Icy slope, toboggan, find coefficient of friction

AI Thread Summary
The discussion revolves around calculating the coefficient of kinetic friction (μk) for a girl sliding down a 30° icy slope, comparing her sliding time on a toboggan versus without it. The girl slides down in time t1=τ without the toboggan and t2=½τ with it, where the friction on the toboggan is negligible. The participants derive equations for acceleration with and without friction, leading to expressions for distance and time. Ultimately, they equate the two scenarios to solve for μk, arriving at a final value of μk=0.43. The calculations emphasize that mass and gravitational acceleration do not affect the outcome, focusing solely on the angle of the slope and the friction coefficient.
marsten
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Homework Statement


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.

Homework Equations


mgsin\alpha-f_k=ma_{x}
n=mgcos\alpha
f_{k}=μ_{k}n=μ_{k}mgcos\alpha

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
 
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marsten said:

Homework Statement


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.

Homework Equations


mgsin\alpha-f_k=ma_{x}
n=mgcos\alpha
f_{k}=μ_{k}n=μ_{k}mgcos\alpha

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
That's good so far. Now that you have the time and acceleration, that might suggest using what other equations?
 
marsten said:
##mgsin\alpha-f_k=ma_{x}##

got ax=4,9 m/s2 since the kinetic friction between the toboggan and surface is negligible.
Please define ax. From the first equn above, it looks like it's the acceleration without the toboggan.
 
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.
 
dean barry said:
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.
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.
 
PhanthomJay said:
That's good so far. Now that you have the time and acceleration, that might suggest using what other equations?

Some equation to find the distance? Like:

s=v_0t+½a_xt^2=½a_xt^2
 
haruspex said:
Please define ax. From the first equn above, it looks like it's the acceleration without the toboggan.

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
 
marsten said:
Some equation to find the distance? Like:

s=v_0t+½a_xt^2=½a_xt^2
Sure, but there are two different accelerations - one with friction and one without.
Please answer my question in post #3.
 
marsten said:
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
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
haruspex said:
Sure, but there are two different accelerations - one with friction and one without.
Please answer my question in post #3.

Sorry. The acceleration that I first calculated must be the acceleration without friction and without toboggan.
 
  • #11
haruspex said:
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!

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
marsten said:
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?
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
haruspex said:
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.

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}=0,29Am I doing right? The result looks good to me.
 
  • #14
marsten said:
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)}}##
I think you lost a factor of 2 in there somewhere. The rest looks ok.
 
  • #15
haruspex said:
I think you lost a factor of 2 in there somewhere. The rest looks ok.
Ok, so it will be something 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
Yes.
marsten said:
Ok, so it will be something 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)}}
 
  • #17
haruspex said:
Yes.

The final answer will be \mu_k=0,14

Thank you very much!
 
  • #18
Nope, not the correct answer. Missed one thing. It's \mu_k=0,43

Thanks again!
 
  • #19
marsten said:
Nope, not the correct answer. Missed one thing. It's \mu_k=0,43

Thanks again!
Right, it's ##\frac 34 \tan(\alpha)##, yes?
 
  • #20
haruspex said:
Right, it's ##\frac 34 \tan(\alpha)##, yes?

Yep!
 
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