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Coeffecient of static friction problem

  • Thread starter AnkhUNC
  • Start date
  • #1
91
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Homework Statement



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).

Homework Equations





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?
 
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Answers and Replies

  • #2
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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

(Rearrange)

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:
 
  • #3
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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.
 
  • #4
91
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That didn't work out unfortunately :(
 
  • #5
238
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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
[tex]mgsin(\theta_{max})=\mu_{static}mgcos(\theta)[/tex]

Now, solve for [tex]\mu_{static} [/tex]. You can see that it is independent of the mass of the the object.
 
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  • #6
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ustatic = mgsin(18)/mgcos(19)? But can I solve this even if I don't have m?
 
  • #7
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ustatic = mgsin(18)/mgcos(19)? But can I solve this even if I don't have m?
No. Think about it. What is the [tex]\theta_{max}[/tex] that prevents it from falling? The thetas must be equal.
 
  • #8
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So its just gsin(19)/gcos(19)
 
  • #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.
 
  • #10
238
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So its just gsin(19)/gcos(19)
[tex]\theta_{max}[/tex] is the angle just as it starts to slide. *Hint: look at your given information. =P
 
  • #11
oh sorry missed a few posts there.
 
  • #12
91
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ic thanks! I end up with .32492 then. Which was correct!
 
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  • #13
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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:
 
  • #14
238
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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:
No, g is the acceleration due to gravity. [tex]g \approx 9.8m/s^2[/tex]
 
  • #15
19
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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

(Rearrange)

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?
 
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  • #16
238
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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.
 
  • #17
19
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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.
 

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