Forces and Kinematics Problem - Am I right?

1. Jan 17, 2006

blockitoff

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.

2. Jan 17, 2006

Hootenanny

Staff Emeritus
Using g as 9.8 I got 0.697(3sf)

3. Jan 17, 2006

blockitoff

4. Jan 17, 2006

Hootenanny

Staff Emeritus
$$F= \mu R \\ \mu R + 25gcos50=26.7 \\ \mu 25gsin50=26.7 - 25gcos50 \\ \mu = \frac{26.7-25gcos50}{25gsin50} \\ \mu = 0.697$$
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: Jan 17, 2006
5. Jan 17, 2006

blockitoff

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

6. Jan 17, 2006

blockitoff

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

7. Jan 17, 2006

Hootenanny

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

8. Jan 17, 2006

blockitoff

ok cause i dunno where you got all the numbers from

9. Jan 17, 2006

Hootenanny

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

10. Jan 17, 2006

blockitoff

R is the normal force?

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

11. Jan 17, 2006

Hootenanny

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

12. Jan 17, 2006

blockitoff

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

13. Jan 17, 2006

Hootenanny

Staff Emeritus
cos50=sin40.

14. Jan 17, 2006

Hootenanny

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

15. Jan 17, 2006

blockitoff

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. Jan 17, 2006

blockitoff

F = uR is, yes.

17. Jan 17, 2006

Hootenanny

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

18. Jan 17, 2006

blockitoff

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. Jan 17, 2006

blockitoff

Yes, I drew a free body diagram.

20. Jan 17, 2006

Hootenanny

Staff Emeritus
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.