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Frictional Force Problem

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    A block of mass 10kg is released on a rough incline plane which is inclined at 30 degrees. Block start descending with acceleretion 2m/s^2. Kinetic friction force acting on the block is:
    A) 10. B) 30. C) 50. D) 50*3^(1/2)

    2. Relevant equations

    f= meu*N
    a=meu*g
    3. The attempt at a solution
    meu= 2/10=1/5
    N=mgcostheta
    = 50*3^(1/2)
    frictional force= 50*3^(1/2) / 5
    which is 10*3^(1/2) and is not in the options.
     
  2. jcsd
  3. Jan 27, 2016 #2

    Suraj M

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    I think you should start off by determining the forces acting on the object first
    Equate through forces, you don't even need to bring in the variable ##μ##
     
  4. Jan 27, 2016 #3
    50-friction=ma
    50-friction= 20
    f=30N
    But why doesn't my initial working give me this answer?
     
  5. Jan 27, 2016 #4

    Suraj M

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    How did you get your coefficient of friction to be 1/5?
    It's not in plane with the ground hence you can't just take 2/10 you have to take their components
     
    Last edited: Jan 27, 2016
  6. Jan 27, 2016 #5
    What do you mean by "planar with the ground"? The f=N*meu formula has conditions? I thought it could be applied to any situation.
     
  7. Jan 27, 2016 #6

    cnh1995

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    Which force will pull the block down and which force will oppose the motion? Draw FBD. As Suraj said earlier, you don't need μ.
     
  8. Jan 27, 2016 #7
    Thank you, the answer matched. However I do not completely understand why my initial method did not give me the same answer.
     
  9. Jan 27, 2016 #8

    Suraj M

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    Not conditions Priyadarshini, your normal force is ?
    You've taken 2/10
    That denominator is the normal acceleration right? But that's not the normal acceleration in this case because the plane is not parallel to the ground
    So you'll have to take a component of this.
    FBD please
    (PS- Look at us 3 Indians at this hour :-) sorry off subject- ignore this)
     
  10. Jan 27, 2016 #9
    Ohhh. I see! Thank you!
    Haha, true. The name is Priyadarshini by the way, not Priya.
     
  11. Jan 27, 2016 #10

    Suraj M

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    Sorry my bad, down here we just omit the "darshini"
    No offence, force of habit
    Hope you got your answer
     
  12. Jan 27, 2016 #11
    None taken, a lot of people do that.
    Yup, thanks a lot.
     
  13. Jan 27, 2016 #12
    mgsin(30)-friction = ma
    10*10*(1/2) - friction = 10*2

    friction = 30 N
     
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