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Equilibrium problem involving wierd static friction

  1. Jan 8, 2008 #1
    This is not necessarily a homework problem, but a sample problem of my book that I do not quite understand.

    A rock climber with mass m = 55 kg rests during a "chimney climb," pressing only with her shoulders and feet against the walls of a fissure of width w = 1.0 m. Her center of mass is a horizontal distance d = 0.20 m from the wall against which her shoulders are pressed. The coefficient of static friction between her shoes and the wall is 1.1, and between her shoulders and the wall is 0.70. To rest, the climber wants to minimize her horizontal push on the walls. The minimum occurs when her feet and her shoulders are both on the verge of sliding. (a) what is the minimum horizontal push on the wall?

    In the explanation of the problem, I get confused about their explanation of static friction:

    "We want the climber to be on the verge of sliding at both her feet and her shoulders. That means we want the static frictional forces there to be at their maximum values. Those maximum values are Fs=U*N"

    The part that muddles my mind is that they say they want her to be on the verge of sliding..so they use the "maximum" value. If the person exerted more force on both walls, wouldn't they still remain in static equilibrium? I would presume that they would start to slide after exerting a force less than U*N. It seems as though the book is implying that the person will start to slide if their force exertions exceed U*N.

    Furthermore, for part B, the question asks "For that push, what must be the vertical distance h between her feet and her shoes?"

    The explanation says you have to use the fact that the net torque must equal 0, and that you should choose an appropriate axis to simplify the problem..but what exactly is the arm to be rotated? I can't see you would find the motion arm of the gravity force as its extended line is straight down, as is an extension downwards of the origin...
     
    Last edited: Jan 8, 2008
  2. jcsd
  3. Jan 8, 2008 #2

    Dick

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    I think the most confusing part is "For that push, what must be the vertical distance h between her feet and her shoes?". That is a typo, right? You mean "feet and her shoulders", right? You know that the normal forces on both walls are equal, right? And the point to the maximum of static frictional force is that the static frictional force will equal the applied force right up until the moment you start to slide. So you can replace the frictional forces with exactly N*mu_s, where N is the normal force. The sum of the frictional forces is mg. Now you just have to balance torques. A good axis is the center of mass of the climber. You can vary the distance between the feet and shoulders of the climber in such a way as to minimize the exerted normal force N. I haven't actually tried to work this out, I've been too busy scratching my head trying to figure out what's really important. And probably won't tonight. Good luck!
     
  4. Jan 9, 2008 #3

    Shooting Star

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    The "maximum" value of static friction. When a force is pushing a body, then the maximum static frictional force F=kN. Here, weight is that force which is pushing the climber down and is being supported by friction. So, any less normal force exerted by the person on the wall will reduce F and she will slide, since it is F which is holding her back. More normal force exerted by the person will definitely prevent the sliding, since F will increase, but we want her to exert just the right N which will make F just large enough to prevent slipping.

    Draw a freebody diagram. The climber is like rod resting between the two walls at some angle. Then use the standard techniques for a Statics problem, that is, the sum of the horizontal and vertical forces must be individually zero, and the moment of all the forces about any point is zero. Label the forces of friction and normal reactions properly at the points of contacts.
     
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