The discussion revolves around the calculation of the coefficient of friction (μ) in the equation F=μmg, exploring how μ is determined experimentally and the implications of varying conditions such as material composition and environmental factors. Participants delve into the complexities of friction in different scenarios, particularly in the context of auto racing and physics problems involving free body diagrams.
Participants exhibit a range of views on the calculation and variability of μ, with no consensus on the best methods for determining it or the implications of its variability in practical scenarios. Additionally, there is ongoing debate regarding the correctness of specific examples from H.Young's text, with some participants expressing confusion and others attempting to clarify the concepts involved.
Limitations in the discussion include the dependence on empirical data for μ, the variability introduced by different conditions, and unresolved questions regarding the accuracy of specific physics examples. The discussion also highlights the complexity of force analysis in scenarios involving multiple forces.
This discussion may be of interest to students studying physics, particularly those grappling with concepts of friction, force analysis, and the application of theoretical principles to practical problems.
luckis11 said:F=μmg. How was μ calculated? Obviously experiments were needed. So, done the experiments, what's the exact conclusionology that μ is that much for that type of ground and that much for the other type of ground?
Please describe the problem. Sounds like example 7.6, not 7.7, where there's a crate being slid up a ramp with friction. If that's the problem you mean, then you are told that the initial velocity is 5 m/s up the ramp. (Don't take the vertical component of that velocity.)luckis11 said:Is H.Young wrong at example 7.7 or am I confused? Shouldn't the upward speed be (sin30)5=2.5? That gives that the box cannot rise above 0.32metres whereas he says it goes up to 0.8metres having faced friction too?
What free body diagram?luckis11 said:Forget it. Do you know where I can find the proofs that the free body diagram is true?
I don't see any problem with those examples. (Assuming we have the same book, the 12th edition.) Did you have a specific issue?luckis11 said:The analysis of forces at H.Young's examples 5-9, 5-17, 5-18.
What do you mean by "proof"? I thought you had a question about how to draw a free body diagram? Is there something that you think is done wrong? Be specific.luckis11 said:No "problem". I asked for the proof.
Sorry, but I have no idea what you're looking for. To do a force analysis one identifies all the forces acting, draws a free body diagram, and applies Newton's laws. You cannot tell whether a particular analysis is correct without actually looking at the details.luckis11 said:By proof I mean that instead of wondering why this force analysis is true (if it is true), the because should be ready somewhere as either logical proof or unexplained experimental evidence.
Again: Be specific. What exactly in those examples is troubling you?luckis11 said:Forget about the constant μ. Regarding H.Young's examples 5-9, 5-17, 5-18, alone:
By proof I mean that instead of wondering WHY this force analysis is true (if it is true), the BECAUSE should be ready somewhere as either logical proof or experimental evidence with no logical explanation.
Let me remind you that that's only useful if the force in question is the only force acting on the mass. When multiple forces are involved, mdu/dt gives the net force.luckis11 said:Let me remind you that Force is the mdu/dt that happens to the mass.
The force of gravity is mg acting down. If gravity were the only force acting, the object would accelerate downward with du/dt = g. But gravity isn't the only force acting on an object on an incline.Thus, e.g. why the mdu/dt because of gravity should be equal to mgsin30?
The normal force is perpendicular to the surface. There's no net force in that direction, so there's no acceleration in that direction. (The normal force and the perpendicular component of gravity add to zero.)Another example: The force n goes upwards, whereas no mass goes toward that direction.
You can use mdu/dt if you isolate the force so that it's the only one acting.luckis11 said:If mdu/dt is only the net force, how are the non-net forces defined?
The components 'act' whenever gravity acts.luckis11 said:n=mgcos30 can "act" (happen) only when mgsin30 happens.
That's because mgsin30 is the net force.But when mgsin30 happens it's the only one which happens as a mdu/dt.
Since the normal force is a 'reactive' force, you can't (at least not in this scenario). But why would you want to?Then how can I isolate n as a mdu/dt?
Huh? Recall that I said that it's the net force on an object that equals mdu/dt. If you can arrange for a given force to be the only force acting, then you can directly apply mdu/dt.luckis11 said:It seems you changed your mind and now you say that mg and n cannot be mdu/dt.
That's a 'generic' definition of force. Not very useful here when multiple forces act. You can obviously have forces acting without a mass accelerating.Then, what are they, since the definition of force is mdu/dt?
Huh?luckis11 said:Which are "obviously", what?
They are forces, specifically electromagnetic and gravitational forces.luckis11 said:n and mg, what are they?
You are getting hung up on mdu/dt as some sort of 'definition' of force; much better to think of mdu/dt (Newton's 2nd law) as a property of force in general.They cannot be mdu/dt as you said, therefore they are vectors of what?