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Pulleys moving down with constant speed 
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#1
Nov1911, 07:49 PM

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1. The problem statement, all variables and given/known data
Please refer diagram attached. The pulleys are moving down with constant speed u each. Pulleys are light , string ideal. What is the speed with the mass moves up? 2. Relevant equations 3. The attempt at a solution Now, I got the answer as ucosΘ, obviously. But the answer is u/cosΘ. the solution too is given. But I feel that the latter is wrong as the velocity goes to to infinity as the mass goes upwards. Please tell me which is correct. 


#2
Nov2011, 12:40 AM

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Why don't you explain your reasoning for the 'obvious' result, u*cos(θ)?



#3
Nov2011, 12:48 AM

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P: 10,344

If the halflength of the rope between the pulleys is L, the mass is at depth y=L*cos(Θ), and dL/dt = u. The rate of change of y is dy/dt=d(L*cos(Θ))/dt. Both L and Θ are changing as the rope is pulled while x = LsinΘ stays constant. If you take these into account you will get the correct result, that the mass raises wit u/cosΘ.
ehild 


#4
Nov2011, 01:13 AM

P: 34

Pulleys moving down with constant speed



#5
Nov2011, 01:17 AM

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#6
Nov2011, 01:25 AM

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P: 10,344

ehild 


#7
Nov2011, 07:26 AM

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#8
Nov2011, 07:30 AM

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#9
Nov2011, 07:36 AM

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#10
Nov2011, 09:52 PM

P: 34

OK, I see now. Using Pythagoras theorem we can relate the legths as
[itex]DA^{2}[/itex]=[itex]DC^{2}[/itex]+[itex]AC^{2}[/itex] then differentiating, [itex]2DA\frac{dDA}{dt}[/itex]=[itex]0+2AC\frac{dDC}{dt}[/itex] And finally the answer will come u/cosθ. Right? 


#11
Nov2011, 11:08 PM

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ehild 


#12
Nov2111, 04:35 AM

P: 34

Yes, thank you.



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