Coefficient of Kinetic Friction on Inclined Plane? [FORCES]

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SUMMARY

The discussion centers on determining the coefficient of kinetic friction (μk) on an inclined plane using a block and cart setup. The key equations involved are ƩF = ma and μk = Fk / Fn, where Fn is the normal force calculated as mgcosθ and the downhill force as mgsinθ. Participants emphasize the necessity of measuring acceleration to accurately derive μk, as simply knowing the mass and angle is insufficient. The consensus is that an experimental approach is required, where the angle is adjusted to achieve constant velocity, allowing for the calculation of μk using the relationship μk = tanθ.

PREREQUISITES
  • Understanding of Newton's Second Law (ƩF = ma)
  • Knowledge of forces acting on an inclined plane (normal force and gravitational force components)
  • Familiarity with the concept of kinetic friction and its relationship to normal force
  • Ability to perform basic trigonometric calculations (sine and cosine functions)
NEXT STEPS
  • Conduct experiments to measure acceleration on an inclined plane using a protractor and a block and cart setup.
  • Explore the relationship between static and kinetic friction in detail, focusing on how to transition between the two states.
  • Learn about the practical considerations of measuring forces and angles in physics experiments to ensure accuracy.
  • Study the derivation of the coefficient of kinetic friction using different experimental setups and conditions.
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone interested in experimental physics related to friction and inclined planes.

harujina
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Homework Statement



There's no specific problem statement but I have to explain how I would find out the coefficient of kinetic friction given the angle (θ) of the inclined plane and mass of the object sliding down, nothing else.

Homework Equations



ƩF = ma
μk = Fk (kinetic friction) / Fn (normal force)

The Attempt at a Solution



I split the Fg (Fg = mg; weight) vector into x- and y- components so I could determine Fn (normal force).
As a result, Fn = mgcosθ.
And the force that moves the object downhill (I don't know what to call this) = mgsinθ.

Now, to determine μk I know I must find Fk (kinetic friction) first. Using μk = Fk/Fn, Fk = μk*Fn which = μkmgcosθ.

So, ƩF = (force that moves object downhill) - Fk, ƩF = mgsinθ - μkmgcosθ ; mg's cancel each other.
Therefore ƩF = sinθ - μkcosθ, and ƩF = ma, so... ma = sinθ - μkcosθ.
I then isolated μk (*I don't know if I did this correctly)
* μk = ma - sinθ / -cosθ ?

After doing all of this, I found out that I have all the variables except for ACCELERATION...
and a = ƩF / m which I can not solve for!
Can I just say that a = 0? I have no idea what to do.
 
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harujina said:
So, ƩF = (force that moves object downhill) - Fk, ƩF = mgsinθ - μkmgcosθ ; mg's cancel each other.
The mg's don't cancel. (But mass will when you apply Newton's 2nd law.)

After doing all of this, I found out that I have all the variables except for ACCELERATION...
You need something additional, measured or given. Just knowing the mass and the angle of the board tells you very little. (I can readily swap boards so that it has a different μ, right?)

Unless there's something special about the angle, you haven't enough information to solve for μ.

Perhaps this is an experimental question and you can measure the acceleration?
 
Doc Al said:
The mg's don't cancel. (But mass will when you apply Newton's 2nd law.)


You need something additional, measured or given. Just knowing the mass and the angle of the board tells you very little. (I can readily swap boards so that it has a different μ, right?)

Unless there's something special about the angle, you haven't enough information to solve for μ.

Perhaps this is an experimental question and you can measure the acceleration?

Oh, right...
Yes, it is an experimental question but it says the only objects I have in hand is the track, object sliding down the track, and a protractor (that measures the angle).
I'm still having trouble understanding... a = ƩF/m and ƩF = gsinθ - μkgcosθ
Aren't I still missing two variables?
 
harujina said:
Oh, right...
Yes, it is an experimental question but it says the only objects I have in hand is the track, object sliding down the track, and a protractor (that measures the angle).
I'm still having trouble understanding... a = ƩF/m and ƩF = gsinθ - μkgcosθ
Aren't I still missing two variables?
It's easily proved you need more information. You could put two objects the same mass but very different surfaces on the same incline. You need information about what happens when you do this. And as Doc Al says, the mass isn't useful anyway. Have you quoted the problem exactly as it was given to you?
 
haruspex said:
It's easily proved you need more information. You could put two objects the same mass but very different surfaces on the same incline. You need information about what happens when you do this. And as Doc Al says, the mass isn't useful anyway. Have you quoted the problem exactly as it was given to you?

"You find a block and cart, a track, and a protractor attached to the track. You find nothing else, no stopwatch, string, spring-scale or masses for hanging. The protractor shows the inclination of the track and was used for another experiment where steepness of the slope was found. Describe how you could measure the coefficient of kinetic friction using the items you found, in detail and with a formula."

This is the full question.
Oh, and the block is beneath the friction-less(?) cart, which is why there is friction.
The mass of the block+cart is 0.734 kg.
 
Last edited:
harujina said:
"You find a block and cart, a track, and a protractor attached to the track. You find nothing else, no stopwatch, string, spring-scale or masses for hanging. The protractor shows the inclination of the track and was used for another experiment where steepness of the slope was found. Describe how you could measure the coefficient of kinetic friction using the items you found, in detail and with a formula."

This is the full question.
Oh, and the block is beneath the friction-less(?) cart, which is why there is friction.
The mass of the block+cart is 0.734 kg.
There is a crucial difference from what you originally posted. It says "Describe how you could measure". That is, you are allowed to conduct an experiment, but using the given equipment only.
 
haruspex said:
There is a crucial difference from what you originally posted. It says "Describe how you could measure". That is, you are allowed to conduct an experiment, but using the given equipment only.

I'm still a bit confused. So could I possible say that I adjusted the track where the cart+block moved at constant velocity (∴ no acceleration)?
 
harujina said:
I'm still a bit confused. So could I possible say that I adjusted the track where the cart+block moved at constant velocity (∴ no acceleration)?
Yes indeed. You should supply a little more detail on how you would do that (think about static friction here), what you would measure, and how you would deduce the coefficient of kinetic friction.
 
haruspex said:
Yes indeed. You should supply a little more detail on how you would do that (think about static friction here), what you would measure, and how you would deduce the coefficient of kinetic friction.

Static friction? When the amount of force surpasses the amount of static friction, that's when it starts moving right? But where and why would I need to include that?

If I just say that I would adjust the track, and then repeat the stuff above, is that applicable?
 
  • #10
harujina said:
Static friction? When the amount of force surpasses the amount of static friction, that's when it starts moving right? But where and why would I need to include that?
You want kinetic friction, which is generally somewhat less than static friction. If you adjust the angle until the block moves, what will happen subsequently?
 
  • #11
haruspex said:
You want kinetic friction, which is generally somewhat less than static friction. If you adjust the angle until the block moves, what will happen subsequently?

Um, the frictional resistance decreases? and force is applied to move it downhill?
 
  • #12
harujina said:
Um, the frictional resistance decreases? and force is applied to move it downhill?
Yes, the resistance decreases, but you have raised the angle far enough to overcome the static friction. So what does the FBD tell you about how it will move once it has started?
 
  • #13
haruspex said:
Yes, the resistance decreases, but you have raised the angle far enough to overcome the static friction. So what does the FBD tell you about how it will move once it has started?

It will move with constant velocity and frictional force will be less (shorter arrow) than the downhill(?) force? I don't know if I'm fully understanding the question, sorry.
 
  • #14
harujina said:
It will move with constant velocity and frictional force will be less (shorter arrow) than the downhill(?) force?
The second part of that is correct. You adjusted the angle high enough to overcome static friction, but immediately the frictional force will reduce to be only kinetic friction. Thus the downhill force will now exceed the frictional force by some margin. So now think again about the first half of your answer.
 
  • #15
haruspex said:
The second part of that is correct. You adjusted the angle high enough to overcome static friction, but immediately the frictional force will reduce to be only kinetic friction. Thus the downhill force will now exceed the frictional force by some margin. So now think again about the first half of your answer.

Oh, its initial velocity will be 0 and then it will move at a constant velocity?
Because it wouldn't be accelerating and it wouldn't have been moving from its initial position, that's the whole idea, right?
 
Last edited:
  • #16
harujina said:
Because it wouldn't be accelerating
Why not? You set the angle so that the downslope component of gravity matched/overcame static friction: ## mg \sin(\theta_s) = \mu_s mg\cos(\theta_s) ##. Once it starts moving we have friction ## \mu_k mg\cos(\theta_s) < \mu_s mg\cos(\theta_s) ##, but the downslope force hasn't changed. So why will it not accelerate?
 
  • #17
I'm sorry but I'm really confused now. When I asked before if I could say that I adjusted the track so that the object moved at a constant acceleration (so no acceleration), didn't you say yes? But then why would there have to be acceleration.

In other words, can't I say I set the track to a minimum angle so that it went downhill at a constant speed?
 
  • #18
Here, this is what I said (readjusted):

I would first adjust the track to an inclination in which the block and cart moves at a constant speed (so, at its minimum with no acceleration) then measure angle θ. After, I would decompose vector Fg into its x- and y- components to calculate normal force which = mgcosθ. I will also determine downhill force by mgsinθ. Then, I will calculate kinetic friction by Fk = μkFn; = μkmgcosθ.

So, total net force would be ma = mgsinθ - μkmgcosθ; divide m on both sides
a = gsinθ - μkgcosθ; a = 0
gsinθ = μkgcosθ; divide both sides by g
sinθ = μkcosθ
μk = sinθ/cosθ
μk = tanθ
---
Is there something wrong with this method?
 
  • #19
harujina said:
I would first adjust the track to an inclination in which the block and cart moves at a constant speed (so, at its minimum with no acceleration)
But there's a practical difficulty in doing that. You need to spell out the details of exactly how you would achieve this. If you simply raise the angle until the block moves, it will accelerate down, so that's too high. It would not be practical to try to readjust it as this is happening.
 
  • #20
haruspex said:
But there's a practical difficulty in doing that. You need to spell out the details of exactly how you would achieve this. If you simply raise the angle until the block moves, it will accelerate down, so that's too high. It would not be practical to try to readjust it as this is happening.
Oh, so it simply can't be eyeballed?
This is the only way I have learned to measure the coefficient of kinetic friction in the lab; I know of no other way.
 
  • #21
harujina said:
Oh, so it simply can't be eyeballed?
This is the only way I have learned to measure the coefficient of kinetic friction in the lab; I know of no other way.
What can't be eyeballed? That it is not accelerating? But unless you do something to avoid it it will be accelerating.
Consider this: static friction will be replaced by kinetic friction as soon as movement starts, regardless of the reason movement started. Does that give you any clues?
 
  • #22
haruspex said:
What can't be eyeballed? That it is not accelerating? But unless you do something to avoid it it will be accelerating.
Consider this: static friction will be replaced by kinetic friction as soon as movement starts, regardless of the reason movement started. Does that give you any clues?
I'm still having trouble understanding...
I know that static friction is greater than kinetic friction and that the block and cart will move downhill once static friction is replaced by kinetic friction but I still don't see where and why I would have to mention acceleration?

http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html
On Friction Plot (near the bottom) -> Kinetic Case.
This is a similar method that our class used in the lab and what I have applied for in my experiment. How come this is not practical? And by eyeball, yes, I meant eyeball when it seems like the cart and block are going downhill at constant speed and then measure that angle.
 
  • #23
harujina said:
I would first adjust the track to an inclination in which the block and cart moves at a constant speed
Yes, sure, but what you are missing is that this is more easily said than done. Exactly how will you do this?
If you place the block on the ramp and raise the angle until it moves, the angle will be ##\theta_s##, where ##\tan(\theta_s) = \mu_s##. This will be higher than ##\theta_k## and the block will accelerate. If you raise it to any lower angle nothing will move. So what will you do to find ##\theta_k##?
 

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