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Question about a 2 mass, multiple pulley problem. 
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#1
Aug304, 07:01 PM

P: 11

I was working my way through a series of problems when I came up against one that has me stuck. If anyone here can assist I would certainly appreaciate it. I found a exact copy of the problem online here: http://www.faqs.org/docs/Newtonian/Newtonian_145.htm
Problem #13 I have tried several differant methods for solving this and most commonly I come up with a upward acceleration of g. Never anything less then g. Am I missing something obvious? 


#2
Aug304, 08:10 PM

P: 699

when I solved it I got g/4. The easy way is to redraw the situation, thing of the weight on the right as pulling on the one on the left, because there is a pulley system in the middle and the rope to the one on the right is going through it entirely, then you need half the force to lift the fulley on the left. You have masses of m, and the force pulling in this case is mg. So, the overall force on the pylley system is 1/2mg since the one on the left is pulling at 1/2mg down, and the one on the left is pulling at 1mg down.
F = 1g  1/2g F = 1/2g m = m + m = 2m Fnet = ma 1/2mg = 2ma 1/2g = 2a g/4 = a, therefore the acceleration in this case should be g/4 towards the left, pulling the weight on the left upward at that acceleration. The answer they say is g/5, I dont know where I went wrong or weather they are wrong. 


#3
Aug304, 11:20 PM

Astronomy
Sci Advisor
PF Gold
P: 23,227

the answer they have in the book let me look at your solution and see what I think then 


#4
Aug304, 11:42 PM

Astronomy
Sci Advisor
PF Gold
P: 23,227

Question about a 2 mass, multiple pulley problem.
nenad i dont understand your notation so instead of me trying to criticise your solution how about you try to find what is wrong in my solution.
they say leftmass experiences upwards accel of x (and therefore ropeforce on leftmass is upwards (g+x)M) but because of pulleysystem when leftmass goes up one inch then rightmass must go down two inches. therefore rightmass experiences downwards accel of 2x (and therefore ropeforce on rightmass is upwards (g2x)M) now we have to equate forces Two times the tension in the right rope equals the tension in the left rope. so (g+x)M = 2(g  2x)M g+x = 2g  4x 5x = g x = g/5 


#5
Aug404, 05:26 PM

P: 699

ya, the way you did it was right, but I dont do it using that method of tension of ropes. I used acceleration of the system. You find the forse on each block, then you fing the total force on the system by subtractiong the two, and then fond the acceleration. I still dont know where I went wrong.



#6
Aug404, 05:54 PM

Emeritus
Sci Advisor
P: 7,630

The easiest way to get a solution IMO is the Lagrangian formulation, same as all pulley problems
http://en.wikipedia.org/wiki/Lagrangian_mechanics Of course they don't teach this formulation till after you have had a bunch of pulley problems, because it makes them too easy Let h be the height of the mass on the left. If we move the left mass up h, the right mass drops twice that distance, it's height lowers by 2*h Then the total kinetic energy T = .5*m*(hdot^2+4*hdot^2) (the mass on the right is always moving at twice the velocity of the one on the left, so it has 4 times the kinetic energy) The total potential energy is V = m*g*h  2*m*g*h/2 = m*g*h So L = TV = 5/2*m*hdot^2 + m*g*h [tex] d/dt (\frac{\partial L}{\partial hdot}) = \frac{\partial L}{\partial H} [/tex] 5*m*hdotdot = m*g so hdotdot, the acceleration of the mass on the left, is positive (upwards) at g/5. 


#7
Aug504, 08:45 AM

Mentor
P: 41,454




#8
Aug504, 10:34 AM

P: 699

yes, thanks Doc Al, my mistake.



#9
Aug904, 12:20 PM

P: 11

Thank you everyone for the assistance, It was definatly the aright=2aleft factor that I was missing, I also ended up having to review more basic Atwood machine problems before I fully grasped it. Thanks again that one was driving me crazy.



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