Finding the Coefficient of Friction(due tommorow, )

  • Thread starter Thread starter DarkOtaku
  • Start date Start date
  • Tags Tags
    Coefficient
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 2K views
DarkOtaku
Messages
1
Reaction score
0
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
 
Physics news on Phys.org
which coefficient you need --- kinetic or static?

for kinetic use 327N
for static use 365N
 
And there's no units on the coefficient. Divide Newtons by Newtons and you get a pure number.