Coefficient of static friction

  • Thread starter huybinhs
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  • #26
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m*v^2/r = N*sin(23.9) + μ*N*cos(23.9) .........(1)
mg = N*cos(23.9) - μ*N*sin(23.9)...........(2)
Divide (1) by (2) and solve for μ.
I solved for [tex]\mu[/tex] = 5.08 ???
 
  • #27
rl.bhat
Homework Helper
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v^2/rg = [sin(23.9) + μ*cos(23.9)]/[cos(23.9) - μ*sin(23.9)]
using v = 25.33 m/s and r = 77 m, find v^2/rg. Substitute this value in the above equation and find μ. What is the expected answer?
 
  • #28
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v^2/rg = [sin(23.9) + μ*cos(23.9)]/[cos(23.9) - μ*sin(23.9)]
using v = 25.33 m/s and r = 77 m, find v^2/rg. Substitute this value in the above equation and find μ. What is the expected answer?
I'm not sure whether it's correct or not. The answer is supposed to be in 0.1 - 0.9. And with your equation, the result excesses the range :(
 
  • #29
rl.bhat
Homework Helper
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I'm not sure whether it's correct or not. The answer is supposed to be in 0.1 - 0.9. And with your equation, the result excesses the range :(
I am getting 0.29
0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]
Now solve for μ.
 
  • #30
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I am getting 0.29
0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]
Now solve for μ.
Therefore, the coefficient of static friction for a car not to skid when traveling at 91.2 km/hr is 0.29 = final answer, right? (because this is my last try :(
 
  • #31
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I am getting 0.29
0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]
Now solve for μ.
You r correct. The correct answer is 0.296. Thanks so much for your help! ;)
 

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