Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Coeffecient of static friction problem

  1. May 31, 2008 #1
    1. The problem statement, all variables and given/known data

    I am given a right triangle with a block on it. At angle 18degrees the block does not move. At angle 19degrees the block starts sliding down the ramp. I am given no other information but I am told to find μS.

    The forces acting on the block are going to be Fn opposite the angle the block is sitting on, mgsin(theta) pointed downwards, F->, and the static friction pointed <- (Opposite the force).

    2. Relevant equations

    3. The attempt at a solution

    When it starts moving fS is going to be at its maximum correct? And fs = μS*Fn. Can I get a number for this considering I am only given the angle? Since the block is moving it has already broken through the maximum static friction force so does μS become 0 and the frictional force all becomes kinetic?
    Last edited: May 31, 2008
  2. jcsd
  3. May 31, 2008 #2
    Haha I asked this exact question, well okay it wasn't exact, but I asked a similar question yesterday.

    I dont know if this will help, but I learnt:

    Fn = m*g

    μS mg cos (angle, so thats 19) = mg sin 19


    μS= sin19/cos19 = tan 19

    Do you have a correct answer to match it with? That's what I was told and it seems to be working but as it's for my assignment I have no answer to check if it's right. I hope this might have helped. :smile:
  4. May 31, 2008 #3
    I don't actually I have three tries to see if its right then it submits :P I'll give this a try and see if it works out.
  5. May 31, 2008 #4
    That didn't work out unfortunately :(
  6. May 31, 2008 #5
    Draw a free-body diagram in terms of x and y components.

    We know that [tex]\Sigma F_{y}=ma_{y}=F_{normal_force}-mgcos(\theta)[/tex]

    Therefore, [tex]F_n=mgcos(\theta)[/tex]

    Since we know that, [tex]\Sigma F_{x} = ma_{x} = 0 [/tex], then

    [tex]f_{static}=mgsin(\theta_{max}) [/tex] and

    Now, solve for [tex]\mu_{static} [/tex]. You can see that it is independent of the mass of the the object.
    Last edited: May 31, 2008
  7. May 31, 2008 #6
    ustatic = mgsin(18)/mgcos(19)? But can I solve this even if I don't have m?
  8. May 31, 2008 #7
    No. Think about it. What is the [tex]\theta_{max}[/tex] that prevents it from falling? The thetas must be equal.
  9. May 31, 2008 #8
    So its just gsin(19)/gcos(19)
  10. May 31, 2008 #9
    yeah that free-body diagram is crucial. make sure the x-axis is parallel to the slope of the triangle. i would use 18 degrees for your theta max, not 19, because the balance is not valid for 19 degrees.
  11. May 31, 2008 #10
    [tex]\theta_{max}[/tex] is the angle just as it starts to slide. *Hint: look at your given information. =P
  12. May 31, 2008 #11
    oh sorry missed a few posts there.
  13. May 31, 2008 #12
    ic thanks! I end up with .32492 then. Which was correct!
    Last edited: May 31, 2008
  14. May 31, 2008 #13
    Wow. This is really interesting. Sorry AnkhUNC, that I couldn't help.

    But now I think all my answers are wrong. I posted a thread yesterday called Static Coefficient HELP or something like that, I'm sure its at the bottom of the Introductory Physics section at the moment... but now I'm really lost as to what's the right answer... I jut posted a new thread too, can anyone help? I'm a little confused.

    Wow I just did that gsin(19) thing... can I used that equation for my angle as well? Is g 9.8N?

    Oh no.. :confused:
  15. May 31, 2008 #14
    No, g is the acceleration due to gravity. [tex]g \approx 9.8m/s^2[/tex]
  16. May 31, 2008 #15
    Oh of course, right. :smile: But okay it is 9.8... Hmmm

    This is what I found out yesterday, with mine and that's my angle, (I have 7 different angles but thats the first) and that's what I did to find μS, but I actually need μK, does it mean because my angle is an angle where my sinker is already rolling, not just starting to, that the same equation will give me the μK?

    Fn = m*g

    μS mg cos 55.9= mg sin 55.9


    μS= sin55.9/cos19 = tan 55.9

    = 1.4769

    Then I put that into the F= μ*N

    And got 173.68

    Is the equation right?
    Last edited: May 31, 2008
  17. May 31, 2008 #16
    Like I've said before, [tex]\mu_{static} = \frac{mgsin(\theta)}{mgcos(\theta)}=tan(\theta)[/tex]
    The thetas must be equal.

    Also,[tex]\mu_{static}<1[/tex] in this scenario.
  18. May 31, 2008 #17
    Okay, hmm, both thetas are 55.9, and both m's are 0.012, are both g's 9.8? Of course, they have to be. Hmm.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook