# Forces and Kinematics Problem - Am I right?

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.

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Hootenanny
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Using g as 9.8 I got 0.697(3sf)

Hootenanny
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$$F= \mu R \\ \mu R + 25gcos50=26.7 \\ \mu 25gsin50=26.7 - 25gcos50 \\ \mu = \frac{26.7-25gcos50}{25gsin50} \\ \mu = 0.697$$
Sorry it took my a while to sought out the LaTex. lol I just wanted you to know I was workin on it.

Last edited:
I don't understand that very much. what's muRmuR? like...lol

Sorry, I'm having a lot of trouble and that whole long equation just confused me more.

Hootenanny
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it shouldn't come out like that. I'm jus correcting it. I Hate Latex

ok cause i dunno where you got all the numbers from

Hootenanny
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$$F= \mu R$$
$$\mu R + 25gcos50=26.7$$
$$\mu 25gsin50=26.7 - 25gcos50$$
$$\mu = \frac{26.7-25gcos50}{25gsin50}$$
$$\mu = 0.697$$
I've tried to lay it out better

R is the normal force?

by the way, thank you for helping me along so much so far :)

Hootenanny
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No worries. Yeah, R is the normal reaction. mu is the co-effiecent of friction.

you used 50 degrees as the angle. it's 40 though? :\

Hootenanny
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cos50=sin40.

Hootenanny
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$$F= \mu R$$ is the equation for maximal frictional force.

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 :(

F = uR is, yes.

Hootenanny
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Okay. Have you drawn a diagram with the slope and all the forces acting?

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.

Yes, I drew a free body diagram.

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.

oh i see! let me work it out real quick :)

Hootenanny
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Sorry I didnt spot it earlier!

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?

Hootenanny
Staff Emeritus