Gareth56 11 Posted December 5, 2014 Share Posted December 5, 2014 I'm having some difficulty understanding why the coefficients of static and kinetic friction are constant e.g. they are the same on the Earth and the Moon. For example if μs = Friction Force/Normal Force then because the Normal Force is acceleration due to gravity dependent (mg) and the acceleration due to gravity on the Moon is less than that on Earth wouldn't this change the value of μs or μk, so they would be different on the Earth and the Moon? Here's the question that's causing my confusion;- A crate of 45.0 kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 313N. After that you must reduce your push to 208N to keep it moving at a steady 25.0 cm/s. B.) What push must you exert to give it an acceleration of 1.10 m/s^2? © Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62m/s^2. (i) what magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)? I would have thought that on Earth where g = 9.81m/s^2 to calculate μs & μk you would use 313 N = μs*m*g = μs*45kg*9.81m/s^2 μs = 313/(45*9.81) = 0.709 208 N = μk*45*9.81 μk = 208/(45*9.81) = 0.471 However on the Moon the value of g = 1.62m/s^2 making μs = 313/(45*1.62) = 5.33 and μk = 208/(45*1.62) = 2.85 However the answer leaves μs & μk the same on both the Earth and the Moon???? Thanks Link to post Share on other sites

studiot 2240 Posted December 5, 2014 Share Posted December 5, 2014 Y Hello Gareth, your difficulty is simply resolved. You will find the push force to overcome friction lower on the Moon. This force may be 313N on Earth, but on the Moon it will be less (wouldn't it be fun to try the experiment?) Remember [math]F = \mu N[/math] The coefficient of friction (within sensible limits) is constant so If N varies, then F must vary. B the way you realise that you should use the normal reaction, not the weight to calculate F? Not doing this is a common mistake, that will lead to errors on inclined planes or pressure connections. Link to post Share on other sites

Gareth56 11 Posted December 5, 2014 Author Share Posted December 5, 2014 Hi Still a tad confused I'm afraid. If the Force (static friction) = μs N (and N = mg) then μs = Force (static friction)/ N = Force (static friction)/ mg However if the value of g is different then surely the value μs will be correspondingly different, same goes for μk. Sorry for coming across as thick on this but it really does confuse me as why the values should be the same on both the Earth and the Moon when μs and μk are dependent on the value of g from the equation μs = Force (static friction)/ N = Force (static friction)/ mg Thanks Link to post Share on other sites

studiot 2240 Posted December 5, 2014 Share Posted December 5, 2014 (edited) Don't worry about static v dynamic friction, the calculations works the same way for both. That is why I just used mu and assumed the user would select the appropriate coefficient. Back to the formula. It says F is a variable that depends upon a constant (mu) and a variable (N) in such a way that F is proportional to N and mu is the constant of proportionality. This is an experimentally verified fact for many situations, including all those in basic courses. Now if we change N by going to the Moon and changing g then we will find that F has changed to maintain this proportionality. If, instead, we remained on Earth and changed N by changing the weight edit : mass ( say for instance we had a box that we gradually filled with sand) and measured F for different values of N we would find the same proportionality. Does this help? Edited December 5, 2014 by studiot Link to post Share on other sites

Gareth56 11 Posted December 5, 2014 Author Share Posted December 5, 2014 Don't worry about static v dynamic friction, the calculations works the same way for both. That is why I just used mu and assumed the user would select the appropriate coefficient. Back to the formula. It says F is a variable that depends upon a constant (mu) and a variable (N) in such a way that F is proportional to N and mu is the constant of proportionality. If I could take what I've underlined. Again apologies for my dense understanding of this matter. You say that mu is a constant but if mu depends on the value of g (as we've established above) then given that g can vary e.g. g on the Earth > g on the Moon how can mu be classified as a constant? I agree that N is a variable because N varies with the value of g. Had mu been experimentally determined on the Moon would it have had a different value from that experimentally determined on the Earth? I appreciate your patience, but I will understand if you give up! Link to post Share on other sites

studiot 2240 Posted December 5, 2014 Share Posted December 5, 2014 (edited) I don't recall saying mu depends upon the value of g. It doesn't, why should it. If it did it would not be a constant. The line you underlined says that F depends upon N so if N varies then so must F, because mu is constant. Edited December 5, 2014 by studiot Link to post Share on other sites

Gareth56 11 Posted December 5, 2014 Author Share Posted December 5, 2014 I don't recall saying mu depends upon the value of g. Isn't it the case that by definition mu = F(friction)/Normal Force and again by definition Normal Force = mg Therefore mu depends on the value of g. Link to post Share on other sites

studiot 2240 Posted December 5, 2014 Share Posted December 5, 2014 That's a very poor definition. What happens if N = 0 : That is the two objects are just touching but not exerting any force one each other? Then you have division by zero. Worse, mu may not actually equal that ratio for static friction. In your example (on Earth) If you only pushed with a force of 100N the box would not move. Since the box remains in equilibrium, the horizontal force of friction (F) must also be 100N But this does not mean that mu = 100/(45*9.81). In fact this is true for any value of push less than 313N, including zero. Far better to allow that mu is the coefficient of proportionality as I already said. Further whoever said that the normal force = mg ? It may do, but then again it may be nothing to do with either m or g. If I press a matchhead against a matchbox to light the match, there is considerable friction. But the force I press the match against the box has nothing to do with the mass of either the box or the match (or g for that matter). It is very convenient in mechanics that we have such a simple relationship between the normal contact forces and the force of friction that can be developed. Link to post Share on other sites

Gareth56 11 Posted December 5, 2014 Author Share Posted December 5, 2014 (edited) It's the definition that's given in my University Physics textbook!! However, many thanks for you patience and explantations. It's appreciated. Edited December 5, 2014 by Gareth56 Link to post Share on other sites

Enthalpy 249 Posted December 5, 2014 Share Posted December 5, 2014 Coefficients of friction depend on the contact pressure, but slowly. Between the Earth and the Moon, it won't make a significant difference. But if the contact pressure changes a few magnitudes, the coefficients of friction change easily by a factor of 3, so the measures should always tell the experimental context to be useable. Link to post Share on other sites

studiot 2240 Posted December 5, 2014 Share Posted December 5, 2014 Coefficients of friction depend on the contact pressure, but slowly. Between the Earth and the Moon, it won't make a significant difference. But if the contact pressure changes a few magnitudes, the coefficients of friction change easily by a factor of 3, so the measures should always tell the experimental context to be useable. Whilst it is good that you have refrained from posting a large departure from convention I do not recommend mixing up Force and Pressure to those just learning the subject. Link to post Share on other sites

Bignose 946 Posted December 7, 2014 Share Posted December 7, 2014 (edited) Isn't it the case that by definition mu = F(friction)/Normal Force and again by definition Normal Force = mg Therefore mu depends on the value of g. It's the definition that's given in my University Physics textbook!! So, I think that the confusion stems from the fact that implicit in the model is that [math]\mu[/math] is a constant. If [math]\mu[/math] is taken to be a constant, than indeed, you can define it to be Friction Force/Normal Force. In the same way you can define [math]\pi[/math] to be Circumference of a Circle/Diameter of a Circle. But, just like [math]\pi[/math], once you've set your model to be a fixed [math]\mu[/math] you get to use it in all difference situations. So, in the example above, if we measure [math]\mu[/math] on earth, under earth's gravitational field, the model assumes that indeed [math]\mu[/math] is the same on the moon, on Jupiter, on the sun, etc. Just like you can use the same [math]\pi[/math] for calculation of a circle's area, or a sphere's volume. Now, all that said, unlike [math]\pi[/math], [math]\mu[/math] is very rarely constant under every situation. There is usually at least some amount of weak normal force dependence on it. There is usually a weak environmental component to it (i.e. vacuum vs. earth's atmosphere vs. pure nitrogen are probably different) and a weak temperature dependence. As mentioned above, it is important to use the term Normal force and not just weight. They are the same in a lot of situations, but not always. I.e. consider the simple situation of 2 blocks stacked on top of each other... In short, I think it is best understood that the use of [math]\mu[/math] above is just a model. A model that is very successful under a good range of circumstances, but not always. And it is important to understand when this model is appropriate and it isn't. Edited December 7, 2014 by Bignose Link to post Share on other sites

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now