Recent content by blockitoff

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!

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) =...

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

Yes, I drew a free body diagram.

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) =...

F = uR is, yes.

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 :(

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

R is the normal force? by the way, thank you for helping me along so much so far :)

ok cause i don't know where you got all the numbers from

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

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

Can you please post your work? :smile:

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...