How Does a Climber Use Friction to Stay Stationary?

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SUMMARY

The discussion focuses on calculating the minimum normal force a 75kg climber must exert to remain stationary in a chimney using friction. The static coefficients of friction are 0.8 for the shoes against the wall and 0.6 for the back against the wall. To maintain equilibrium, the frictional force must equal the gravitational force acting on the climber. The correct approach involves recognizing that both friction forces must be considered, leading to the conclusion that the minimum normal force can be derived from the equation F_f = F_g, where F_f is the total friction force and F_g is the gravitational force.

PREREQUISITES
  • Understanding of static friction coefficients
  • Knowledge of Newton's laws of motion
  • Ability to apply force equilibrium equations
  • Familiarity with basic physics concepts related to gravity
NEXT STEPS
  • Study the principles of static friction and its applications in climbing scenarios
  • Learn how to derive equations of motion for objects in equilibrium
  • Explore the effects of different friction coefficients on climbing techniques
  • Investigate real-world applications of force equilibrium in various sports
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Climbers, physics students, and educators looking to understand the mechanics of climbing and the role of friction in maintaining stability.

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Homework Statement



A 75kg climber is supported in a "chimney" by the friction forces exerted on his shoes and back. The static coefficients of friction between his shoes and the wall, and between his back and the wall, are .8 and .6, respectively. what is the minimum normal force he must exert? Assume the walls are vertical and that friction forces are both a maximum.


Homework Equations



F=ma


The Attempt at a Solution



This one confuses me. If the climber is to stay in place, its the lower static friction coefficient that matters, isn't it? That being the case, in order to stay in place acceleration must be zero so Ff + Fmg = 0? Working that out though gives .6Fn + 75*9.8 = 0, which comes out to 735, which seems too high.

I have to assume I am completely missing something, can someone point me in the right direction?
 
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in order for the climber to stay in place, the frictional force must at least be equal to the force due to gravity. If you write out all the forces (best way to start EVERY problem), you will see that F_g is aimed down, F_f is aimed upward, So, using F_f=F_g and solving for N gives the minimum normal force needed.

remember that since the climbers shoes and back are touching in different places there is more than one F_f to consider
 
Yea, that's what I thought I was doing with f_f+f_mg=0. F_f being a function of F_n * .6 (friction coefficient). F_n is equal and opposite the force of the climber's push, right? Doesn't that give you .6Fn + 75kg *9.8 = 0?
 

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