
#1
Oct807, 09:08 PM

P: 36

1. The problem statement, all variables and given/known data
If you use a horizontal force of 30.0 N to slide a 12.0kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor. 2. Relevant equations Fkf=μFn 3. The attempt at a solution Im not sure how to find the μ 



#2
Oct807, 09:28 PM

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#3
Oct807, 09:48 PM

P: 36

Humm I dont think i know how to do it still but i dont know ill try my attempt at the problem and post what i get.




#4
Oct807, 10:02 PM

P: 36

what is the coefficient of kinetic friction
Ok well i got 117.6 when i did the problem not sure if its right or not.




#5
Oct807, 10:05 PM

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#6
Oct807, 10:09 PM

P: 36

Well i have no idea how i did it or if i even did it right. But this is what i did. I kinda just copied a similar problem they did in the book and filled in the differences.
Fp= μmg fp= μ(12kg(9.8) fp117.6 



#7
Oct807, 11:08 PM

P: 210

no 117.72 N is the Normal force on the crate.




#8
Oct807, 11:11 PM

P: 36





#9
Oct807, 11:54 PM

P: 2

Think about newtons second law. The object is in constant velocity, so there is no NET force on the object. It's in equilibrium.




#10
Oct807, 11:58 PM

P: 36

Kinetic friction force Ff kinetic =μkFn Now μk wood on wood equals 0.20 so Ff kinetic=0.20(30) Ff kinetic=6 does this sound right. 



#11
Oct807, 11:59 PM

P: 2

Well.. the question says your trying to find the coefficient of kinetic friction, not the force of kinetic friction.




#12
Oct907, 12:01 AM

P: 36

Man looks like its back to the drawing boards again. Thank you all for the help and tips so far though.




#13
Oct907, 12:06 AM

P: 634

Cheers 



#14
Oct907, 12:07 AM

P: 36

Ok now i think i might have something or its just my bizarre attempt at something.
The coefficient of kinetic friction is defined as the ratio of the kinetic friction force (F) between the surfaces in contact to the normal force: Ff/N. So Ff/N which i found out earlier the friction force is 6 and the normal force is 117.72 so 6/117.72=.0509 



#15
Oct907, 12:12 AM

P: 36

then divide by 30 on both sides so 6/30=.2μk 



#16
Oct907, 12:29 AM

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#17
Oct907, 01:12 AM

P: 36

Ok now i think i might have something or its just my bizarre attempt at something. The coefficient of kinetic friction is defined as the ratio of the kinetic friction force (F) between the surfaces in contact to the normal force: Ff/N. So Ff/N which i found out earlier the friction force is 6 and the normal force is 117.72 so 6/117.72=.0509 



#18
Oct907, 01:44 AM

P: 634

sorry to tell you but frictional force is 30 N, exactly equal to the horizontal force applied (since the object is moving with constant velocity) and normal force is exactly equal to gravitational force (since object is not moving in vertical direction). Their ratio is what you need.



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