Coefficient of Kinetic Friction on Inclined Plane? [FORCES]

In summary, the question asks for a description of how to measure the coefficient of kinetic friction using only a block, cart, track, and protractor. The mass of the block and cart is given as 0.734 kg and the protractor measures the angle of inclination of the track. To measure the coefficient of kinetic friction, the experiment could be set up to ensure the cart and block move at a constant velocity, indicating no acceleration. This could be achieved by adjusting the track and utilizing static friction to keep the block and cart in place until the force applied to move them surpasses the force of static friction. The coefficient of kinetic friction could then be calculated using the formula μk = Fk/Fn, where Fk is the kinetic
  • #1
harujina
77
1

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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  • #2
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?
 
  • #3
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?
 
  • #4
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?
 
  • #5
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.
 
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  • #6
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.
 
  • #7
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)?
 
  • #8
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.
 
  • #9
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?
 
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  • #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##?
 

FAQ: Coefficient of Kinetic Friction on Inclined Plane? [FORCES]

What is the definition of coefficient of kinetic friction?

The coefficient of kinetic friction is a measure of the amount of resistance between two surfaces in contact when one of the surfaces is in motion.

How is the coefficient of kinetic friction on an inclined plane different from a flat surface?

The coefficient of kinetic friction on an inclined plane is affected by the angle of the incline, as well as the weight and surface material of the object sliding down the plane. On a flat surface, the coefficient of kinetic friction remains constant regardless of the direction of motion.

What factors affect the coefficient of kinetic friction on an inclined plane?

The angle of the incline, the weight of the object, and the nature of the surface material all affect the coefficient of kinetic friction on an inclined plane. Additionally, the presence of any external forces, such as air resistance, can also impact the coefficient of kinetic friction.

How is the coefficient of kinetic friction on an inclined plane calculated?

The coefficient of kinetic friction on an inclined plane can be calculated by dividing the force of friction by the normal force. The force of friction can be found by multiplying the coefficient of kinetic friction by the weight of the object, while the normal force is equal to the weight of the object multiplied by the cosine of the angle of the incline.

What are some real-world applications of the coefficient of kinetic friction on an inclined plane?

The coefficient of kinetic friction on an inclined plane is relevant in many real-world scenarios, such as calculating the amount of force needed to push a heavy object up a ramp, determining the stopping distance of a car traveling down a steep hill, and understanding the mechanics of skiing and snowboarding down a slope.

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