Force on an Anchor Continued:

1. May 19, 2010

rambo5330

1. The problem statement, all variables and given/known data
So after determining the way an anchor setup will handle the high tension forces created when at different angles, here is the next step in the question.
attached is a rough diagram that can be refered to for some clarity (hopefully).

So assuming we have a 68kg climber going up a wall being belayed (belayer keeps the slack on the rope to a minimum as the climber ascends the wall, thus reduceing his fall distance) by a 70 kg person (not sure if this matters right now). If the climber were to slip off and fall a distance of 3m before the belayer arrests the fall ... at which point the belay device kicks in and stops the fall in time of roughly 1 second.. I get something like this

Vf2 = Vi2 + 2(a)($$\Delta$$y)
giving a final velocity of 7.67 m/s then I believe I can use an impulse formula here for when the belay device arrests the fall...
Fnet = $$\Delta$$P / $$\Delta$$t
lets say there is a spring like effect and the climbers final velocity is then upward at 0.32 m/s after impact.
so..
Fnet = ( (68Kg)(-0.32m/s) - (68Kg)(7.67m/s) ) / 1 second
= 540 Newtons?

I'm thinking there must be more than this because that is less force than the weight alone of the person hanging on the anchor... the anchor would experience a force = to the weight of the person PLUS whatever force due to acceleration from falling ?

in a dynamic belay situation the person belaying will actually jump up right before the falling climber tensions the rope completely in effect increaseing the amount of time it takes to bring the falling climber to a complete stop thus reducing the impact... just curious how one would go about calculating these forces on the anchor etc.. it may be more advanced but not quite sure.. in the situation where the belayer is also lifted off the ground then the anchor must support the weight of both the belayer and the climber..plus whatever impact forces are involved im assuming, however maybe this is less than if the belayer were to statically absorb the fall? I am also not sure how the angle between the belayer and the climber(angle theta in diagram) will affect this? any thoughts?

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2. May 19, 2010

PhanthomJay

On Belay.................!
This presumably is the time it takes for the climber to come to a momentary stop once the belay device is engaged?
yes, that's the climbers speed just prior to the belayer arresting the fall
yes, that's the net force
but the climbers speed is 0 at the bottom point of the fall, before he/she springs back up again....
I'd look at the force at the bottom of the fall using the 1 second interval to come to a stop; F_net = 68(7.67)/1 = 520N
if F_net = 520 N, F_net consists of the the climbers weight down and the tension in the rope up..the tension must exceed the climbers weight since the climber is accelerating upwards (or decelerating, if you will)...So since F_net = T-W = 520, the rope tension is 520 plus the climbers weight (W = mg = 68(9.8) = 666 N) or T = 1200 N in round numbers.
now we're going from the complex to the very complex....
Assuming a static belayer, if theta is 0 and T = 1200 N, the force on the ring is 2400 N; as theta increases, this ring force decreases. Disclaimer: Don't take this for gospel, a life is at risk....I have climbed a 'rock' wall once in an indoor setting...The belayer was not paying attention when the climber fell...good thing the climber was not very high up the wall......

3. May 19, 2010

rambo5330

Thanks very much for your help, I should tell you im not useing what is said on this forum to determine how I will build my anchor, just very curious. I've been climbing for many years now and feel very solid, however I just read that the way I've been belaying a second climber on larger multipitch routes, is now considered an "old school" way of doing it, as it increases force exactly like what you were explaining how the 1200 N of force is doubled to 2400 N of force on the anchor ring the now recommended method of belaying is directly off that ring itself so the force is 1 to 1 rather than 1 + whatever percentage based on angle of the climber... This is what sparked my interest in these questions.. I love physics.. and rather than just takeing someones word for it I thought it would be cool to sit down and try and solve some of these situations for myself, which in turn helps me better understand physics. so I really appreciate the help.

The picture that sparked my interest is attached.
Now I have to ask it says holding .8 (80%) is this merely because of the angle the belayer is on or perhaps an average number?
And I must say we covered Tension very poorly in my course and I guess this is why I seem confused in these problems. But to clarify the concept of the tension and the anchor system etc... if the force is 1200 N at the ring based on the climbers fall.. how is it doubleing to 2400 N ... is this because that tension is now redirected on the belayers side of the rope ?

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4. May 19, 2010

PhanthomJay

First, I need to correct that tension force of 1200 N in the rope..I forgot that although the weight force is constant, the elastic (stretchable) rope tensile force of 1200 N is an average tensile force in the rope during the impact time...the maximum tension is twice that force , or 2400N at the bottom of the fall, which implies a max force on the ring of 4800 N for theta =0. Sorry about that. Good disclaimer, no? Now why is the force on the ring doubled? Assuming the ring is frictionless, (like an ideal pulley), the tension in the rope on both sides of the ring are the same (the direction changes, but not the magnitude). So that gives you 4800 pounds acting down on the ring (which transfers to the V above the ring, and ultimately to the anchors, where the force on the anchors is per the method in your prior post). If the theta your figure in post #1 was say 60 degrees, the downward force on the ring , found by calculating the resultant force of the 2 tension vectors, would be 2T cos30, or about 1.7 T = 1.7(2400) = 4100 pounds. I don't quite understand the last photo.

5. May 19, 2010

rambo5330

awesome I love your replies, So im guessing this is the same principle used in a haul system (pulley system). Hypothetically speaking if the medium the anchor was attached to was able to move, and we had the rope end where the belayer is attached to completely fixed then a person could move a 200 Lb object with 100lb of force? creating a 2:1 advantage ?

also, I was only taught to determine averages in impulse scenarious etc, so when you say the max force was double the 1200 N is there a formula for this? or whats the reasoning behind this? you are absolutely correct to assume the rope has a good amount of stretch, hence the 1 second stop time, (it may be even greater)

the second picture is one off of a climbing site, and it shows the climber as "force 1" and the belayer as "holding .8" this represents a hypothetical fall of force 1 and then on the belayer side I guess due to losses and maybe the angle 80% of the force create by "force 1" and the anchor system in that picture is a traditional 3-point anchor set up.. so there are three piece's of gear in the rock.. and slings run down to an equalized point creating a \|/ type pattern so it is a more advanced anchor than the one i drew in the pictures the " V " so the load is then distributed evenly over each of the 3 bolts on that anchor.. assuming perfectly equalized I suppose... but it is just there to show you that a set-up like this actually magnifies the force at the anchor as opposed to belaying directly off the anchor itself.....

can I touch on the dynamic belay now as well? im assuming if the belayer would jump uppon arresting the fall, it takes a longer time to bring the climber to a rest, therefore reducing the impulse and overall tension on the rope on the climber side.. that tension is also transfered to the belayers side but since the belayer is already accelerating upward, then does this also decrease the tension on the belayer side even more so? or must belayer side tension = climber side tension?

6. May 20, 2010

PhanthomJay

I wish I had paid more attention when I atended that 2 hour training session on rock climbing 2-digit years ago, because now I have more questions than answers. I'm not familiar with the equipment and methods used, but I wish I was, as I find it very interesting, and I would be more comfortable in responding.
mechanical advantage is usually associated with multiple pulleys, but if you are inquiring, for example, that if you had a 200 pound object on the ground, attached a pulley to it, wrapped a rope around the pulley and attached one end of the rope to a fixed anchor above, and pulled from above on the free end of the rope, you'd have to exert just 100 pounds of force to slowly lift the 200 pound object.
for a linearly varying force (as per Hookes law F=kx), F_avg = (F_max + F_min)/2. The min force is 0 at 0 displacement, and kx at max displacement. Thus, F_max = 2(F_avg). As an example, if you had a 100 pound weight attached to an unstretched spring hanging vertically above it, held it there, then slowly lowered it to its resting (equilibrium) position, there would 100 pounds of force in the spring when you let go. But if instead of slowly lowering it, you released it, the spring would stretch twice as much, and the force in the spring would max out at 200 pounds at the bottom before rebounding. Or from a conservation of energy standpoint, where the gravitational potential energy is mgx, and the spring potential energy is 1/2kx^2, and the object has no speed at the top and bottom of its fall, and since energy is conserved, then mgx = 1/2kx^2, or 100 = kx/2, thus kx = max spring force = 200 pounds.
this is where I get confused..isn't the belayer below the climber?
As I see it, you are correct that overall tension is reduced due to the longer impulse time, but the belayer side tension and climber side tension are still equal, regardless.

Since this discussion is not homework related, I am going to ask the moderators to move it to another sub-forum, where maybe others can respond to your thoughtful questions.

7. May 20, 2010

Gokul43201

Staff Emeritus
I too am not sure where exactly the "0.8" comes from... it may be from the cosine of the angle, but it looks like that would be a lot smaller.

8. May 20, 2010

rambo5330

Thanks again for your well put answers, it's begining to clear things up a lot more. I will mess around with some questions involving conservation of energy and spring potential energy which will hopefully clear that aspect up for me.. the rope stretch makes this quite interesting.
Anyways in regards to the belayer being below the climber your are absolutely correct in a top roping situation, which is the way most novice climbers will start out, especially if you climb in a gym. the belay technique represented by these pictures is actually used on larger multi-pitch climbs. Where you have 2 or more pitchs (rope lengths). one person will lead the the first pitch (climb and place protection), get to the anchor (in this situation he built his own) and then belay the second climber up. at which point they will both be at the 1st anchor (roughly 50m off the ground) , exchange gear and the second climber can now lead the second pitch up to the next anchor (roughly another 50 m up). or the original leader can lead the second pitch as well, it just takes slightly to change everything over. the on and on until you reach the top of the climb.

In this type of climbing, placeing your own protection and building your own anchors, rock can be of varying quality on each climb you do, so minimizing forces on anchors and gear becomes the name of the game. most of the gear these guys are placeing is rated anywhere from 6KN to 25 KN this is with perfect placement.

The website I found these pictures on, which I realise now may be of some great interest to you is http://climbinglife.com/tech-tips/tech-tips/what-s-the-force-on-your-anchor.html [Broken]

I built this thread because all of these years I've been useing the last belay technique shown on that site, re-routing through the anchor (for comfort and ease of belaying), which in effect almost doubles the force at the anchor. I've never had an issue, which is a good confidence builder in the quality of anchors I've built/ used but, if I can do anything to minimize these forces I will. I find this very interesting. It's very rare a climbing accident happens due to gear failure or belaying a climber up and having the anchor fail.. it's usually on the rappel. which most forces are minimized during a rappel already due to the lack of any falls shock loading the anchor. However they do happen quite frequently due to inadequitely sized anchors for the load. knowing how to properly equalize an anchor and the way the load will be dispersed on each piece of protection when building an anchor is essential.

Thanks for the input and help! I encourage anyone to throw there own thoughts at this. I'm sorry it ended up in the homework forum, I initially posted it in general physics, got 99 views and no one touched it. I know its a PF no no but as far as im concerned the first parts were almost homework based, they definetly will help me with my course work and haveing a better understanding of concepts.

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9. May 20, 2010

PhanthomJay

I'm surprised no one touched your post in the general physics section, it must have got lost in a black hole somewhere, as it is an extremely interesting topic on the physics (and hobby) of rock climbing, with some well thought out questions. I'll look at that site you referenced and try to get back to you with more answers (or rather, more questions!). Thanks.

10. May 21, 2010

PhanthomJay

Nice site, thanks. I still don't quite get the ".8", but essentially, you're near doubling the force on the anchors, but you did have 3 of them in a row, so each sees a third of the 'doubled' load (that's the '.6 ' value in the figure). Not as good as the .33 direct belay w/ 3 anchors, but still much much better the redirected belay with a 2-anchor V, though , and much much much much better than the "American Triangle" anchor arrangement (ever hear of it?...... Stay away from it!). (I've been doing some googling ).

11. May 21, 2010

rambo5330

hey! thanks! thought you might find that interesting... yes I've heard of the american triangle, never used it. not a good situation at all.. gotta love the physics involed with the whole sport of climbing!

12. May 21, 2010

PhanthomJay

Outdoor rock climbing must be quite a thrill, my best wishes to you. I've climbed Mt Washington in the Appalachians of NH several times....by foot...dry conditions early fall before the snow flies...up the headwall of the feared Huntington Ravine Trail...very challenging by foot....I've several times noticed some rock climbers scaling the Pinnacle...now that's really challenging...awesome sight...made me envious...worst weather in the world atop the peak..once when climbing the Huntington with my son in late September, we got halfway up and the clouds started to thicken...temp was below freezing at the summit...snow likely...now what to do, the trailbook guide said never to go down..had no choice..left the backpack on the cliff and scaled down on our butts...what a rush...and although i've grown old now, that mountain still calls......sorry to digress, but this mountain talk brings back some fond memories...thanks

13. May 25, 2010

rambo5330

Hey, back again ended up working all the way through the long weekend, but thats great to hear you get out into the mountains. keep it up... it really is a great stress reliever. to me it's what life is all about. Anyways I revisted this question with some new knowledge (my spring skills were/are quite weak). So heres what I got
I used a 70 kg climber, he falls 3 meters before the belayer arrests his fall
in this 3 meters he gets up to a speed of 7.67 m/s
so once the belayer begins to arrest his fall here is what I did

PEgi+ KEi = PEsf

(i also found online in a lab experiment someone doing something vary similar and they determined the spring constant for there climbing rope to be roughly 200 N/M, so I use this)
so solving for displacement x I got

1/2(kx2) - mgx - 1/2(mvi2)

this gave me a quadratic, solving for it I ended up with two answers
x = 9.11m or x = -2.255m , -2.255 meters made the most sense for me since 9 meters is a lot of stretch! and in an upward direction? are both answers valid or whats going on here?

anyways I used -2.255m for my calculation for how long it would take to stop...
$$\Delta$$x = 1/2(Vo + Vf)(t)
solving for t i got roughly 0.6 seconds
so I used this value in the impulse formula

F_net = (MVf -MVi)/t = -((70kg)(7.67m/s))/ 0.6s = 895 N

so this is where i get lost... so that is the the net force, in an upward direction... decellerating your fall. I think maybe where im getting messed up is on a free body diagram...

mg is downards, tension is upwards, $$\sum$$F_y = T - mg = 895 N
so T = 895N + (70kg)(9.8) = 1580 N

so now we have tension... But we are neglecting the spring aspect...

so the formula i have for spring forces are
F_avg = (F_min + F_max)/2

so since F_spring = -kx
we are useing 200 N/m for k here
and when my spring is displaced by a maximum -2.255m then isnt my F_spring_max simply F_max = -200N/m(-2.255m) = 450 N
?? where does this even come in? why up above are you useing the force determined by the impulse equation and substituting it into the spring equation.. this part is where i get lost... because when i goto draw a free body i have a Tension force acting upwards, a spring force acting upwards, and mg acting downard and im not sure wether my sum of forces should be the Tension force i first calculated on impact, PLUS the spring force - mg ... or what? please help!

14. May 25, 2010

rambo5330

So I found a report that pretty much sums up everything I've been trying to create only they do a wayy better job and it's part of an actual course with the title (go figure) "the physics of rock climbing"
definitely check this out!!

a very cool project and it just may excite you even more into the physics of rock climbing...
I'm sure I will have some more questions now

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15. May 26, 2010

PhanthomJay

Ignore the negative answer in these problems. If 9 meters seems high, maybe the assumed 'k' value of the rope is too small?
t is 2.4 seconds using x=9.11 m
yes, but the net force using t=2.4 is 225 N
yes
This is the average tension, T is not constant, and using the corrected values, T_avg = 910N
yes, F_max = 1820 N
that's 200 (9.11) = 1820 N
No, in the free body diagram, the tension force IS the spring force.....