Archive: 11 posts

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Ok so before i even got to uni they gave me maths questions! Now before you try and work these out, you will probabaly have needed to have done some kind of maths or engineering degree or further maths a-level. You will also have to like mechanics. Ok here is one of them, as written: a) The frictional resistance to the motion of a train is 'k' times the combined weight of the engine and its coach. An engine of mass 'M' works at a constant power 'P' and attains a maximum speed 'U' when pulling a coach of mass 'M' up a hill of inclination 'alpha' to the horizontal. Down the same hill with a coach of mass 'M/2', the engine can attain a maximum speed '2U'. Show that 'P' is '12kMgU/5' b) The engine pulls a coach of mass '3M/2' on level ground. Find the acceleration of the train when the speed is 'U/5' and show that the maximum speed is '(24/25)U' If anyone manages to find the solution, I might be asking you more!

2009-09-16 13:47:00 Author: ladylyn1 Posts: 836

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Part 1 looks like normal Maths A-Level to me, bit of Mechanics 1. Unless there's something nasty after you formulate the equations. Anyway, you are going to Cambridge right? What exactly did you expect

2009-09-16 14:07:00 Author: rtm223 Posts: 6497

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Oh dear... I'm not going to have to learn these am i Here's my blind stab at it, you have to work out the power needed to get up the hill with Kx(2M) (seeing as the coach and the engine both weigh M at this point). Now from the talks with the headmaster I know Power = Strength.Speed so your P will equal U or 2U x the strength needed to overcome your friction. Then once you've got that you do the same for going down and then maybe do a similtaneous equations or something Tell me how I would've done lol

2009-09-16 16:14:00 Author: Shermzor Posts: 1330

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You missed out taking into account the effect of gravity, based upon a slope of angle 'alpha', although the trig functions will cancel out at some point (really helps when they give you the answer lol). All you really need to do for 1 is balance the engine, frictional and gravitational forces at the equilibrium points described and then rearrange. I may actually have a stab at this later on this evening. I've not done any mechanics in 6 years, so it could be interesting. I assume you are giving them a shot yourself lyn?

2009-09-16 16:28:00 Author: rtm223 Posts: 6497

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Oh silly me I'm printing off this question to see if Garry can work it out

2009-09-16 17:51:00 Author: Shermzor Posts: 1330

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Sounds like a fun problem. The hard part is not doing the math, but sorting out the mechanical relations between all terms. The frictional force (which opposes motion) is equal in magnitude to kMg*cos(a) for just the engine, and k(m+M)g*cos(a) for the train+coach. They are proportional by the constant k to the normal force of the object/system, which is found by finding the perpendicular component of the force of gravity [hence Mg*cos(a) and (m+M)*cos(a)] I really don't feel like solving the problem because I really can't be bothered, but try to find the net force of the train in the direction of it's motion, etc. Try drawing free-body diagrams to organize your forces.

2009-09-16 18:00:00 Author: comphermc Posts: 5338

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Hm well i've done up to Mechanics 3 so unless im having a real blond moment i still cant work it out... Its all this Little Big Planet over summer thats drained my maths ability lol! Anyway, ill start with the start... the question states that 'The frictional resistance to the motion of a train is 'k' times the combined weight of the engine and its coach' so on a hill of angle 'alpha' it is 'k(M+M)gcos(a)' therefore '2kMgcos(a)'... Ok then the weight acting down the hill: 2Mgsin(a) Power = Force*velocity... therefore P = (2kMgcos(a) + 2Mgsin(a))U for the uphill section. So i think thats right... Now downhill (this is where im getting stuck) weight acts downhill but friction acts uphill. Force of engine also acts downhill. Total weight now is 3Mg/2. Weight acting downhill: (3Mg/2) * sin(a) Friction acting uphill: (3Mgk/2) * cos(a) Force of engine powering downhill = ? Power = force x velocity and the forces must balance... hm stuck

2009-09-16 22:24:00 Author: ladylyn1 Posts: 836

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Actually, I'm stuck on the same problem! From the force balance, I assumed the downhill equation could be set up as: P + (3/2)mgusin(a) = (3/2)mgukcos(a) This pairing should be legit, as the units match (kgm^2s^-3) Down: P = (3/2)mgkucos(a) - (3/2)mgusin(a) And going up, P = 2kmgucos(a) + 2mgusin(a) The problem is that from those simultaneous equations, I can't remove the angle, so I've done something fundamentally wrong. I ended up with P = (8/5)mgkucos(a), not (12/5)mgku. Its amazing how holidays turn me into an idiot! D:

2009-09-17 02:00:00 Author: Sraaz Posts: 1

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Too ... complex ... for ... my .. 12 ... year ... old ... mind ... *splodes*

2009-09-17 02:18:00 Author: TheMarvelousHat Posts: 542

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Ok i got the solution! Thanks for the help guys!

2009-09-17 13:38:00 Author: ladylyn1 Posts: 836

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I've always hated math. I don't like it. I can't calculate. I can't I won't I WILL NOT have anything to do with this! ARRRGH!

2009-09-17 17:32:00 Author: BasketSnake Posts: 2391

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