How Do I Calculate the Coefficient of Friction in a Ramp Experiment?

  • Thread starter HappyScientist
  • Start date
  • Tags
    Physics
In summary, the conversation discusses an experiment conducted to find the coefficient of friction on rubber using a ramp and photo-gates to record the time and velocity of a car-like object. The experimenter is trying to determine the correct variables to use in their calculation and is struggling to find a reasonable value for the coefficient of friction. They have considered the forces of gravity and rolling friction, but are unsure about how to solve the system using torques. They are also considering measuring the wheels to find the average moment of inertia.
  • #1
HappyScientist
2
0
This is not exactly a homework question, but rather a question I pose to a set of experiments I preformed as an extension after my original assignment. I would like to know the answer to this question as I have been asked to demonstrate to my fellow classmates the experiment.

An experiment was conducted to find the coefficient of friction on rubber, the following was the set-up;

- Ramp was set at a θ angle, using a step (to create an incline plane)
- Two Photo-gates were set up in a row(at the base of the incline plane)
- A sort of car-like object -with a card to trigger the gates, was placed on the ramp and released
- The Photo-gates record the time triggered and released.

I can work out the values myself, I am asking If I have the correct variables, if so, how do I use them? and if not what am I missing and how would I use the variables if I had them

Homework Statement


I cannot find a reasonable value for the coefficient of friction.

Variables:
Theta Angle of ramp = θ
Length of card on car-like object = l
Gate 1: Entry 1 = n1
Exit 1 = x1
Gate 2: Entry 2 = n2
Exit 2 = x2

Time = t
Gravity = g
Coefficient of friction = μ
Velocity = v
Mass = m
Displacement = s
Force = f
Acceleration = a

Homework Equations



Equations of straight line motion,
K.E = 1/2mv2
Work = f*s
s = v2/2μg
a = vfinal - vinitial/t

The Attempt at a Solution



I worked out the velocity the car-like object moved though both gates:
v1 = Before the car-like object is released
v2 = Whilst passing through the first gate
v3 = Whilst passing through the last gate

I then figured that the deceleration between gate 1 and 2 could be used to calculate the stopping time(the time it would take for the object to come to rest) by;
v3/deceleration
I then used this number to try and calculate the stopping distance through the equations of straight line motion and then mu through (s = v2/2μg), but kept getting a number of mu around 0.009, which seemed too small to be reasonable.

EDIT 1: The car-like object is a "Trolley" commonly used in physics experiements.

Please share your thoughts.

HappyScientist
 
Last edited:
Physics news on Phys.org
  • #2
Does the car have wheels?
 
  • #3
CWatters said:
Does the car have wheels?
Yes, I recall another student calling it a trolley.
 
  • #4
It looks like what you have going on is appropriate, seeing as how (assuming negligable air drag) the only forces on the trolly are gravity and (rolling) friction, which are both independent of velocity; hence they are constant forces with respect to velocity ≠ 0.

The only thing I'm not sure of is this:
Your actually measuring rolling friction between the tires and rubber, and I haven't done a whole lot with rolling friction. I know that it is (significantly) lower than both static and kinetic friction, but I don't know if you solve the system linearly and angularly, if you'll get the same result. That is: solve for mu using torques and solve for it using forces (or energies) that you will get the same mu.

I would test this out before I put it out there. Regardless, solving this system using torques should get you the correct mu.

What you'll have then is this:

Assume no friction. What happens? The car slides, with the wheels not spinning at all, as the only torque acting on the wheels is from friction. This means that if you can find the torque, you can find mu rather easily.

So what do you have to work with?
Well you have to points, and you know the velocities, right? If you measure the wheels, then you can come up with an average moment of inertia, and also a standard deviation. This would probably be the route I would take, but then again it may be overkill for what you're trying to accomplish.
 
  • #5


As a fellow scientist, I am happy to provide some insight into your experiment and help you find the correct value for the coefficient of friction.

Firstly, it is important to note that the coefficient of friction is a measure of the resistance between two surfaces in contact. In this case, it is the resistance between the rubber ramp and the car-like object. Therefore, the variables you have listed are correct and can be used to calculate the coefficient of friction.

To calculate the coefficient of friction, you will need to use the following equation:

μ = (mgsinθ)/(mgsinθ + ma)

Where:
μ = coefficient of friction
m = mass of the car-like object
g = acceleration due to gravity
θ = angle of the ramp
a = acceleration of the car-like object

In your experiment, the car-like object is released from rest at the top of the ramp, so the initial velocity (v1) is 0. The final velocity (v3) will also be 0 as the object comes to a stop at the bottom of the ramp.

Therefore, using the equation for acceleration (a = (v3-v1)/t), we can calculate the acceleration of the object as it moves down the ramp.

Next, we can use the equation for work (W = fs) to calculate the force (f) acting on the object as it moves down the ramp. This force is equal to the weight of the object (mg) plus the frictional force (μmgcosθ).

Finally, we can substitute the values we have calculated into the equation for the coefficient of friction to find the correct value.

I hope this helps you in your experiment and demonstrates the importance of understanding the variables and equations involved in finding the coefficient of friction. Good luck with your presentation to your classmates!
 

1. What is "mu" in a physics problem?

"Mu" is the symbol used to represent the coefficient of friction in a physics problem. It is a dimensionless quantity that describes the amount of resistance between two surfaces in contact.

2. How do you find the value of mu in a physics problem?

To find the value of mu, you need to know the normal force acting between the two surfaces and the frictional force acting on the object. The coefficient of friction can then be calculated by dividing the frictional force by the normal force.

3. Why is the coefficient of friction important in physics?

The coefficient of friction is important because it helps us understand the forces acting on an object and how it will behave. It is also a fundamental concept in mechanics and is used in many real-world applications, such as designing machines and predicting the motion of objects.

4. How does the value of mu affect the motion of an object?

The value of mu directly affects the amount of friction between two surfaces. A higher value of mu means there is more resistance between the surfaces, which can cause an object to slow down or stop. On the other hand, a lower value of mu means there is less resistance, allowing the object to move more easily.

5. Can the value of mu be greater than 1?

Yes, the coefficient of friction can be greater than 1 in certain circumstances. This usually occurs when the surfaces in contact have a high degree of roughness or when there is an external force acting on the object. However, values greater than 1 are not common and usually indicate a high level of resistance between the surfaces.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top