Dot Physics

Parkour Physics: Wall Climb

Forgive me if I don’t know the official parkour term for this move. This is where you have two walls that are close to each other and you vertically climb them. Here is a shot of Mark Witmer (from Ninja Warrior) doing the wall climb.

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Doesn’t look too hard, does it? Well, I think it depends on how far apart the two walls are. This is actually one parkour move that my kids like to do (Hey kids! Don’t do that! Let me get my camera though because this will be perfect for my blog)

i-fae649a21b5de4580fbe3784b0c47774-wallclimb.jpg

I am going to start with this second kind of wall climb. Simply because it is easier due to symmetry. So, what is the question? (that’s a question) How about – how hard do you have to push on the walls to stay up? I will assume using only two feet as this is what happens when moving up (you move your hands and keep your feet still then switch). Here is a free-body-diagram.

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Maybe this wasn’t the best pic to use for this. Anyway, there are a couple of important points.

  • First, notice that I am drawing the forces acting ON the object (kid). I wouldn’t be surprised to see a student draw this diagram that included the force the child is pushing on the wall. That would be wrong.
  • Here I am diverting from the normal free-body-diagram where every force is acting at the same location. Actually, it won’t make a difference here (since I will assume the two normal forces and the two frictional forces are the same magnitude). However, if I were doing the other kind of wall climb, this might matter.
  • The other cool thing is that it is friction that prevents the person from falling, not the normal force. Of course, the way the friction force works, the greater the normal force, the greater the friction.

This doesn’t look too difficult. The sum of the forces in the vertical direction must be zero at equilibrium. This means:

i-a8a32674346a22ef127156a0fd76fa2b-la_te_xi_t_1_22.jpg

In the horizontal direction, the two normal forces must add up to zero, but since they are the only forces, that is easy. Using the model for friction then (assuming maximum static friction):

i-44f2c222f64d0fa8dcfe01faf546be17-la_te_xi_t_1_51.jpg

This gives an expression for how hard the wall pushes on the person in the perpendicular sense. However, the persons legs must push the opposite of the net force exerted on it (friction plus normal force) for each leg. So, how hard does the climber’s legs have to push? Each leg would have to push down and out with a magnitude of:

i-2b21b1d3e2c5514e638ff46c2438ae68-la_te_xi_t_1_64.jpg

This is the force each leg would have to exert on the wall. If the coefficient of static friction is 0.8 and the climber is just at the point where he/she is about to slip (max friction), then each leg would have to push with a force 0.8 times his/her weight. Compare this to 0.5 times the weight for each leg standing on the ground. So, it is doable.

But how can I take into account the distance between the walls? If the walls are really far apart, it is pretty tough (I imagine). If the walls are too close together, I think it is also tough – maybe even not possible (if you can’t fit). I think I can best model this by assuming that your legs can only push in the direction parallel to the line of the leg (or arm). I know this isn’t quite true, but it’s the best I have. So, if your leg makes an angle theta with respect to the horizontal, then the following must be true for the friction and the normal force: (note that this is not a free body diagram)

i-c4ca6df2563a6e506d6d744448a94e66-untitled_7.jpg

If the combo of these forces must be along that line then:

i-c6e49002cfcd2be4c567c4601a76925a-la_te_xi_t_1_42.jpg

So, if the climber does not fall, then the friction force (on one foot) must be half the weight. And if the friction and normal force are along the shown line, then:

i-0719f7aba218630ed988b46de8f08f2f-la_te_xi_t_1_71.jpg

Does this make sense? Well, if you were standing on the ground, then theta (in this case) would be pi/2. This would give a needed normal force of zero. What about the case where the legs are horizontal? This would be theta = 0 and the normal force would be infinite. Of course you can probably be completely horizontal, but this is because the human body isn’t exactly a single line. Also, the point of contact is not a point.

How about I make a graph, I like doing that. Suppose the climber is a height h with a mass m. Also, I will assume that the climber bends in the middle. This is for the case of the hands on one wall and the feet on the other. Here is a diagram:

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Now, I want theta as a function of d and h where d is the distance between the two walls. Then theta can be expressed as:

i-2af75f580f53236fb2a994aa112267a5-la_te_xi_t_1_111.jpg

But I really want the tangent of theta.

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Putting this together, I get:

i-10c67e78989e112b292ec334fb19c0fa-la_te_xi_t_1_10.jpg

So, if a climber is 1.5 meters with a mass of 70 kg then the force needed to exert as a function of distance between the walls would look like:

i-409dc4728c1b49cc449c32287c36cdcc-figure_11.jpg

I know I made some possibly not valid assumptions here, but something is ok. What happens when the walls get close together? Then the climber’s legs are almost vertical. In this case, each would have to exert just half the weight. The graph at least agrees with that.

Comments

  1. #1 Joseph Smidt
    August 22, 2009

    This website is amazing. I will be promoting it, as well as following it myself, for sure!

  2. #2 rob
    August 24, 2009

    i used to do that in doorframes when i was a kid. as i got taller, i could do it in hallways. then later i used the technique, called stemming, to help during one pitch ascending Devil’s Tower.

  3. #3 María
    August 25, 2009

    I don’t understand this part

    “So, if your leg makes an angle theta with respect to the horizontal, then the following must be true for the friction and the normal force:”

    I can’t “see it”, i don’t understand why you put the angle between the vectors that way. I’ve always seen it puting origin with origin and measuring that angle. Maybe it’s too late and my mind is switched off, tomorrow i’ll read it again. Anyway, i hope you can explain it to me.

    bye!

  4. #4 María
    August 26, 2009

    I understand everything now, you were only using another way to express the sum of vectors Ff+Fn = leg :)

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