- #26

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I solved for [tex]\mu[/tex] = 5.08 ???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.

- Thread starter huybinhs
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- #26

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I solved for [tex]\mu[/tex] = 5.08 ???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.

- #27

rl.bhat

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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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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 :(

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?

- #29

rl.bhat

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I am getting 0.29I'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 :(

0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]

Now solve for μ.

- #30

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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 :(I am getting 0.29

0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]

Now solve for μ.

- #31

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You r correct. The correct answer is 0.296. Thanks so much for your help! ;)I am getting 0.29

0.85 = [ 0.405 + μ*0.914]/[0.914 - μ*0.405]

Now solve for μ.

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