studiot

These are reasonable explanations of the differnce between outcomes and events and general probability terminology. http://www.mathgoodies.com/lessons/vol6/intro_probability.html

No, I answered the general problem. To specifically calculate metacogitans figures you need to calculate the value of y at x = 0.5, say Y, from the equation for the circle, to obtain a second side to the triangle. The angle subtended at the centre is then twice the angle whose tangent is given by Y/X , or you can obtain the centre angle by calculating the chord and then using the cosine rule. So the equation of a circle the circle is (x-a)2+(y-b)2=r2. In this case a = b = r = 2 Solving for y leads to the two values, we want the lower or smaller one of 0.67712434. This leads to the tangent of the deflection angle (in radians) being 0.67712434/1.5 = 0.45141623 The deflection angle is thus 0.424031 radians The angle at the centre (theta) is twice this or 0.848062 radians The sine of this is 0.75 Thus the area of the segment is 2(0.848062-0.75) = 0.19724 area units. The area of the triangle is 1/2 base x height = 0.5*1.5*0.677124 = 0.507843 area units The difference is (0.507843 - 0.19724) = 0.311 area units. This method is accurate for all values and does not require calculus. Does this answer your question, Tar?

Daedalus, I answered this question, which I think is not the same as the original since that was not for a quarter circle

We acknowledge two types of probability distributions: Discrete and Continuous. For a discrete variable the summation sign (using capital sigma) is appropriate. For a continuous variable (which can but need not run from minus infinity to plus infinity but only between certain numbers say x=a and x=b) the integral is appropriate. Do you understand what the terms outcome and event mean in statistics?

@Tar For the quarter circle you describe The pesky LaTex bug has struck again so here is a scan.

The idea is that local vibration has ceased so there is no kinetic energy associated with this. This says nothing about general solid body translational kinetic or potential energy. This is why we cannot account the absolute internal energy of a body, only internal energy changes.

You need to be clear in your mind which is the dependent and which is the independent variable. The probability is the dependent variable. x is the independent variable in Bignose's case. Function's case is more complex since there are two independent varaiables (x and y) to get the area. Does this help?

There are some at Science Forums who seem to have elevated mathematics to the state of a religion with the catechism. Mathematics is necessary and sufficient for all Physics. I have challenged several of this view and each has so far failed to meet that challenge for the circumstance proposed. For instance no one has yet offered a mathematical formula for the production of concrete of guaranteed specific properties, whilst I can offer a practical Physics procedure to achieve this. Mathematics and Physics are different, if they were not there would be no point separating the disciplines, and there is not doubt that good mathematics is of the utmost importance in good Physics. But is it either necessary or sufficient? (Do you know the difference?) Well consider the number 10.3572 (any number will do) but I like that one. A simple contrast might be: Physics is about process, Mathematics is about result. Mathematics doesn't care about Physics cares about how you got there, but less or naught about the result, Mathematics cares about the result and its difference from 10.3571 or 10.3573. Together that is a powerful combination. That is why it is easy to construct examples where there is no mathematical definition/description of the process, but there is a Physics one. Similarly Physics couldn't care less if there is a difference between the the empty set of sequences and the sequence of a single term '0'.

Two comments. There are also a lot of real world measurements of similar experiments these days as well as substantial hard engineering based on them. One of the weirdest conclusions of relativistic time dilation is the necessity to abandon simultaneity as an absolute. This means that two observers will not necessarily agree that two events they observe are simultaneous. ie one may see the two events as simultaneous, the other may not.

Others have already observed that you bijection statement is unsupported. For some examples see my post#8 here 1+1=2 and hypothesis - Speculations - Science Forums http://www.scienceforums.net/topic/84274-112-and-hypothesis/ Yeah +1 The edit button doesn't seem available at the moment but you might like to look at Professor Ian Stewart's semi mathematical book From Here to Infinity. Go well

Congratulations on seeing immediately the missing point. +1 ps Collisions between particles are elastic or inelastic.

My apologies, I seem to have swopped density for mass half way through my analysis. I will put that right when I can.

Sometimes, but not always. For example see post#4 Rule 4 here, where 1+1=1 http://www.scienceforums.net/topic/84417-boolean-equation-truth-table/ There are other principles and processes. For example "The triangle is the smallest closed polygon." "The Hairy Ball Theorem" etc

Not to me it doesn't indicate the above conclusion. What it indicates is that c is an asymptotic value to some curve. There are many other such examples in applied maths. All travelling entities follow the same laws of motion; At one end of the velocity curve (the low velocity end) the curve may be approximated by a straight line (as with many curves), which is equivalent to saying that the laws of motion are Newtonian (or Galilaean). At the other (high) end the velocity curve becomes asymptotic to a line defining c and Einstinian (or Lorentz) laws of motion hold. That's all there is to it - It's the same for all cats.

I liked your analogy delta1212 +1

I am glad you are beginning to see the light, but I sorry to tell you that my comments only addressed the beginning of the problem and you have quite a way to go to fully sort it. It would make your life a lot easier for future problems if you tried out the questions I was suggesting, in the order I suggested them. There are three and only three forces acting on the body. All three of these may be replaced by a single resultant. Two of these are a right angles. Your question asks for the coefficient of friction, which is the ratio of two of them. It is, however very, very important to realise that you either have the resultant acting or the system of three forces, but not both. It is common to show the resultant dashed or in another colour or distinguished in some other way because of this. The resultant is the sum of all the forces acting (in this case 3) it replaces them as an alternative. This leads to the theorem that if the resultant is zero then the forces are in equilibrium, or that the acceleration of the body is zero, by Newton’s second Law. You need this fact to complete the problem. Turning to your use of x and y directions. It is conventional to use x for the horizontal direction and y for the vertical. So if an examiner sees them, that is what he will think. The acceleration given is parallel to the plane, not vertical ie not along the x or y axis. Can you tell me what is the acceleration perpendicular (normal) to the plane? You then have two accelerations, one parallel to the plane and one perpendicular to the plane and you can write two Newton’s second law equations to obtain the two forces you require to find the coefficient of friction. How are we doing ?

So Duhem didn't understand Maxwell, which is probably why he ends up as a footnote if mentioned at all. Duhem had many mistaken ideas. http://en.wikipedia.org/wiki/Pierre_Duhem You seem to have a remarkable reluctance to acknowledge my contributions to your thread. Why is this?

Perhaps the word resultant force instead of net force might make it more clear?

Since there is only one temporal dimension, any proposal can only look at it fractally in a Cantor's dust sort of way. https://www.google.co.uk/search?hl=en-GB&source=hp&q=cantor+dust+fractal+dimension&gbv=2&oq=cantors+dust&gs_l=heirloom-hp.1.3.0i10j0i22i30l9.1265.3640.0.6422.12.12.0.0.0.0.156.1408.1j11.12.0....0...1ac.1.34.heirloom-hp..0.12.1408.5-hF1xKtqq4 does this help?

Yes and the body is accelerating So it is not in equilibrium So your free body diagram should not be an equilibrium diagram This is what swansont (and your book) means by 'net force' So can you write Newton's second law for the body in the direction of down the plane? Remember that Newton's second law is a vector equation, so the force and the acceleration have direction.

Consider the two equations x = 3 2x = 6 These equations are really one and the same equation, since we can get the second by multiplying the first through by 2. We say they are dependent. Dependent equations may or may not be solvable. The ones above are since there is one equation and one unknown but x+y=3 2x+2y=6 cannot be solved since there are now two unknowns and only one equation. The second is still double the first and still dependent. Now consider x=3 x2= 1 These two equations are not consistent since either can be solved, but there is no solution that satisfies both.

Hi Nicholas If you do more maths, you will find that it is not always so easy. But yes you have detailed two approaches to solutions. Did you also have any thoughts on the actual question?

Imatfaal you may wish to know that Chikis is also working on this question in this thread, or at least the presentation. (This is not a complaint.) http://www.scienceforums.net/topic/84559-how-do-i-show-the-underline-when-am-solving-equations/

The equations are dependent if one is a multiple of the other That is if there is one number I can multiply one equation through to get the other. What happens if you multiply the second equation by 3? Can you now solve the system?

Where does this force come from. That is : what provides it or why does it exist? And how does this fit in with Newton's Laws? That is : which of Newton's laws says that a body 'has a force because it is moving'? Swansont has already provided one strong hint. IMHO it is more important that you properly understand this point than that you get the correct numerical answers to this problem (which you will do easily once you get the point).