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

  • Thread starter Thread starter HappyScientist
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
To calculate the coefficient of friction in a ramp experiment, the setup involves a ramp at an angle θ, two photo-gates to measure the car-like object's velocity, and various physical parameters such as mass, displacement, and acceleration. The experiment aims to determine the deceleration between the gates to find the stopping distance and subsequently calculate the coefficient of friction using the equation s = v²/2μg. The discussion highlights that the object is a trolley, and the friction being measured is rolling friction, which is typically lower than static or kinetic friction. It suggests that using torques in the calculations might yield a more accurate value for μ. Overall, the experiment's methodology appears sound, but the results may require further validation or adjustments.
HappyScientist
Messages
2
Reaction score
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
Does the car have wheels?
 
CWatters said:
Does the car have wheels?
Yes, I recall another student calling it a trolley.
 
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top