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Finding coeffecient of friction

  1. May 23, 2013 #1

    462chevelle

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    1. The problem statement, all variables and given/known data

    A 3 kg block slides down a 30 degree inclined plane with constant acceleration of 0.5 m/s2. The block starts from rest at the top. The length of the incline is 2 m.


    2. Relevant equations
    Vf^2=Vi^2 +2ad velocity is 1.414
    Fg=3 x 9.8=29.4
    Fn= cos(30) (9.8 x 3) = 25.5
    f//=sin (30) (9.8 x 3) =14.7
    f=ma = 4.242=3 x 1.414
    Ff= F//-f = 10.458=14.7-4.242
    Ff=mu x (Fg) = 10.458= mu x 29.4 = .356



    3. The attempt at a solution
    Ff=mu x (Fg) = 10.458= mu x 29.4 = .356
    this is incorrect. I feel like im missing a simple step. I just cant figure out where im messing up. any direction here? thanks
     
  2. jcsd
  3. May 23, 2013 #2
    I am having a hard time following your equations.
    These are correct and applicable:
    Fg=3 x 9.8=29.4
    Fn= cos(30) (9.8 x 3) = 25.5
    f//=sin (30) (9.8 x 3) =14.7

    Ff = u X Normal Force = u(Fn)

    Set up an equation relating Ff, F//, m, and a and then solve for u
     
  4. May 23, 2013 #3

    SammyS

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    Hello 462chevelle. Welcome to PF !

    From what you have given it's difficult to know what it is that you are trying find out.

    Please state the complete problem, word for word as it was given to you.
     
  5. May 24, 2013 #4

    462chevelle

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    mass = 3
    acceleration = .5 m/s
    velocity = 1.414
    Fg = 29.4
    Fn = 25.5
    F// = 14.7
    Ff= 10.458
    and what I got for mu
    10.458 = mu x 29.4
    solve it and get .356 which is the wrong answer.
    and the question I posted is the question. the only thing not stated is that im trying to find the coefficient of friction.
    Sammy s . thanks. most of the time when im having trouble with a physics problem. if I google it. it always pops this forum up for good explanations.
     
  6. May 24, 2013 #5
    Ff = mu X Fn
    mu is what you are trying to find. You cannot find mu until you find the net force on the mass.
    You must use net force = ma to find mu
    I would like to give you the complete solution but this is prohibited.
     
  7. May 24, 2013 #6

    462chevelle

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    oh. I though it was Ff = mu x Fg so if I need the Fn that's 25.5.
    so I have 10.458 = mu x 25.5 I get .41 but that is still wrong. am I doing anything wrong that's obvious or skipping a step?
     
  8. May 24, 2013 #7
    Not quite. You are given the mass and acceleration. What is the net force on the mass? The net force is the component of the weight down the incline minus the friction force. Have you drawn a diagram of this situation? A diagram often helps.
     
  9. May 24, 2013 #8

    462chevelle

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    would that be the F// minus the force of friction. so that is 4.242, and yes I have drawn out a diagram. that helps a lot in understanding whats going on
     
  10. May 24, 2013 #9
    I think yes. The net force is F// - Ff. Where did you get 4.242?
    You can get the net force from the given mass and acceleration.
     
  11. May 24, 2013 #10

    462chevelle

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    maybe my force of friction is incorrect. cause if you do f = ma you get 1.5 = 3 x .5
     
  12. May 24, 2013 #11

    462chevelle

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    so I could take 1.5 from 14.7 and get 13.2 as force of friction
     
  13. May 24, 2013 #12
    F// - Ff = ma

    You said that F// = 14.7 and that ma = 3 X .5
    what is Ff?

    Ff = (mu)Fn
    and you said that Fn = 25.5
     
  14. May 24, 2013 #13
    so I could take 1.5 from 14.7 and get 13.2 as force of friction . YES

    now, use Ff = (mu) X Fn to get mui
     
  15. May 24, 2013 #14

    462chevelle

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    that would be .51 but that is incorrect to. on my lunch break im going to get a new page and start over to see if im just writing something down wrong or skipping something
     
  16. May 24, 2013 #15
    Check your precision. .51 might be slightly off.
     
  17. May 24, 2013 #16

    462chevelle

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    FINALLY its .52. when should I decide which to round to when im doing an equation like this?
     
  18. May 24, 2013 #17

    462chevelle

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    I appreciate the help
     
  19. May 24, 2013 #18
    When you do the calculations, keep the numbers in your calculator if you can to have the highest precision possible. In this problem, since the mass was 3 kg one might think that only one digit of precision is necessary but obviously this is not the case. So, keep as many digits of precision as you can. I assume you are entering this answer into Quest or some other online program, yes? These programs have a small tolerance on the answers but I guess .51 is not close enough.
     
  20. May 24, 2013 #19

    462chevelle

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    yes its thinkquest. ive found that out that sometimes I have to redo the problem a few times rounding to different decimals to get it right.
     
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