Question regarding coefficient of static friction

  • #1
Hi all, I encounter an un-explainable question during my lab, so I figured it would be a good shot asking here. Also, my English isn't perfect, so if there is anything unclear, say it out and I'm happy to clarify it.

In the lab, we placed a sled on a track and slowly lifted one end of the track up by arms until the sled started to slide. Once it showed any sign of moving, we stopped lifting and record the height. The trial was repeated multiple times with the same sled, same track but with more and more mass on the sled. Then we used the resulted height and the mass to calculate the coefficient of static friction.
In the calculation, we simplified the equation of finding the coefficient of static friction down to :
μ = tan(θ) , at which θ was obtained by arcsin(height of the lifted end / length of the track)
This equation is generally approved online and by teachers, and we also derived ourselves to that, so we believe the equation is correct.
When we substituted the numbers into θ, we get different outcome, since we got different height.
This is where I am confused.
By theory, the coefficient of static friction does not change unless the material or the pressure changes, which none happened in our case, meaning the it should not change.
By my logic, larger mass led to larger normal force. Larger normal force means larger friction. Larger friction means a need of larger inclination for the sled to slide. However, larger inclination will, using the equation, cause the coefficient to vary.

My only suspicion is a possible flaw in the logic, which states "a need of larger inclination for the sled to slide." If the mass of the sled was increased, the weight also increased. When the weight increases, under the same angle, the sliding force, the force that causes the sled to slide down because of gravity, of the sled downward increases. So it is possible for all the trials to have the same angle, if the measuring process is precise enough.

I'm not that sure about my own suspicion, but it seems like the only valid explanation for this problem. Can anyone verify that or explain the problem here? Thanks for any help provided.
 
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Answers and Replies

  • #2
CWatters
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By my logic, larger mass led to larger normal force. Larger normal force means larger friction. Larger friction means a need of larger inclination for the sled to slide.

No, the larger mass does increases the normal force and friction BUT it also increases the force acting down the slope. These two effects cancel out which is why the mass does not end up in the equation.


Perhaps see..
 
  • #3
berkeman
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μ = tan(θ) , at which θ was obtained by arcsin(height of the lifted end / length of the track)

You probably know this, but just to clarify -- it is the height of the start of the track divided by the horizontal distance, not the hypotenuse length of the track. :smile:
 
  • #4
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I have trouble understanding what you are describing. Did you get about the same angle for all the masses that you used in the experiments or not?

Chet
 
  • #5
No, the larger mass does increases the normal force and friction BUT it also increases the force acting down the slope. These two effects cancel out which is why the mass does not end up in the equation.

So my suspicion was on point. Thanks for the clarification, really appreciate it. :smile:
 
  • #6
I have trouble understanding what you are describing. Did you get about the same angle for all the masses that you used in the experiments or not?

Sorry for making it unclear. I did not get the same angle, and that was why I was confused because it meant the coefficient of static friction was changing. At the time I see your comment, CWatters already answered my question, but I appreciate your intention to help. :smile:
 
  • #7
You probably know this, but just to clarify -- it is the height of the start of the track divided by the horizontal distance, not the hypotenuse length of the track. :smile:

Yes I'm fully aware of that. Thanks for your help. :smile:
 
  • #8
CWatters
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You should get roughly the same angle how much did it vary?
 
  • #9
You should get roughly the same angle how much did it vary?
The angles, all in radians, I got were:
0.2544, 0.2424, 0.1954, 0.1886, 0.1776 and 0.1641
the materials were all the same except of the mass of each.
 
  • #10
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The angles, all in radians, I got were:
0.2544, 0.2424, 0.1954, 0.1886, 0.1776 and 0.1641
the materials were all the same except of the mass of each.
What was the mass that went along with each of these?

Chet
 
  • #11
sophiecentaur
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What was the mass that went along with each of these?

Chet
And what were the materials? Were they 'ordinary' materials and were the surfaces clean etc? The friction law relies on the materials following Hooke's Law (at a microscopic level), which keeps the actual contact area proportional to the normal force.
 

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