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Coefficient of Friction. Please Help?

  • Thread starter mparsons06
  • Start date
  • #1
61
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Homework Statement



A skier traveling 12.6 m/s reaches the foot of a steady upward 15.2° incline and glides 12.0 m up along this slope before coming to rest. What was the average coefficient of friction?

Homework Equations





The Attempt at a Solution



D = 12m
Velocity initial = 12.3 m/s
V final = 0 m/s
Angle = 15.2° incline

Use the equation Vf^2 = Vi^2 + 2*a*d
Find a

0 = 12.3^2 + 2*a*12
2*a*12 = -12.3^2

Divide both sides by 2*12
a = 6.30375 m/s^2


Force parallel = mass * g * sin θ
Force normal = mass*g* cos θ
Friction Force = µ * mass*g* cos θ

Going up an incline 2 forces decrease your velocity
Force parallel and Friction, so we add these 2 forces to find the total force decreasing your velocity. Since these forces decrease your velocity they are negative.
a = 6.30375 m/s^2


∑ Forces = mass * acceleration
(-mass * g * sin θ) + -(µ * mass*g* cos θ) = mass * 6.30375
Notice mass cancels


(- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.30375
6.30375 = -2.57 + -9.457 µ
6.30375 = -12.027 µ
µ = 0.524

But the answer is incorrect. Can someone please tell me where I messed up?
 

Answers and Replies

  • #2
430
2
By how much? Your initial velocity suddenly went from 12.6 m/s to 12.3m/s.
 
  • #3
61
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I just recalculated my mistake, and I got [tex]\mu[/tex] = 0.549.

Does that seem right to you?
 
  • #4
430
2
(- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.30375
6.30375 = -2.57 + -9.457 µ
6.30375 = -12.027 µ
µ = 0.524

But the answer is incorrect. Can someone please tell me where I messed up?[/QUOTE]

In your original calculation you have put the net acceleration +ve which isn't the case as forces are along the acceleration.
 
  • #5
61
0
So, I'm slightly confused. Do you mean it should be:

(- g * sin 15.2°) + (-µ * g* cos 15.2° ) = 6.615
6.615 = -2.57 + 9.457 µ
6.615 = 6.887 µ
µ = 0.961

I'm drawing a blank. =/
 
  • #6
430
2
No I meant the opposite. It should be
-g*sin15.2 - µ*g*cos15.2 = -6.615.
I had only quoted you.
One more thing, why are you always adding the the terms on left when there is a unknown
µ with one of the terms. Like here(this is yours):
6.615 = -2.57 + 9.457 µ
6.615 = 6.887 µ​
[/CENTER]

I noticed it in the first post too.
 
  • #7
61
0
No I meant the opposite. It should be
-g*sin15.2 - µ*g*cos15.2 = -6.615.
I had only quoted you.
One more thing, why are you always adding the the terms on left when there is a unknown
µ with one of the terms. Like here(this is yours):
6.615 = -2.57 + 9.457 µ
6.615 = 6.887 µ​
[/CENTER]
So I redid it:

-g*sin15.2 - µ*g*cos15.2 = -6.615.
-2.572 - 9.457 µ = -6.615
-9.457 µ = -4.043
µ = 0.4275

Is that correct?
 
  • #8
430
2
Yes looks good to me.
 

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