# Icy slope, toboggan, find coefficient of friction

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1. Jun 15, 2015

### marsten

1. The problem statement, all variables and given/known data
A girl, with the mass m, slides down an icy slope at the time t1=τ. The slope is inclined at 30° below horizontal. The coefficient of kinetic friction between the girl and icy slope is μk. The girl notices that she can slide down the slope significantly faster if she sits on a toboggan. The time to slide down on a toboggan (which also has the mass m) proves to be t2=½τ. The friction between the toboggan and icy slope can be assumed to be negligible. Determine μk.

2. Relevant equations
$mgsin\alpha-f_k=ma_{x}$
$n=mgcos\alpha$
$f_{k}=μ_{k}n=μ_{k}mgcos\alpha$

3. The attempt at a solution
I did a free-body diagram of the girl on toboggan.
Then I tried to solve ax with the equations above and got ax=4,9 m/s2 since the kinetic friction between the toboggan and surface is negligible.

Not sure what to do next or if I'm doing wrong from the beginning?

Thanks for any help!

/Marcus from Sweden

2. Jun 15, 2015

### PhanthomJay

That's good so far. Now that you have the time and acceleration, that might suggest using what other equations?

3. Jun 16, 2015

### haruspex

Please define ax. From the first equn above, it looks like it's the acceleration without the toboggan.

4. Jun 16, 2015

### dean barry

Id be inclined to add a value for m ( i assume g = 9.8) and maybe set it up on an excel spreadsheet, then you can quickly detemine if altering m alters the result.

5. Jun 16, 2015

### haruspex

It's not hard to see that the mass and g are both irrelevant. One way is through dimensional analysis. We are given two times and an angle. Suppose we have m and g as well. How could you combine those algebraically into an equation in which the dimensions (M, L, T) match up? There is no term with an M component to balance with m, and no term with a length component to balance that in g. Therefore m and g cannot feature in the equation.

6. Jun 16, 2015

### marsten

Some equation to find the distance? Like:

$$s=v_0t+½a_xt^2=½a_xt^2$$

7. Jun 16, 2015

### marsten

Ahh, should I use 2m? Using the same equation gives me:

$2mgsin\alpha-\mu_k2mgcos\alpha=ma_x$

$a_x=2g(sin\alpha-\mu_kcos\alpha)$

$a_x=2*9,81*sin(30)$

$a_x=9,81 m/s^2$

8. Jun 16, 2015

### haruspex

Sure, but there are two different accelerations - one with friction and one without.
Please answer my question in post #3.

9. Jun 16, 2015

### haruspex

No, you don't know the mass of the toboggan and you don't care. If you increase the mass on the left of that equation you need to increase it on the right too. They will always cancel. The acceleration is obviously never as much as g!

10. Jun 16, 2015

### marsten

Sorry. The acceleration that I first calculated must be the acceleration without friction and without toboggan.

11. Jun 16, 2015

### marsten

So as long as I put $\mu_k=0$ I will always get the same result (4,91 m/s^2). Could this acceleration help me anywhere? Or is it not to any use for me?

12. Jun 16, 2015

### haruspex

That is useful. If you assume the distance is s then it allows you to write an expression for the time taken without friction. But please, don't put a numerical value for g, just leave it as 'g'.
Then repeat the exercise for the case where there is friction. Same s, same g, different acceleration, different time. Use different symbols for the two accelerations and times.

13. Jun 17, 2015

### marsten

The different accelerations that I have come up with so far is:
Acceleration with friction: $a_{x1}=g(sin\alpha-\mu_kcos\alpha)$
Acceleration without friction: $a_{x2}=g(sin\alpha)$

And the given times:
Time with friction: $t_1=\tau$
Time without friction: $t_2=\frac{1}{2}\tau$

And by using the equation that I mentioned in #6, $s=½a_xt^2$, these are the two times that I've come up with:

Time with friction:
$s=\frac{a_{x1}t_1^2}{2}⇔s=\frac{g(sin\alpha-\mu_kcos\alpha)\tau^2}{2}⇔2s=g(sin\alpha-\mu_kcos\alpha)\tau^2⇔\tau=\sqrt{\frac{2s}{g(sin\alpha-\mu_kcos\alpha)}}$

Time without friction:
$s=\frac{a_{x2}t_2^2}{2}⇔s=\frac{g(sin\alpha)\tau^2}{4}⇔4s=g(sin\alpha)\tau^2⇔\tau=\sqrt{\frac{4s}{g(sin\alpha)}}$

Then I put the times equal each other and solved for $\mu_k$ like this:

$$\frac{2s}{g(sin\alpha-\mu_kcos\alpha)}=\frac{4s}{g(sin\alpha)}$$
⇔​
$$2s(g(sin\alpha))=4s(g(sin\alpha-\mu_kcos\alpha))$$
⇔​
$$g(sin\alpha)=2g(sin\alpha-\mu_kcos\alpha)$$
⇔​
$$sin\alpha=2(sin\alpha-\mu_kcos\alpha)$$
⇔​
$$sin\alpha=2sin\alpha-2\mu_kcos\alpha$$
⇔​
$$2sin\alpha-sin\alpha=2\mu_kcos\alpha$$
⇔​
$$sin\alpha=2\mu_kcos\alpha$$
⇔​
$$\mu_k=\frac{sin\alpha}{2cos\alpha}$$

Then put in the values

$$\mu_k=\frac{sin30}{2cos30}=0,29$$

Am I doing right? The result looks good to me.

14. Jun 17, 2015

### haruspex

I think you lost a factor of 2 in there somewhere. The rest looks ok.

15. Jun 17, 2015

### marsten

Ok, so it will be someting like this?

$s=\frac{a_{x2}t_2^2}{2}=\frac{g(sin\alpha)\tau^2}{2\cdot2^2}=\frac{g(sin\alpha)\tau^2}{8}$

$\tau=\sqrt{\frac{8s}{g(sin\alpha)}}$

16. Jun 17, 2015

### haruspex

Yes.

17. Jun 17, 2015

### marsten

The final answer will be $\mu_k=0,14$

Thank you very much!!

18. Jun 17, 2015

### marsten

Nope, not the correct answer. Missed one thing. It's $\mu_k=0,43$

Thanks again!

19. Jun 17, 2015

### haruspex

Right, it's $\frac 34 \tan(\alpha)$, yes?

20. Jun 17, 2015

Yep!