Thinker Posted October 9, 2003 Share Posted October 9, 2003 (edited) I spent the last couple of weeks working near Salmon, ID...way up in the hills. As I was relaxing one evening in the motel hot tub (with a great little buzz from some mediocre single malt), I realized that the same principle that allowed my arm to 'stick' to the nearly vertical side of the hot tub and thus support my arm that was holding my book is the same force that holds a climbing shoe on nearly vertical rock. Based on my (sometimes faulty) memory, that force is defined as static friction. The interesting thing about static friction is that it can be overcome, and then the rules of kinetic friction take over. That's when you get that sickening feeling of your shoes buttering off the rock...or your arm starts sliding down the wall of the hot tub...or your tires break loose and start skidding. A Google search yielded this little figure to illustrate the point: Note the instantaneous dip in friction resistance once the threshold of motion has been reached. This would indicate that (theoreically) it's better to find a way to reset that sliding shoe than to hang on to the bitter buttery end...or at least find a way to stop the sliding motion for a brief instant to allow the static friction to resume it's meager grip on the rock. (I know this all sounds a little too Cartesian, but it endured the scotch buzz and so it must be meant to assist someone struggling with slippery shoes. Come to think of it, I was kind of like a brain in a vat that night....) edit: oh yeah, context: http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html Edited October 9, 2003 by Thinker Quote Link to comment Share on other sites More sharing options...
klenke Posted October 9, 2003 Share Posted October 9, 2003 Yep, definitely a thinker. Yep, we all know that when that shoe slips, no use thinking it's gonna restick itself. Nope, better pick it up and reapply it or hope it slides down to a suitable foothold. Or you can leave it and scratch like hell. Quote Link to comment Share on other sites More sharing options...
gslater Posted October 9, 2003 Share Posted October 9, 2003 Yup, the static coefficient of friction is always higher than the kinetic/dynamic/sliding coefficient. That little college factoid always seems to pop back into my head whenever my foot starts to slide. Quote Link to comment Share on other sites More sharing options...
Dan_Harris Posted October 9, 2003 Share Posted October 9, 2003 When you start to slide what needs to increase is the normal force in order to increse the force of friction, i.e. press harder perpendicular to the rock surface. Most people usually push along the rock in their struggle to stop slipping. Easier said than done. Quote Link to comment Share on other sites More sharing options...
yeti Posted October 9, 2003 Share Posted October 9, 2003 If the normal force is perpendicular to the rock and the friction force is parallel to the rock how do you think that more normal force will stop you from sliding? Wouldnt more friction be needed to counter the slide due to gravity. Gravity points down, while friction points up along the rock. More normal force would just mean an equal force from the rock pushing back at you, otherwise your foot would go into the rock along the direction of the normal force. I geuss if you could crush rock thereby creating steps in the rock, friction climbing would rather trivial so in that sense more normal force may help you get up the climb. Yeti Quote Link to comment Share on other sites More sharing options...
gslater Posted October 9, 2003 Share Posted October 9, 2003 yeti said: If the normal force is perpendicular to the rock and the friction force is parallel to the rock how do you think that more normal force will stop you from sliding? Wouldnt more friction be needed to counter the slide due to gravity. Gravity points down, while friction points up along the rock. More normal force would just mean an equal force from the rock pushing back at you, otherwise your foot would go into the rock along the direction of the normal force. Yeti More normal force (if you can find a way to get it) will increase the total friction, thus slowing the slide. Friction = the coefficient of friction times the normal force (aka the component of the weight in the direction into the rock). The trick is to increase the normal force. I can see it now.... Quick!!! I'm sliding!!! Throw me some and !!! Quote Link to comment Share on other sites More sharing options...
Al_Pine Posted October 9, 2003 Share Posted October 9, 2003 If you're on a 45 degree slab, adding weight will add equal weight perpendicular to the rock as to that pushing you down the slab. I don't know what range friction coefs live in, but it seems like for the right combination of friction and slab angle you would stick better if you were wearing a backpack full of rocks! The bad thing to do though is to drag a rope. In that case you have almost no downward (and thus no toward-the-rock) component, but a sizeable component pulling you back down the slab. Quote Link to comment Share on other sites More sharing options...
Dr_Flash_Amazing Posted October 9, 2003 Share Posted October 9, 2003 Thinker said: Come to think of it, I was kind of like a brain in a vat that night....) The above reminds DFA of the below (from Alk3's classic 'Bleeder'): it's one thing that I never said, "I'm truly happy in my heart and in my head" a lonely liver suspended in liquid one thing that I never did was smile, missing a case, lacking a lid" Quote Link to comment Share on other sites More sharing options...
specialed Posted October 9, 2003 Share Posted October 9, 2003 So how exaclty is that beta going to help me get up 11c slab? Quote Link to comment Share on other sites More sharing options...
Al_Pine Posted October 9, 2003 Share Posted October 9, 2003 Just visualize that the slab is a hot-tub wall, and if you skate, you'll slide down into the tub with Thinker Quote Link to comment Share on other sites More sharing options...
klenke Posted October 9, 2003 Share Posted October 9, 2003 Engineering with nothing better to do finds his way to a keyboard and types away... Forgive me but I can't seem to get Greek symbols to show up so I've tried to mark them in different colors instead. q = theta phi = phi ms = mus and mk = muk Increasing your weight while slipping, be it by eating rapidly or picking up a barbell, will not necessarily cause you to stop slipping. Here's why: The assumption here is for an ideal case where there are no external forces to the system. There is no wind force or muscular force or a pull force from a rope; "things of that natua" as A'nold would say. All that is present is one's weight due to gravity [W]. This weight causes a normal force [N=Wcosq], a surface-parallel force [F=Wsinq], and a static frictional force [f = (ms)N], where q is the angle of inclination of the rock from the horizontal. Now, for the sake of argument, let's assign a value to the coefficient of static friction. This coefficient can range from 0.0 to 1.0. A null value denotes a perfectly frictionless surface, something that is generally impossible to achieve in practice. A value of 1.0 would correspond to being riveted or epoxied to the mountain. Let us therefore choose ms = 0.6. I have no idea if this is a good ballpark figure for rock. (Some rock is lichen covered, some is very rough.) But for the sake of argument the exact value doesn't matter. Furthermore, mk must be less than ms, so let's choose mk = 0.4. Now, for any given value of ms (in the absence of forces other than one's weight, W) there is an angle of repose, phi. For sliding to be impending (about to happen), F must be exactly opposed by f. This requires Wsinq = (ms)N = (ms)Wcosq. W cancels out. With ms known, the angle of repose at which impending motion is about to occur can be calculated as phi = q = arctan (0.6) = 31 degrees. This means that for rock inclinations <= 31 degrees, no motion will occur. For inclinations >31 degrees, motion will occur and the magnitude of the frictional force will instantaneously drop to f = mkN = mkWcosq. It may appear, based on this last equation, that increasing W (your weight) will increase the friction force and you can therefore use this increased force as an impetus to stop sliding. The problem is that W is also a factor in the sliding force F (i.e., F=Wsinq). For equal values of W for both equations, it can be seen that F increases as q increases; but f decreases as q increases. This is true for all values of W. The result is that F gets bigger while f gets smaller. Not a good thing for avoiding slipping or the continuance thereof. Conclusion: (scenarios) 1. If you're slipping because the inclination of the rock you were trying to climb up exceeded the angle of repose, then you cannot regain a static condition simply by increasing your weight. 2. However, if you're slipping because you unintentionally lowered the available friction by way of decreasing your own normal force (because you leaned into the rock too much instead of keeping your butt out as is required), then you can possibly stop sliding just by getting your butt back out where it's supposed to be. In this case, the inclination of the rock is less than the angle of repose. Getting your butt back out simply increases N, thereby increasing f. 3. Scenario 2 can also be remedied by increasing your weight, which would increase f quicker than it increases F. 4. If the rock inclination is greater than the angle of repose then it really doesn't matter where your butt is. That is, even at maximum N value (maximum butt extension), you're going to slide and won't be able stop yourself. Limits: (asymptopes) 1. In theory, mk can have a value of unity (1.0) like ms, but this is really not practicable. If mk could be 1.0, however, then this would result in a maximum angle of repose of phi = arctan (1.0) = 45 degrees. This means that for rock inclinations > 45 degrees friction alone will not and cannot save you and you therefore better come up with a different constraining force. A pro placement above you or a hand or foothold would work well for this. 2. For a ms value of 0.0 (hence mk = 0.0) the contact surface must be horizontal. Any slight tilt will induce sliding. Furthermore, since this is a frictionless surface (ms = 0.0), no value of W will stop the sliding. The only way to stop the sliding would be to introduce an external force such as a shepherd's crook. Speaking of shepherd's crook, I see one coming in from my side. Later, boys and gir.... Quote Link to comment Share on other sites More sharing options...
yeti Posted October 9, 2003 Share Posted October 9, 2003 In order for you to stay in a static position on slab of rock the friction force must be equal to the amount of downward force applied. on a vertical slab this would be your entire weight, however on a slab that is not vertical friction force must only equal the portion of your weight in the downward ( Parallel to the rock) direction. gslater said:"Friction = the coefficient of friction times the normal force (aka the component of the weight in the direction into the rock). For forces to be equal to each other must both be in opposite directions on the same plane. However Normal force points into or out of the rock while friction force is along the rock. To increase your normal force you would have to get heavier or push harder against the rock. Al_Pine said:"for the right combination of friction and slab angle you would stick better if you were wearing a backpack full of rocks!" Do you really think that it would be a good idea to climb a delicate slab with a super heavy pack? Doubt it!! You cannot increase the coefficient of friction of a certain material by adding weight you can however increase the force exerted by the material until failure as seen in the first post drawing, so no reason to add more weight you wont climb harder just slip sooner. So as we all know when climbing slabs stick your ass out to 1) get more contact of your sole to the rock 2) To reduce the amount of weight in the downward direction therby requiring less friction. Yeti Quote Link to comment Share on other sites More sharing options...
Gordonb Posted October 9, 2003 Share Posted October 9, 2003 Normal force on a 45 degree wall is 45 degrees, In other words it is the force going straight into the wall, not towards the ground. Wearing a bag of rocks won't help because it will just add more force into the ground not normal to the wall. you want normal force not tangential force. To add more force normal to the surface, you may have to undercling or stem off of something so that the force pushes into the wall. Quote Link to comment Share on other sites More sharing options...
yeti Posted October 9, 2003 Share Posted October 9, 2003 Klenke said: "However, if you're slipping because you unintentionally lowered the available friction by way of decreasing your own normal force (because you leaned into the rock too much instead of keeping your butt out as is required), then you can possibly stop sliding just by getting your butt back out where it's supposed to be. In this case, the inclination of the rock is less than the angle of repose. Getting your butt back out simply increases N, thereby increasing f. " Particularly klenke said: " Getting your butt back out simply increases N, thereby increasing f. " this doesnt seem correct to me, im no engineer however. Getting my but out will put more of my normal force, which remains constant throught my climb, perpendicular to the rock therfore less in the downward direction therefore reducing friction allowing me to "walk" back down the graph away from the dip that marks slipping, or where kinetic friction comes into play. Yeti Quote Link to comment Share on other sites More sharing options...
Ursa_Eagle Posted October 9, 2003 Share Posted October 9, 2003 klenke said: Forgive me but I can't seem to get Greek symbols to show up so I've tried to mark them in different colors instead. q = theta phi = phi ms = mus and mk = muk well mu is easy: µ = Alt + 0181 Phi can be approximated with "Latin letter O with Stroke", either small or large: ø = Alt + 0248 Ø = Alt + 0216 Theta is non-existant in the Ariel realm... gotta love Character Map Quote Link to comment Share on other sites More sharing options...
klenke Posted October 9, 2003 Share Posted October 9, 2003 Yeti: a climber's normal force N is NOT constant throughout the climb. Only his or her weight W is constant. The normal force is highly dependent on the location of one's center of gravity and the applied line of force to the wall (usually through the legs). Lean too close to the wall and you are in effect changing the line of application of the normal force AND decreasing N because more of your weight is directed ALONG the wall. In the limit as your center of gravity coincides with the wall (i.e., you're hugging the wall), there is no longer a normal force into the wall, only a downward force onto whatever foothold(s) you're standing on. Someone else made a comment about getting your butt out to increase the contact surface of your shoes against the wall. This is only partially correct. In friction analysis, the total area of contacting surface does not play into the friction force. That is, it neither increases nor decreases the available friction force. Getting your butt out simply increases the all-important normal force. However, if you lean out, you may be able to get more of that sticky stuff that makes up the sole of your rock shoes on the wall whereby it can "grip." I use grip here as something other than friction alone. Grip would imply a shoe deforming around a nubbin, such as is possible when you flex your toes. Quote Link to comment Share on other sites More sharing options...
klenke Posted October 9, 2003 Share Posted October 9, 2003 Ursa_Eagle said: well mu is easy: µ = Alt + 0181 Phi can be approximated with "Latin letter O with Stroke", either small or large: ø = Alt + 0248 Ø = Alt + 0216 Theta is non-existant in the Ariel realm... gotta love Character Map Hey, thanks for that advice. Let me try those: µ ø Ø Cool! Can you send me a complete list of stroke commands? That would be much appreciated. Even so, I'm not going to go back and edit my main analysis post. Quote Link to comment Share on other sites More sharing options...
Ursa_Eagle Posted October 9, 2003 Share Posted October 9, 2003 I just used character map. Now when you install windows you have to tell it to install it manually, I don't think it comes automatically anymore. Quote Link to comment Share on other sites More sharing options...
Thinker Posted October 10, 2003 Author Share Posted October 10, 2003 klenke said: However, if you lean out, you may be able to get more of that sticky stuff that makes up the sole of your rock shoes on the wall whereby it can "grip." I use grip here as something other than friction alone. Grip would imply a shoe deforming around a nubbin, such as is possible when you flex your toes. I see reality similarly. We intuitively take advantage of every nubbin or dish on a slab, effectively decreasing the angle of the slab in that locale. In addition, the soft rubber on shoe soles can actually deform and allow the individual grains of rock/grit to press into it, again with the effect of decreasing the slab angle and increasing the normal force (envision the grains of rubber as tiny toes all gripping the grains of rock as the shoe is pressed against it.) Since the characteristics of rubber change with temperature, there should be an optimal range where the rubber deforms enough to grip well, but not enough to start decreasing the coefficient of static friction. It would actually be interesting to get Mike Garrison's reaction to all this...anybody familiar with his stuff on rec.climbing? Quote Link to comment Share on other sites More sharing options...
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