How to Derive the Coefficient of Friction from Time and Acceleration?

  • Thread starter Thread starter Mille89
  • Start date Start date
  • Tags Tags
    Expression
AI Thread Summary
To derive the coefficient of friction (COF) from time and acceleration, the experiment involves measuring the time a plastic box takes to travel up a 0.93m inclined plane at a 30-degree angle. The average time recorded was 1.335 seconds, leading to a calculated acceleration of 1.04 m/s². The discussion emphasizes the need to analyze the forces acting on the box, including gravitational force components along the slope. A Free Body Diagram is recommended to clarify the forces involved, particularly for both the box and the suspended mass. The goal is to establish a relationship between the measured time, acceleration, and the COF for further analysis of kinetic friction.
Mille89
Messages
5
Reaction score
0

Homework Statement


Hi! I am working on a lab-report where we have measured the time a body use up a inclined plane.
i am going to find a expression for the coefficient of friction expressed by the time and acceleration that i have found.

This expression do i need for the next task where i am going to find the kinetic friction.

But i have no idea how i can find the expression:oops:

Homework Equations


9fd47b2a39f7a7856952afec1f1db72c67af6161
= f/N

The Attempt at a Solution


9fd47b2a39f7a7856952afec1f1db72c67af6161
= 1/2*at2
 
Physics news on Phys.org
Mille89 said:
going to find a expression for the coefficient of friction expressed by the time and acceleration that i have found.
Please describe the experiment in more detail, in particular what measurements were taken.
What forces acted on the body? Which of them had components in the direction of the acceleration?
What equation (ΣF=ma) can you write?
 
This lab exercise was about Newtons 2 law.
Where we where going to find:
  • The angle of the plane (30)
  • Find the lengst of the plane (0.93m)
  • Measure the time up the plane with different underlay, 5 times each.
  • The constant acceleration and compare it with the sum of the forces on the box
  • And Then find the expression for the COF...
I have measured the time a plasticbox(0.0959 kg) use up the plane (length 0.93m). I did this 5 times and found out that the mean was 1.335 s.

Then i used s=vt+1/2*at2
0.93m=0*1.335s+1/2a(1.335s)2
a=2*0.93m/(1.335s)2
a=1.04m/s2
Hmm I think this is wrong i see now that its the constant acceleration i am going to find.

Its like 10 years since i had physics last, so i have forgoten a lot of it:oops: and we where two on this task but right now its only me working on it:confused:

IMG_20190303_101557.jpg

m1(plasticbox) =0.0959kg
m2(Weight, power source) = 0.2kg
 

Attachments

  • IMG_20190303_101557.jpg
    IMG_20190303_101557.jpg
    49.9 KB · Views: 340
Mille89 said:
I think this is wrong
No, your estimate of the acceleration looks fine.
What you need now is the force versus acceleration equations for the two masses.
Draw a separate Free Body Diagram for each.
Write out the ΣF=ma equation for the suspended mass.
For the mass on the ramp, what components of forces act along the slope?
 
The acceleration for the two masse is the same am1m2=2*s/t2 = 2*0.93m/(1.335s)2= 1.04m/s2

The force:
ΣF=ma
Gx= mgsin(30) = 0.470N
Is ΣF Then 0.470N on the mass 1?
IMG_20190305_072043.jpg
 

Attachments

  • IMG_20190305_072043.jpg
    IMG_20190305_072043.jpg
    35.6 KB · Views: 357
Mille89 said:
Gx= mgsin(30) = 0.470N
Is ΣF Then 0.470N on the mass 1?
No, that's just Gx. There are two other forces parallel to the ramp.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top