What is the Smallest Coefficient of Friction Between Crutches and Ground?

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SUMMARY

The smallest coefficient of friction between crutches and the ground is determined to be 0.153, based on the analysis of forces acting on the crutches supporting a 170 lbs individual. Each crutch supports 42.5 lbs, which converts to 189 N, while the normal force calculated is 202 N. The calculations utilize the equation fk = uk(n) to derive the coefficient of friction and confirm the compression force in each crutch as 202 N.

PREREQUISITES
  • Understanding of free body diagrams
  • Knowledge of basic physics equations, particularly F=ma
  • Familiarity with trigonometric functions for force resolution
  • Concept of coefficient of friction in physics
NEXT STEPS
  • Study the principles of static and kinetic friction
  • Learn about force resolution in two dimensions
  • Explore applications of free body diagrams in real-world scenarios
  • Investigate the effects of angle on force distribution in physics
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Students in physics, engineers analyzing support structures, and anyone interested in biomechanics related to mobility aids like crutches.

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Homework Statement


The person in This picture weights 170lbs. A seen from the front, each light crutch
makes and angle of 22degrees with the vertical. Half of the persons weight is
supported by the crutches. The other half is sipported by the vertical forces
of the ground on the person's feet. Assuming that the person is moving with
a constant velocity and the force exerted by the ground on the crutches acts
along the crutches, determine (a) smallest possible coefficient of friction
between crutches and ground and (b) the magnitude of the compression force
in each crutch


Homework Equations


F=ma


The Attempt at a Solution


I have first drew a free body diagram for one side and I get my
forces to be this.

x forces are -fk
y forces are -mg, n

I determined that each crutch has 42.5 lbs of force
I converted the 42.5 lbs to 189n

I solved the n force to be 202n by drawing a triangle and then using sin22(189)
= 70.8n. Then getting 202n using pyth. theraom.
Am I starting this correctly?
Thanks,
Kevin
 

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Which force components do you plan to use for determining the coefficient of friction?
 
I used fk=uk(n)
70.8n = uk(461.8n)
uk= horizontal force/total normal force
I get uk to be .153 for part a

part b I get the compression force to be 202n from the findings before.

This seems to be correct.
Thanks,
Kevin
 

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