Wikipedia:Reference desk/Archives/Mathematics/2013 November 12

= November 12 =

Question about angles ( Probably badly worded Apologies)
Sorry if I explain this badly. Is their a certain degree angle on an inclined plane that halves the distance an object could travel if it was moving along a flat surface ? For example. If a certain force can move a 20 kilogram object 20 feet along a flat surface, Is there a specific angle of of degree ( e.i pushing the same weight, with the same force but up an inclined plane) that would halve the distance that the object is moved. So using the example, what degree angle of plane would the object have to be pushed up, with the same amount of force and have it only moved 10 feet. Is there a specific angle or is it different in every case?

Apologies again if badly worded. regards

Rob


 * I'm thinking it would vary according to the coefficient of friction. BTW, the science Desk would be a better place for this Q.  Also, for the distance on the inclined plane, do you mean the total distance, or just the horizontal component ? StuRat (talk) 01:36, 12 November 2013 (UTC)
 * Possibly 30° or 60° as sin(30°)=0.5 and cos(60°)=0.5.--User:Salix alba (talk): 09:43, 12 November 2013 (UTC)
 * Suppose an object with weight W initialy has kinetic energy E and is moving along a horizontal plane with coefficient of friction &mu;. Then it will come to rest after travelling a distance a such that
 * $$E = \mu Wa$$
 * Now suppose the same object with the same initial kinetic energy is moving up a plane inclined at an angle &theta; to the horizontal. It will come to rest (possibly just for an instant) after travelling a distance b such that
 * $$E = \mu Wb \cos \theta + W b \sin \theta$$
 * We can eliminate E and W from these equations to give:
 * $$\mu a = \mu b \cos \theta + b \sin \theta$$
 * If we assume that a = 2 b cos &theta; (i.e. the object on the inclined plane travels half the horizontal distance of the object on the flat plane) then we have
 * $$2 \mu \cos \theta = \mu \cos \theta + \sin \theta$$
 * $$\Rightarrow \mu = \tan \theta$$
 * So, yes, the angle does depend on the coefficient of friction. Gandalf61 (talk) 10:17, 12 November 2013 (UTC)

Ok then, Is there such a thing as std amount of friction? If Standard or heavy friction coefficient, what would the angle be then?

Many thanks

Rob


 * No, there is no standard coefficient of friction. It really does depend on the surfaces, or the wheels being used.  You need to try an experiment to calculate the coefficient.  In your example above, with your 20 kg object coming to rest in 20 feet, you need to know the speed at which it started, then you can calculate the coefficient of friction for that surface.  (By the way, force times time is what matters to determine the starting speed.)  If the coefficient of friction is low, perhaps 0.1 for roller skates, then the angle would be about six degrees.  For wood or cardboard sliding on a rough surface, with coefficient of friction around 1.0 the angle would be 45 degrees.  You can find the angle very easily by putting the object on a sloping surface and adjusting the angle until the body is on the point of sliding down (or slides down at constant speed).  This is the angle you require.  All of the calculations assume that the coefficient of friction on the slope is the same as that on the horizontal surface.    D b f i r s   11:03, 15 November 2013 (UTC)

Can &pi; Be Expressed As a Finite Combination Of Algebraic functions(*) & Algebraic numbers ?
(*) Including exponentiation and/or radicals.

The natural logarithm is obviously not an algebraic function, otherwise we'd have $$\pi=-i\ln(-1),$$ or $$\pi=\Gamma^2(\tfrac12).$$ And although many expressions involving infinite sums and infinite products consisting solely of algebraic terms for this constant do exist, they are obviously not finite. Definite integrals are also excluded. — 86.125.207.235 (talk) 04:50, 12 November 2013 (UTC)
 * Doesn't the transcendentalism of &pi; preclude that?--Jasper Deng (talk) 06:02, 12 November 2013 (UTC)
 * No. The Gelfond–Schneider constant $$2^\sqrt2$$ is transcendental, for instance. — 86.125.207.235 (talk) 06:32, 12 November 2013 (UTC)
 * cx is a transcendental function for c not 0 or 1. If f is an algebraic function and a an algebraic number, f(a) is an algebraic number. This is all covered in Transcendental function, see the section under exceptional sets.Phoenixia1177 (talk) 07:57, 12 November 2013 (UTC)
 * Dam' ! Thanks for the tip ! :-) — 86.125.207.235 (talk) 10:01, 12 November 2013 (UTC)

Intuition and the Continuum Hyp.
I'm not sure if there is a formal answer to this, but I have a curious case of my intuition going back and forth, and I'd greatly appreciate any insight others might have. The issue is regarding what the value of |c| tells us about how "rich" the system we are working in is. For example, suppose |c| = $$\aleph_2$$ (so CH fails). On the one hand, it seems like there would be "more" sets than if CH held since we would need to enrich the universe enough to cover the $$\aleph_1$$ sized case. On the other hand, I can just as easily imagine that we are talking about a more restricted universe in which we have taken away enough sets to get rid of witnesses that would make the sets of size $$\aleph_1$$ countable or |c|. I do realize that this is a vague and half-assed way of construing the situation, but I'm trying to sort out which way I feel in a vague gut-feeling kind of way. At any rate, any insight would be welcome. Thank you for any help:-)Phoenixia1177 (talk) 20:26, 12 November 2013 (UTC)
 * There is one argument that ontological maximalism cuts in favor of CH. It goes something like this:  At stage &omega;+1 of the iterative hierarchy, the reals are already "frozen".  When we go to stage &omega;+2, where we get sets of reals (well, sets of sets of naturals, which comes to the same thing even if you might need a few more stages to get equivalence classes of Cauchy sequences etc), we want to have as many sets of reals as possible (for example, if you have two transitive models with the same reals, but the reals of M1 are a subset of the reals of M2, then M2 is closer to being correct).
 * But you'll easily see that taking more sets of reals, for the same underlying reals, can change CH from false to true, but never from true to false.
 * Another way to put it: Suppose there are two transitive models that both contain all the reals, and that disagree on the truth value of CH.  Then CH is true. --Trovatore (talk) 23:36, 12 November 2013 (UTC)
 * You might want to read "Believing the Axioms" parts 1 & 2 by Penelope Maddy. It's freely (legitimately, it seems) available. Just Google it. YohanN7 (talk) 00:34, 13 November 2013 (UTC)
 * I like that, your arguement/reply has an immediate intuitive appeal about it (a variant appears in the linked paper as well). Thank you both:-) I still have the second part of the linked article to read through, but it's very interesting. (I feel like this is a really weak reply, but I spent the better part of the day I set aside for sleep repairing my furnace in a very cold basement- every elaboration I've tried turns into philosophical ramblings...) Phoenixia1177 (talk) 08:30, 13 November 2013 (UTC)