Finding the Coefficient of Friction(due tommorow, need help)

  • Thread starter DarkOtaku
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  • #1
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A 50.0 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the chair is in motion a 327 N horizontal force keeps it moving at a constant velocity. Find the coefficient of friction between the chair and the floor. (In this problem use the "327 N" force, but just remember, because of static friction, it always takes a little bit greater of a force to "Get" an object moving.)




  • FF=[tex]\mu[/tex]FN
  • [tex]\Sigma[/tex]Fv=FN+(Fg)=ma
    FN=Fg=mg
  • [tex]\Sigma[/tex]Fh=Fpush+(-FF)=ma
    [tex]\Sigma[/tex]Fh=FF=ma

  • FF=[tex]\mu[/tex]mg



FF=[tex]\mu[/tex]FN
[tex]\mu[/tex]=FF[tex]/[/tex]FN
[tex]\mu[/tex]=327 N[tex]/[/tex]365 N=0.896 N
 

Answers and Replies

  • #2
1,137
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which coefficient you need --- kinetic or static?

for kinetic use 327N
for static use 365N
 
  • #3
gneill
Mentor
20,925
2,866
And there's no units on the coefficient. Divide Newtons by Newtons and you get a pure number.
 

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