Imagine, for example, that you are climbing without benefit of equipment other than a safety belay line handled by a companion above you. You face a tilted slab of rock that is met at an obtuse angle by another slab of similar tilt, the juncture forming a ridge something like an edge of a pyramid. (The juncture is not sharp, though; it is worn as smooth as the faces of the slabs.) Where is the easiest climb: directly up one of the slabs or along the ridge?

Again, suppose that the grade on a slab is steep but would nevertheless h allow you to stand and manage to walk up the rock. Is it safer to lean forward and use your hands? What if you are on a sloped ledge and find a similarly sloped ledge at about waist level: is it safer to lean forward and partially . support yourself with your hands on the higher ledge?

Consider a third situation. A fairly wide vertical crack might allow you to "chimney": to climb into the crack, push your back against one side and press your feet against the opposite side, bending your knees to make the fit. To move upward you might essentially walk up one side of the crack while shoving your back up along the other side. (You gain additional, but minor, support if you press your hands against the wall at your back or, if the crack is narrow enough, against the opposite wall.) Or you might place one foot on each wall and press down while raising your body. Because you must exert constant pressure against the walls, a long chimney is tiring. If you stop to rest, is there an optimal location for your feet, one that would minimize the necessary push on the rock? When you climb, how exactly should you move your feet and back to decrease the chance of slipping? Suppose that as you climb you find the rock becoming slick. Should you place your feet higher or lower with respect to where your back pushes on the rock?

If, to cite a final example, the crack is quite narrow and the rock on one side of it juts out more than the other, you might try a "lie-back." The technique requires that you be on the side of the crack opposite the section that juts out. You insert your fingers into the crack and pull against its side. Then you place your feet on the projecting rock on the opposite side and press hard. As you pull with your arms and push with your legs, the tension you create stabilizes you, but the effort is draining. How should you position your arms and legs to lessen the fatigue, and how should you move them to shift yourself upward safely?

Some of the principles involved in these varied techniques of climbing were examined in 1976 by R. R. Hudson of the

Darlington College of Technology and W. Johnson of the University of Manchester Institute of Science and Technology. The techniques rely on the friction between the climber and the rock. Friction, of course, is the force that opposes the sliding of one surface over another, such as a climber's shoe over rock. A simplified | explanation of friction involves the h small-scale roughness of the surfaces. Even if two surfaces, such as flat rock k and the sole of a shoe, appear to be smooth, they are likely to be covered with tiny hills and valleys. When a force tries to slide the surfaces over each other, the hills on one surface catch on the hills and in the valleys of the opposite surface. The collective resistance to the sliding motion is the friction.

Before I get to the tricky cases of rock climbing explored by Hudson and Johnson, I shall examine the forces involved in some common situations of standing and walking. When you stand upright on a floor, two forces operate on you: your weight pulls down, and the floor pushes up just as much. The forces can be represented by vectors that indicate size and direction [see Figure 1]. The weight vector is assigned to the center of mass, which is approximately behind the navel when you are upright. The force from the floor is called the normal force, "normal" indicating that the force is perpendicular to the support surface-in this case the floor.

If you begin to step forward with one foot, your leg muscles push backward on the other foot, which experiences friction that must exactly counter the push if it is not to slip. The friction is represented by a vector parallel to the floor. If you push harder and the trailing foot still does not move, the friction must then be larger and, again, an exact match to the push. If you push too hard, so that the push exceeds some upper limit on the friction, the hills will yield and your trailing foot will slip. The upper limit on the friction is determined by the product of the normal force and the "coefficient of friction," a measure of the roughness of the sole and the floor.

Because the normal force equals your weight, the upper limit on the friction depends primarily on the coefficient of friction. When the coefficient is large, as it is on dry concrete, the upper limit is also large; you may be able to push quite hard without slipping. When the coefficient is small, as it is on slippery ice, even a small push makes the trailing foot slide.

An equivalent way to examine the role of friction is to redraw the friction vector so that it and the normal force form the legs of a right triangle. The procedure does not change the value of the friction, because the essence of the vector-its length and orientation-is retained. The hypotenuse of the triangle, also a vector, is said to be the sum of the normal force and the friction; it is called the reaction force.

The angle between the reaction force and the normal force-what I shall call the reaction angle-is important. As you push harder and increase the friction, the reaction angle increases. When the angle reaches a certain critical value, the friction is at its maximum value and the trailing foot is on the verge of slipping. The critical case arises when the tangent of the angle is equal to the coefficient of friction. In some examples of rock climbing the reaction angle is easier to study than the friction on the climber, and so I shall be considering it further.

Now consider the forces involved when you stand upright on a ramp [see Figure 2]. Your weight still pulls you toward the support surface and leads to a normal force that is perpendicular to the surface, but now your weight also pulls you down along the surface. That new feature creates friction on the feet even if you do not push backward. An easy way to picture the forces is to separate the weight vector into its components. One component is perpendicular to the ramp and is matched by the normal force. The other component points down the ramp, and if you are stable it is matched by the friction. Here again the friction can be redrawn so that it, the normal force and the reaction force form a right triangle. In this example the reaction angle happens to be the same as the tilt of the ramp.

Imagine what happens if the tilt (and so also the reaction angle) increases and you remain upright. The weight component directed down the ramp increases and so, automatically, does the friction that counters it. The weight component perpendicular to the ramp decreases, as does the normal force that opposes it. When the reaction angle reaches some critical value, the friction reaches its upper limit and you are on the verge of slipping. That critical value sets a limit on the pitch you can withstand without slipping.

If the pitch is moderate and you push backward on one foot in order to step up the ramp with the other foot, the push is directed down the ramp and requires that the friction on the trailing foot increase. This tilts the reaction force forward from the vertical and increases the reaction angle [see Figure 3]. The stability of the foot can be described in terms of either the friction or the reaction angle. The foot is stable if the friction does not exceed its upper limit or, equivalently, if the reaction angle does not exceed its critical value. If you carelessly push too hard, your foot can slip even if the pitch is only moderate.

Armed with these examples, I now return to the specific questions about rock climbing. First was the question of whether to scale a tilted slab directly or along its juncture with another, similarly tilted slab. On either route 1t you always need to keep the reaction angle below its critical value, which is set by the coefficient of friction between your shoes and the rock. The danger of slipping is greatest when you step up the slope and push backward on the trailing foot. To be safe you should choose the route with the least pitch, even when you might be able to stand stably on another route with a steeper grade. The least pitch is along the ridgelike juncture. One way to see this is to run a line straight down a slab so that it is the leg of a right triangle, the hypotenuse of which is the ridgelike juncture [see Figure 4]. The leg, being shorter than the hypotenuse, must have the steeper pitch.

When you can stand upright on a steep slope, should you bend over to use your hands? Most rock-climbing instructors teach that you should always stand upright if possible and that bending over greatly increases the chance your feet will slip. Are the instructors right? Friction must still counter the weight component that is directed down the slope, but now there is friction on both hands and feet. Would that not be safer than standing upright and depending only on the friction on the feet?

Hudson and Johnson sided with the instructors for several reasons. Your hands will encounter some friction, but the coefficient of friction between skin and rock is small, and so the upper limit on that friction is also small. If your hands slip the only force that can stop a full slide down the slope is the friction on your feet. Although there is enough friction to hold you if you stand upright, when you lean over there very well may not be enough.

To explain the result I must review the requirements of stability and introduce a new one. So far I have satisfied three requirements. The forces up and down the slope must balance; the forces perpendicular to the slope must balance; and friction must not exceed its upper limit or, equivalently, a reaction angle must not exceed its critical value. The new requirement is that the torques on a climber must balance. (A torque is the tendency of a force to make an object rotate about some pivot point.) For a climber to be stable the torques from the weight and the reaction forces must balance for every possible choice of pivot.

The requirement is easily simplified. For example, suppose the climber has leaned forward to put his or her hands on a

uniform slope [Figure 5]. The torques are guaranteed to balance if extensions of the weight vector and the reaction forces on the hands and feet pass through a common point. To picture this condition, start with the weight vector. It is vertical and passes through the climber's center of mass, which lies somewhere near the navel (the actual location depends on the configuration of the body). Mentally draw a vertical line through the center of mass.

Now consider the reaction forces. If the climber is stable extensions of those forces must meet at some point on the vertical line-a point I shall call the zero-torque point, because the torque from each force is zero if the pivot is chosen to be there. If the forces do meet at a zero-torque point, then the climber is stable against rotation for any other choice of pivot. I do not mean to imply that the climber consciously runs through a calculation to see if this requirement is met. Rather, I mean that if it is not met, the climber feels the instability and must alter the distribution of weight, the extent of leaning or the tension in muscles to reattain stability.

Is the leaning orientation shown in the illustration safer than an upright stance? Notice that for stability against torques the reaction force on the feet must tilt forward if it is to pass through a zero-torque point. Hudson and Johnson argued that the reorientation of the reaction force increases the reaction angle at the feet and that, if the slope is steep, the angle may exceed its critical value. And so to lean forward is folly, even if there is friction on the hands.

I thought about the matter further. Suppose the hands are positioned a certain distance up the slope from the feet. The sum of the friction on hands and feet must counter the weight component down the slope, but exactly how much friction is on the hands and how much is on the feet? Without knowing details about the muscular forces inside the body the question cannot be answered. There are four different variables involved in the example-two normal forces and two frictional forces-but we have only three equations in which forces and torques are balanced. Three equations are insufficient to determine four variables. If you plop your hands down on the slope, there is no practical way to predict how much friction will be on your hands or feet.

I identified still another danger in leaning forward onto a slope. Suppose the hands are placed well up the slope. If the reaction force on them is to pass through a zero-torque point, the friction on them must be directed down the slope. That in turn requires that the friction on the feet be even larger than when you stand upright, for now it must balance the weight component down the slope plus the friction on the hands. The situation seems no less dangerous if the climber leans well forward to put the hands on a second sloped ledge at about waist level. In either case the danger lessens if the hands are kept low and the lean small, but as a rule the climber should obey the instructor and avoid leaning well forward onto the rock.

I next consider the chimney climb [Figure 7], which as a youth I did routinely in various caves of West Texas. For stability in this maneuver the sum of the friction on the feet and the shoulders balances the climber's weight, and the normal forces on the feet and the shoulders match. (This time there are only three different variables, and the equations of balance can be solved.) The climber's exertion is reflected by the size of the normal force, and so during a rest the goal is to reduce that force as much as is safely possible. Hudson and Johnson stated that the normal force can be minimized if the feet are positioned at a certain distance below the shoulders and if both feet and shoulders are on the verge of sliding. As you might guess, the distance depends on the coefficients of friction at the feet and the shoulders. If either coefficient should decrease- because of wetness, say-the climber must decrease the vertical distance between feet and shoulders.

To check Hudson and Johnson's conclusion I went through the following mental exercise. First, how does the friction on the feet depend on where the feet are placed? Consider the climber's weight and the reaction forces on the feet and the shoulders [right]. Again, stability requires that extensions of the forces pass through a zero-torque point. Imagine what happens to the reaction force on the feet if they are moved and the normal force is unchanged. If the feet move downward the friction on them must increase if the reaction force is still to extend up toward a zero-torque point. If the feet move upward the converse happens. Also, because the sum of the forces must balance the climber's weight, any change in the friction at the feet must be met by an opposite change in the friction at the shoulders.

Suppose the climber places the feet at some low but stable point on the chimney and relaxes the push against the rock until the feet are on the verge of slipping. The reaction angle at the feet is then at its critical value. If the climber were to move the feet downward, the friction on the feet would increase; to keep the reaction angle from exceeding its critical value, the normal force would have to be increased. And so moving the feet down is not a good idea.

What about moving the feet upward? The friction on the feet then decreases, and the reaction angle can be kept at its critical value while the normal force is safely decreased. Moving the feet up may allow the climber to relax somewhat, but because of changes in the forces at the shoulders, the feet cannot be lifted too high. As the feet are raised and the climber relaxes, the friction at the shoulders edges toward its upper limit. The normal force and the push by the climber can be minimized when the friction at the shoulders reaches the upper limit. If the feet are lifted any higher, the normal force and push must increase.

To move up a chimney a climber must push downward on one foot, but the push requires additional friction on the foot and additional normal force if the foot is not to slip. To be safe the climber should prepare for an upward move by first raising one foot and bracing it against the wall. If the lower foot happens to slip during the ascent, support by the higher foot can prevent a fall. The same technique can make a lie-back safer: when the climber must push downward on a foot to raise the body, the other foot should be poised higher in case it must take over if the lower foot slips.

A lie-back is tiring primarily because of the great tension on the arms [see illustration left]. If the climber wants to rest and minimize the tension, should the feet be high or low? You can explore the question by balancing forces and requiring that the reaction forces and weight extend through a zero-torque point.

In their own determination of the best location for the feet, Hudson and Johnson made several simplications. (They assumed that the friction on the hands is small enough to ignore and that the friction on the feet fully supports the climber's weight. They also assumed that the reaction force on the hands is horizontal. The zero-torque point is then at the intersection of a vertical line through the center of mass and a horizontal line through the hands. If the climber raises the feet and they are still to supply the same amount of friction, what must happen to the normal force to keep the reaction force at the feet aimed at the zero-torque point?

Bibliography

ELEMENTARY ROCK CLIMBING MECHANICS. R. R. Hudson and W. Johnson in International Journal of Mechanical Engineering Education, Vol. 4, No. 4, pages 357-367; 1976.

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