If the superconductor starts off with no initial current, and a magnet is brought near, then one will be induced in the superconductor. (see Faraday's law of induction). This current that is now set up in the superconductor will persist with little to no resistance (a property of a...
Yes, I see how the F parallel comes out to be larger than mg if one is allowed to construct that triangle. Perhaps that is the reason? Mg and F parallel are vectors, whereas the bottom of that triangle is not. Maybe since the vectors are in vector space and the bottom of the triangle is not...
Thank you I will look into this a little more. As I understand geometry, as long as you can validly construct the triangle, it cannot be a 'wrong' triangle. Any and all triangles that I could construct validly using the information given should lead to the same result. I know there's an error...
Yes, I see how using other triangles gets you to the correct result of mgsin theta for F parallel. I suppose my question is why can one not use the triangle in the image below? It is certainly just as valid of a triangle as any other, and all rules of trigonometry to my knowledge have been...
I agree, sin theta is equal to opposite divided by hypotenuse. But opposite of theta in the picture is mg, and the hypotenuse is F parallel. So that gives us Sin theta = mg/F parallel. Which makes F parallel equal to mg/sin theta...do you see the dilemma?
I know that force parallel (the force of the block down the inclined plane) in this case should be mgSin(theta), but why is the derivation in the picture wrong? According to that triangle could it not be mg/(Sin theta)?