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Forces and Kinematics Problem - Am I right?

  • Thread starter blockitoff
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  • #1
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Can someone tell me if I'm right? 0.98 seems a bit high for the coefficient of friction?

A 25kg sled is pulled at a steady 1.3 m/s by a rope inclined 40 degrees above horizontal. If the pulling force is 26.7 N, what is the coefficient of friction between the sled runners and the snow?

m = 25 kg, v = 1.3 m/s, Fp = 26.7 N, mgcos40 = Fn = 187.7 N,

Ftotal = ma (Ftotal = 0 because of no acceleration).
Fp + mgsin40 - Ff = 0,
26.7 + 157.5 = Ff, Ff = 184.2.

u = Ff/Fn = 184.2/187.7 = 0.98.
 

Answers and Replies

  • #2
Hootenanny
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Using g as 9.8 I got 0.697(3sf)
 
  • #3
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Can you please post your work? :rofl:
 
  • #4
Hootenanny
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[tex]F= \mu R
\\
\mu R + 25gcos50=26.7
\\
\mu 25gsin50=26.7 - 25gcos50
\\
\mu = \frac{26.7-25gcos50}{25gsin50}
\\
\mu = 0.697[/tex]
Sorry it took my a while to sought out the LaTex. lol :blushing: I just wanted you to know I was workin on it.
 
Last edited:
  • #5
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I don't understand that very much. what's muRmuR? like...lol
 
  • #6
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Sorry, I'm having a lot of trouble and that whole long equation just confused me more.
 
  • #7
Hootenanny
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it shouldn't come out like that. I'm jus correcting it. I Hate Latex
 
  • #8
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ok cause i dunno where you got all the numbers from
 
  • #9
Hootenanny
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[tex]F= \mu R[/tex]
[tex]\mu R + 25gcos50=26.7[/tex]
[tex]\mu 25gsin50=26.7 - 25gcos50[/tex]
[tex]\mu = \frac{26.7-25gcos50}{25gsin50}[/tex]
[tex]\mu = 0.697[/tex]
I've tried to lay it out better
 
  • #10
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R is the normal force?

by the way, thank you for helping me along so much so far :)
 
  • #11
Hootenanny
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No worries. Yeah, R is the normal reaction. mu is the co-effiecent of friction.
 
  • #12
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you used 50 degrees as the angle. it's 40 though? :\
 
  • #13
Hootenanny
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cos50=sin40.
 
  • #14
Hootenanny
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[tex]F= \mu R[/tex] is the equation for maximal frictional force.
 
  • #15
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numerically, that's correct. but i don't understand that concept at all. can you demonstrate it with 40 degrees? i apologize for being so lost :(
 
  • #16
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F = uR is, yes.
 
  • #17
Hootenanny
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Okay. Have you drawn a diagram with the slope and all the forces acting?
 
  • #18
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F(total) = ma

F(total) = 0

F(normal) = (m)(g)(cos40) = (25)(9.8)(cos40) = 187.7

F(pulling) - F(friction) + 25(9.8)(sin40) = 0

26.7 + 25(9.8)(sin40) = F(friction)

26.7 + 157.5 = F(friction)

184.2 = F(friction)

F(friction) = F(normal) * mu

mu = F(friction)/F(normal) = 184.2/187.7 = 0.98

The coefficient of friction equals 0.98.
 
  • #19
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Yes, I drew a free body diagram.
 
  • #20
Hootenanny
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blockitoff said:
F(total) = ma
F(pulling) - F(friction) + 25(9.8)(sin40) = 0
Third line down. it should read: F(pulling)-F(Friction)-25(9.8)(sin40)=0 because the component of the weight it acting in the opposite direction to the pulling force.
 
  • #21
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oh i see! let me work it out real quick :)
 
  • #22
Hootenanny
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Sorry I didnt spot it earlier!
 
  • #23
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F(total) = ma

F(total) = 0

F(normal) = (m)(g)(cos40) = (25)(9.8)(cos40) = 187.7

F(pulling) - F(friction) - 25(9.8)(sin40) = 0

26.7 - 25(9.8)(sin40) = F(friction)

26.7 - 157.5 = F(friction)

-130.8 = F(friction)

F(friction) = F(normal) * mu

mu = F(friction)/F(normal) = -130.8/187.7 = -0.697

Ok I got it! MY ONLY QUESTION that remains is how did I know to just make -0.697 positive to be 0.697?
 
  • #24
Hootenanny
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The co-efficent of fiction is always greater than or equal to zero. The negative sign just indicates that it acts in the opposite direction to the pulling force which we decided was positive. You would have got a positive answer if we took the frictional force and weight to be positive with the pulling force being negative. Always give a positive answer for co-effiecent of friction as it is dimensionless (i.e. is scalar because ithas no direction).
 
  • #25
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Right, because it can only range between 0 and 1. 0 is like slippery ice with no friction, and 1 is like extremely rough surface. But you explained it perfectly. THANK YOU!
 

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