# Homework Help: Physics help to measure friction

Tags:
1. Dec 5, 2017

### rhiana

<Moderator's note: Moved from a technical forum and thus no template.>

Here’s the question
Design a simple expirament that you could carry out in your own home to
i) determine coefficient of friction between your object and surface
ii) prove that coefficient of static friction is only dependent on the surfaces in contact, and is not affected by any changes in mass of your object

So first, I designed my expirament to pull a 2kg book across a flat surface. I don’t know what variables I need but I think that I need the formula:
Ff=mu*Fn (mu is coefficient of friction)

So I have mass, and I could figure our force applied (which I would say is 3N). What else do I need for this?

2. Dec 5, 2017

A spring balance is useful for measuring the force. Basically, one that can measure small amounts like 2 or 3 pounds accurately.

3. Dec 5, 2017

### rhiana

How do I know that the mass doesn’t impact the coefficient of friction?

Because if
Ff=mu*Fn
Ff=mu*mg (since it’s on a flat surface)
Ff=mu*(3)(9.8)
Ff=mu*29.4

So even if I don’t have all the variables needed, I can already tell that the mass DOES impact the mu

4. Dec 5, 2017

### rhiana

Sorry I meant (2)(9.8) not (3)(9.8)

5. Dec 5, 2017

If you double the mass and the force that is needed to pull the object across the surface doubles, then $\mu$ is constant. You can draw of graph of $F_f$ vs. mass/weight=$F_N$, and see if you get a straight line. The slope is the coefficient of friction. You can also use the same spring balance to weigh the objects, giving you $F_N$.

6. Dec 5, 2017

### rhiana

Ok so if
Ff=Fn
Then
Mu*fn=fn
mu*mg=mg

I got this far but I don’t think that cancelling it out is a good idea or else mu= 0

Is there something wrong I’m doing here ?

7. Dec 5, 2017

If you found $F_f$ very nearly equal to $F_N$, then $\mu=1.0$ (approximately). Another suggestion: Try changing the surface: e.g. on slippery ice, or even a well-polished floor: $\mu$ should be much smaller.

8. Dec 5, 2017

I’m getting confused over the $you are using 9. Dec 5, 2017 ### Charles Link Presumably you are pulling the object across a surface, and$ F_f $is the force required to pull it.$ F_N $is the normal force between the surfaces, which is the weight of the object. Typically, you will find for a rough surface, that$ F_f $may be as large or larger than$ F_N $,(making$ \mu \geq 1.0 $), while for a smooth surface$ F_f $is very small , and thereby$ \mu $can be quite small. 10. Dec 5, 2017 ### Merlin3189 In England we call that "begging the question". You are taking a formula based on the assumption that mass doesn't affect the coeft of friction and using that formula to prove the very same thing. So rephrase the question, how do you know that Ff = μ * Fn where μ is constant for different masses Notice, the formula is always true if we let μ vary, but then it would not help us. Maybe it would be helpful to you if we rewrite the formula as μ = F/N or Coeft = Friction / Normal (I'm using F for Ff and N for Fn for simplicity) Now calculate μ for different masses and see if it is affected by the changes. 11. Dec 5, 2017 ### rhiana Okay! So if I change the surface to be more slippery, how would that impact my formula? Can ff=fn? (I just want it to look more like a coefficient force that I see in the textbook like 0.8 or something) would I need to know the force of friction to be able to make it more slippery? 12. Dec 5, 2017 ### Charles Link$ F_f $is the force needed to pull the object across the surface at a constant speed. You really need to measure$ F_f $experimentally, but you can also imagine the result you will get. Is it easier to pull a sled on ice than it is across a concrete sidewalk? Also, if you put two people on the sled, instead of one, do you need to pull twice as hard? 13. Dec 5, 2017 ### rhiana Okay so if I need to find the coefficient of friction I did this Ff=fn Mu*(mg)=mg mu*3*9.8=3*9.8 Mu(29.4)=29.4 mu=29.4/29.4 mu=1 And if mass was 1000 Mu*(mg)=mg mu*1000*9.8=1000*9.8 Mu(9800)=9800 mu=9800/9800 mu=1 Is this the right way of thinking about it? 14. Dec 5, 2017 ### Merlin3189 15. Dec 5, 2017 ### Charles Link Did you find$ F_f =F_N $experimentally for both weights on the surface that you used? That could be the case for a relatively rough surface, but there are many, many surfaces that will give$ F_f \neq F_N $. If you found$ F_f =F_N $for both cases, then yes,$ \mu=1.0 $(approximately) for the coefficient of friction between the object and the surface it is sliding on. 16. Dec 5, 2017 ### rhiana If ff does not = fn , there is no equation to calculate it without knowing the mu , right? So if I wanted it to be on a slipperier surface I could say that I measured the Ff = 10N Then, I could use the formula Ff=mu*fn 10=mu*(3)(9.8) 10=mu*(29.4) 10/29.4=mu 0.340=mu But, my only issue is that my mass now is a variable and if mass is 1000, then mu=0.0010 not 0.340 and then mass matters now 17. Dec 5, 2017 ### Charles Link I think you need to perform the actual experiment: You would find that very helpful, because your guesses on what you might get aren't accurate enough to give meaningful numbers to work with.$ \\ $I could give you a simple example or two: Suppose you weighed mass1 and it was$ F_N=2 $lbs. You then found$F_f=1.5 $lbs. (measured). With the same surface, for mass2 made of the same material but with a weight attached$ F_N=3 $lbs., you find$F_f $to pull the object across the surface was measured to be$ F_f=2.25 $lbs. Compute$ \mu $for the surface.$ \\ $For the same two masses on ice,$ F_f=0.2 $lbs. and$ 0.3 $lbs. respectively. Compute$ \mu $for the material when it slides on ice.$ \\ ## Perhaps these two examples will be helpful.

Last edited: Dec 5, 2017
18. Dec 5, 2017

@rhiana Please see my edited post 17.

19. Dec 5, 2017

### Merlin3189

Can I just keep using F for the Friction force and N for the Normal force please? It makes typing quicker and my eyesight is not good on the subscripts.

Ok , remember F=μN means μ = F/N
So if you make F = N then μ = F/F or N/N = 1 that means the friction force is equal to the weight and you don't usually get such a big μ
If μ = 0.8 say, as you want, then F/N = 0.8 or F= 0.8 * N ie. friction force is 80% of the weight, still prety high.

You pulled a 2kg book with 3N , so for your book on your surface, F=3 Newton and N = 20 Newton so μ = 3/20 = 0.15 not bad.
(You could say N= 2kg * 9.8 Newton per kg = 19.6 Newton, then μ = 3 / 19.8 and I need a calculator to get μ = 0.152 )

The question was to show that μ does not depend on mass, though it might depend on the surfaces.
So changing the surfaces will not help us. Then μ might change, but that's expected. And it might not change, but that might be expected as well.
The thing we need to know about is the mass and μ.
So you must change mass and see what happens to μ.

So you have a book weighing 20 Newton (about) and you measured F and calculated μ
Now you want to change the mass (and therefore the weight), measure F again and calculate μ again. But you must not change the surfaces, because then you would not know whether any change was due to the mass / weight or the surfaces. So how can you do that?
How could you, say, double the mass of the book while keeping everything about the surfaces the same?
And then treble the mass, 4x the mass, ... each time measuring F and calculating the new μ to see if it comes out the same or not.

20. Dec 5, 2017

### rhiana

Okay, so I did preform the experiment on different surfaces, but I’m not able to find out the force of friction since I don’t have any tools to figure it out.

Also, I did your examples and on ice, but they don’t have the same mu which the question is asking me to have

It wants to know why the mass isn’t important in the question and changing the mass does nothing to the mu.

I did this

Mass NOT on ice
Ff=mu(fn)
2=mu(1.5)
1.3=mu

Same mass ON ice
.2=mu(1.5)
.13=mu

And the second NOT on ice
2.25=mu(3)
0.75=mu

The same second mass ON ice
.3=mu(3)
.1=mu

Is .1 close enough to .13 to be considered the same?

Does this prove that mu is not dependant on mass?