your speed is increasing. In this case you need to know the stopping distance, which is essentially dependant on the rate at which your speed decreases.

Think about it...

If you are travelling at 100 meters per second and you decellerate at 1m/s/s (so for every second that goes by, you go one meter per second slower) then the force is going to be really low (picture yourself sitting in a car slowing down at that rate, how much would you be pushed into your seatbelt?) and you'll come to a stop in 100 seconds.

Now picture the same situation except you are decellarating at 100m/s/s (so for every second, you go 100 meters per second slower, so you are coming to a complete stop within one second) and imagine how much you'll be thrown into your seatbelt in this case. it's a SLAMMING ON THE BRAKES thing instead of slowly pressuring the brakes.

[Flutterguy]

Just make sure your backpacks don't have anvils in them.

[vep]

Acceleration is meters per second per second

Thanks for the explanation!
This quoted part, tho, has always been fairly confusing to me. I can understand meters per second. But what does "per second per second" or per second squared actually mean? I mean if it's 9.81 meters per second, then I understand that but what happens and what does it mean when it's second squared?

[Deleted]

Thanks for the explanation!
This quoted part, tho, has always been fairly confusing to me. I can understand meters per second. But what does "per second per second" or per second squared actually mean? I mean if it's 9.81 meters per second, then I understand that but what happens and what does it mean when it's second squared?

Meters per second per second is exactly what it sounds like.

So, 9.81 meters is a distance.
9.81 meters per second is a speed. That is, the speed at which you a travelling when you cross a distance of 9.81 meters in a time of one second
9.81 meters per second per second is an acceleration. So for every second, your speed will increase by 9.81 meters per second. So 1 second into your journey your speed will be 9.81 meters per second. Two seconds into your journey, your speed will be 19.62 meters per second, and so on.

When you have things like meters per second squared it's just an easy way of writing the unit. To get a speed, you multiply meters (m) by time in seconds (s), so speed is measured in ms (which is the same as saying m*s or m x s). To get an acceleration you multiple meters, seconds, and then more seconds all together, so that would be mss, but when you get a thing, and you multiply it by itself, then you are squaring it. so like 2x2 is the same as 2 squared. Seconds times seconds is the same as seconds squared. So meters x seconds x seconds becomes meters x seconds squared, or ms^2

[Deleted]

i have a masters degree in maths and physics and i cant answer your question.
go team uk educaton!

[vep]

i have a masters degree in maths and physics and i cant answer your question.
go team uk educaton!

Look on the bright side, at least you understand a lot more than I do

Thanks Nixx and Erin for your explanations!

[Deleted]

i have a masters degree in maths and physics and i cant answer your question.
go team uk educaton!

You can't answer it because he didn't provide all the relevant information. I got an E in maths with mechanics at AS level and then bombed out cause I was awful and can't answer it either

Edit: feeling well clever now though :3

[Deleted]

i cant answer it as i finished my degree..um....13 years ago (yay maths!)
not used any of it since

I do know a lot about double entry though <3

[Deleted]

I do know a lot about double entry though <3

Bow chicka wow wow.

[Deleted]

Bow chicka wow wow.

accountancy ftw!!!1!

[Deleted]

Did you guys just make accounting sound hot? :|

Accountancy is hot. Ohh I have some receipts here I need to go through with you before I refill petty cash. Oh there's only one chair at your desk, i'll just sit on your knee and we can go through them together now.

Oh no, our purchase ledger is stored in the bottom drawer of this filing cabinet. Looks like I have to hitch up my skirt and bend down to pick it up.