# Coeffecient of static friction problem

AnkhUNC

## 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).

## 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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Razza
Haha I asked this exact question, well okay it wasn't exact, but I asked a similar question yesterday.

I don't know if this will help, but I learnt:

Fn = m*g

μS mg cos (angle, so that's 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. AnkhUNC
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.

AnkhUNC
That didn't work out unfortunately :(

konthelion
Draw a free-body diagram in terms of x and y components.

We know that $$\Sigma F_{y}=ma_{y}=F_{normal_force}-mgcos(\theta)$$

Therefore, $$F_n=mgcos(\theta)$$

Since we know that, $$\Sigma F_{x} = ma_{x} = 0$$, then

$$f_{static}=mgsin(\theta_{max})$$ and
$$mgsin(\theta_{max})=\mu_{static}mgcos(\theta)$$

Now, solve for $$\mu_{static}$$. You can see that it is independent of the mass of the the object.

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AnkhUNC
ustatic = mgsin(18)/mgcos(19)? But can I solve this even if I don't have m?

konthelion
ustatic = mgsin(18)/mgcos(19)? But can I solve this even if I don't have m?
No. Think about it. What is the $$\theta_{max}$$ that prevents it from falling? The thetas must be equal.

AnkhUNC
So its just gsin(19)/gcos(19)

genghiskron
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.

konthelion
So its just gsin(19)/gcos(19)
$$\theta_{max}$$ is the angle just as it starts to slide. *Hint: look at your given information. =P

genghiskron
oh sorry missed a few posts there.

AnkhUNC
ic thanks! I end up with .32492 then. Which was correct!

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Razza
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.. konthelion
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.. No, g is the acceleration due to gravity. $$g \approx 9.8m/s^2$$

Razza
Oh of course, right. 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 that's 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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konthelion
Like I've said before, $$\mu_{static} = \frac{mgsin(\theta)}{mgcos(\theta)}=tan(\theta)$$
The thetas must be equal.

Also,$$\mu_{static}<1$$ in this scenario.

Razza
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.