How to find coefficient of Friction? - Please help, due tomorrow.

[rum2563]

The question:

The driver of a 1.2X10^3 kg car traveling 45 km/h [W] on a slippery road applies the brakes, skidding to a stop in 35 m. Determine the coefficient of friction between the road and the car tires.

Equations:

- Ff = uFn
- d = 0.5 (Vi + Vf) * time
- Fn = mg
- F = ma


My attempt:

To find normal force:
Fn = (1.2 * 10^3) (9.81)
= 11772 N


To find acceleration:
d = 0.5 (Vi + Vf) * time
35 = 0.5 (45 + 0) * time
35 = 0.5 (45) * time
35 = 22.5 * time
35/22.5 = time
time = 1.6 seconds

acceleration = (Vf - Vi) / time
= (0 - 45) / 1.6
= -28.12 m/s^2


To find force using Newtons' law:

F= (1.2*10^3)(.28.12)
= -33744 N

u (coefficient of friction) = -33744 / 11772
= -2.87

Is the answer correct?
Basically we have to find the coefficient of friction, but I am confused as to how to find it out. The main problem is that acceleration is negative, can anyone please help.
This is due tomorrow for me.

Thanks.

 

[21385]

watch out for 45km/h...

 

[rum2563]

watch out for 45km/h...

Could you please elaborate a bit more? Thanks.

 

[21385]

divide by 3.6 to get meters per second

 

[rum2563]

divide by 3.6 to get meters per second

But I am confused. Shouldn't I divide by 1.6 because that is the time? Or is 3.6 another number? Thanks for your help, please explain.

 

[Doc Al]

To find acceleration:
d = 0.5 (Vi + Vf) * time
35 = 0.5 (45 + 0) * time
35 = 0.5 (45) * time
35 = 22.5 * time
35/22.5 = time
time = 1.6 seconds

acceleration = (Vf - Vi) / time
= (0 - 45) / 1.6
= -28.12 m/s^2

Careful with units! Convert the car speed to m/s.

The main problem is that acceleration is negative, can anyone please help.

Nothing wrong with a negative acceleration--the car is slowing down. The sign merely specifies direction--when computing the coefficient, consider magnitudes.

 

[21385]

The standard units for velocity is m/s. The velocity used in the question is in km/h. The other units used in the question is in meters and naturally, you would want meters per second. You have to convert the velocity to get a correct answer. To convert 45 km/h (1000m/1km)(1 h/3600 seconds) ===> 45 km/h divided by 3.6 would get you 12.5 m/s, which is what you want.

 

[rum2563]

So when I convert 45 km/h to m/s, I get 12.5 m/s

So do I use 12.5m/s to get the acceleration just like I did in my question?

 

[21385]

yep, that should give you the correct answer

 

[rum2563]

One more question, when I do all the calculations, I get a negative acceleration. So therefore, I would get a negative value for coefficient of friction. Is that OK, or would it not make sense because friction cannot be negative.

This is what I am doing:

u = -9378 / 11772
= -0.8

Please help, thanks.

 

[Doc Al]

Friction, a force, can certainly be "negative". Remember: sign just refers to direction.

But the coefficient of friction cannot be negative: Ff = uFn describes the magnitude of the friction force, not the direction or sign.

 

[Feldoh]

The main problem is that acceleration is negative, can anyone please help.
Thanks.

I believe that a negative acceleration just means you are slowing down in whatever direction you called positive or, you are speeding up in the direction you called negative

 

[rum2563]

Arite, thank you very much. But can you tell me that do I have to plug in 12.5m/s OR 45km/h in the formula which I am using to find time? Because I use 12.5m/s I get a lower value for u, and if I use 45km/h I get a higher value for u.

Thanks.

 

[21385]

plug in 12.5 m/s, and that will get you the correct answer

 

[rum2563]

Yep, I got it.

Thank you very much for all the people who have helped me in this topic. I cannot believe how quick the responses and the help was. I think PhysicsForums are a great resource and I hope I can contribute more to it. Thanks to all the people.

 

[ychathun]

And what was your answer?
I got an answer and I need to make sure about it.. Can u please help me??

 

[mpheng]

Road Friction Testing Machine
Test are carried out periodically on major roads to assess their friction (accumulations if oil and rubber can seriously reduce the coefficient of friction).
You are required to design a suitable portable rig to carry out such tests that can be deployed from the back of a van or a estate car.
Your design should include a range of capabilities, i.e. easy change of test tyre – assuming you are using a test tyre; the ability to artificially wet the surface in a controlled manner and flexible measuring capabilities.
As well as the basic mechanical design of the system, you will need to consider how undulations and bumpiness in the road surface can be treated to avoid inappropriate results.
You will need to specify the electronics of the measurement system and how the coefficient of friction will be calculated and reported.

Anyone can help me?

 

[Spinnor]

For a start look here:

http://www.google.com/search?hl=en&..._gc.r_pw.&fp=e60174e408cdd637&biw=800&bih=417

http://www.google.com/search?um=1&h...gc.r_pw.&biw=800&bih=417&ie=UTF-8&sa=N&tab=iw

http://en.wikipedia.org/wiki/Road_slipperiness