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Two pulley and three mass system

  1. Sep 2, 2016 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A light pulley can rotate freely about its axis of symmetry which is fixed in a horizontal position. A light inextensible string passes over the pulley. At one end the string carries a mass 4m, while the other end supports a second light pulley. A second string passes over this pulley and carries masses m and 4m at its ends. The whole system undergoes planar motion with the masses moving vertically. Find the acceleration of each of the masses.

    2. Relevant equations

    Newtonian/ lagrangian approach

    3. The attempt at a solution
    The lagrangian approach is probably easier but I thought I'd try to crank out the answers using Newtonian mechanics. I would show a picture of the set up but the menu above does not seem to have an option unless I provide a URL. For the far left mass, I have $$T_1 - 4mg = 4m \ddot{x} = 4ma_1,$$ where ##a_1## is the acceleration of this mass relative to an inertial frame. Similarly, for the other masses attached to the second pulley, we have $$T_2 - mg = m_{2}a_2\,\,\,\,\text{and}\,\,\,\,T_2 - 4mg = 4ma_{3}$$ where ##a_{2,3}## are the accelerations of masses 2 and 3 relative to an inertial frame. Now, ##T_1 = 2T_2##. Write ##a_{i,P}## as the acceleration of the masses relative to the mobile pulley. So $$a_{i} = a_{i,P} + a_{P}$$ where ##a_P## is the acceleration of the mobile pulley relative to an inertial frame which is equal to -a_1. Rearranging the two equations for masses 2 and 3 above give $$T_2 = \frac{T_1}{2} = m(a_{2,P} - a_1 + g)\,\,\,\,\,\,\,\text{and}\,\,\,\,\,\,T_2 = \frac{T_1}{2} = 4m (a_{3,P} - a_1 +g)$$ Subsequently subbing in ##T_1 = 4m(a_1+g)## gives me 2 eqns with 3 unknowns.

    A final constraint comes from the fact that the strings are all inextensible so that ##a_{2,P} = - a_{3,P}## Then ##a_{3} + a_1 = -(a_2 - a_1)## so that ##a_3 + a_2 + 2a_1 = 0##. We now have three eqns and three unknowns so can solve.

    My answers are ##a_1 = -g/9, a_2 = -5g/9## and ##a_3 = 7g/9##. My answer for ##a_1## is correct but I seem to have the answers for ##a_2## and ##a_3## mixed up. Could be a mistake in the book but I just wanted to check that my reasoning above is sound - especially regarding what I have done relating accelerations from inertial frame to the non inertial accelerating frame of the mobile pulley.

    Thanks!
     
  2. jcsd
  3. Sep 2, 2016 #2

    TSny

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    I believe everything is OK up to where you say:
    I don't see how you got the first equation. Also, the second equation does not follow from the first equation.
    (I think you probably just made a typo in the first equation. The second equation looks correct. When I solve your equations, I get that a2 and a3 are switched from what you got.)
     
    Last edited: Sep 2, 2016
  4. Sep 3, 2016 #3

    ehild

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    @CAF123 You can draw the picture, copy it and paste into the post. Or you can click on "UPLOAD" and choose file from your computer.
    upload_2016-9-3_7-1-36.png
     
    Last edited: Sep 3, 2016
  5. Sep 3, 2016 #4

    CAF123

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    Thanks TSny, I see that I made a typo in the first equation and an algebraic error later on, so I now agree that my initial answers were indeed mixed up. I just wanted to ask a few more conceptual things about the equations I wrote down if that is ok -

    I measure x,y,z to be the displacements of the three masses relative to the ground (inertial frame, see picture). Is it correct to say that since in an inertial frame the only forces that come into play here are tension and gravity (i.e real forces, not pseudo/fictitious) the equations I write for the masses, namely ##T_1 - 4mg = 4m \ddot{x} = 4ma_1,## ##T_2 - mg = m \ddot{y} = ma_2\,\,\,\,\text{and}\,\,\,\,T_2 - 4mg = 4m\ddot{z} = 4ma_{3}## are all valid to an inertial frame? I initially took the latter two equations to be correct relative to the mobile pulley but changed my mind based on the reasoning above - just wanted to check if it is ok.
    Since I am making a transformation between non inertial frame and an inertial one, I don't think what I write goes under the remit of a 'galilean transformation' only that if the mobile pulley is moving at constant acceleration relative to the inertial frame (which it must if the strings are inextensible) , it takes the 'form' of a galilean transformation.

    The equation ##a_2 = -a_3## does not hold because if, say, the far left 4m mass goes up then the mobile pulley must go down. If the m mass attached to the second pulley goes up then the 4m mass attached to the second mass must go down. So, in a sense this 4m mass shares a common direction of velocity as the direction the mobile pulley is moving giving rise to the following series of equations:
    $$ T_1 - 4mg = 4m a_1, \,\,\,\,\, T_2-mg = m (a_2-a_1),\,\,\,\,4mg - T_2 = 4m(a_3+a_1),\,\,\,\,2T_2=T_1$$ (or any equivalent series of equations in which I assume the far left mass goes down say and mobile pulley up etc...) I noticed that in the latter two equations, ##(a_2-a_1)## and ##(a_3+a_1)## have a different sign of ##a_1## which is not present in my transformation law ##a_{i} = a_{i,P} + a_{P} = a_{i,P} - a_1##.

    Thanks for any comments!
     

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  6. Sep 3, 2016 #5

    TSny

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    Yes, the equations as written assume that you are using an inertial frame. The frame of the moving pulley is not an inertial frame.

    The acceleration equation ##a_i = a_{i,P} + a_P## has the form of a Galilean transformation of velocities. But it is just a kinematical relation that does not require any of the frames to be inertial. For a particle ##p## moving relative to two arbitrary frames ##A## and ##B##, you would have the relation ##a_{p, A} = a_{p,B} + a_{B,A}##. And this would be valid even if ##A## and/or ##B## are noninertial frames. The only requirement is that you assume Newton's absolute time which is the same for all frames.

    I'm not sure what to comment on here. Your reason for why ##a_2 \neq -a_3## sounds right. Your first two equations are written with positive directions upward; whereas, the third equation is written as though you are taking downward as the positive direction. This could have something to do with the difference in sign of ##a_1## in these two equations. In the second equation, did you mean to write ##a_{2,P}## instead of ##a_2##? In the third equation, did you mean to write ##a_{3,P}## instead of ##a_3##?
     
  7. Sep 3, 2016 #6

    CAF123

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    I see, thanks. So in the special case where A and B are both inertial (i.e moving at constant velocity wrt each other) then the transformation is said to be a galilean transformation.

    Ah yes, sorry it was for the second equation but I think I made an error in the third. Take upwards to be positive throughout and assume the mass m goes upwards and 4m downwards - then the first mass m attached to the moving pulley has equation ##T_2-mg = m(a_{2,P} - a_1)## and for the second mass, ##T_2 - 4mg = 4m(a_{3,P} -a_1) = 4m(-a_{2,P} - a_1)##.
     
  8. Sep 3, 2016 #7

    TSny

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    Yes. If the frames are both inertial, then the acceleration of the particle will be the same relative to the two frames (since the frames have zero acceleration relative to each other).

    That looks good.
     
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