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Coefficient of friction on an inclined plane

  • Thread starter Coco12
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  • #1
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Homework Statement



a block with a mass of 0.310kg at a constant velocity , calculate the coefficient of kinetic friction on an incline plane

Homework Equations



f=ma
fg=mg
fparallel= mg* sin30 degrees


The Attempt at a Solution


I found the FN which is 2.60 and the f parallel to be 1.52. i found force applied to be 2n as the block moved up the incline and the as the block moved downwards to be 1N which is a 1/2 of that. I understand that you have to calculate net force going up and down the incline then since they will be equal to one another, try to isolate μ.
which will give you something like μ= x tan 30 degree. however how do i find out what the x is?
Is the force applied = to the f parallel
This is what I did:

F1 (block moving up) = 2F2 (moving downward)
fnet (for box going upward): fparallel + F1 + Ff(mu*Fn)
Fnet (for box going downward) : F2- Ff+ Fparallel
Then i made them equal to each other since they have a constant velocity, then i subtracted one from the other to get f parallel (m*g*sin31) = mu* mg Cos31 ,
then by dividing the fparallel by the mu*m g cos31, i got
mu= tan 31 degrees..


Am i doing this right?
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
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...try to isolate μ. which will give you something like μ= x tan 30 degree. however how do i find out what the x is?
It will be obvious from the way you substituted the expressions for the forces into ƩF=ma. The final expression need not be in that form. Work through the derivation first, see where that leads you.

Note: 30deg is a very nice angle.
sin(30)=1/2
cos(30)=(√3)/2
tan(30)=1/√3
... use the RHS versions without converting to decimal.

Your problem statement does not give the angle of the incline however.

F1 (block moving up) = 2F2 (moving downward)
fnet (for box going upward): fparallel + F1 + Ff(mu*Fn)
Fnet (for box going downward) : F2- Ff+ Fparallel
Then i made them equal to each other since they have a constant velocity, then i subtracted one from the other to get f parallel (m*g*sin31) = mu* mg Cos31 ,
then by dividing the fparallel by the mu*m g cos31, i got
mu= tan 31 degrees..
... where did you get 31deg from?
Apart from that you appear to have answered your own question (above).
In your case it turned out that x=1.
 
  • #3
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Sorry I forgot to say that the angle given is 31 degrees
 
  • #4
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... where did you get 31deg from?
Apart from that you appear to have answered your own question (above).
In your case it turned out that x=1.[/QUOTE]

So apart from that did I do it correctly?
 
  • #5
Simon Bridge
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Do you have any reason not to be confident about it?
How could you check it?
 
  • #6
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Thanks I think I got it
 

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