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Force does work -- Two masses and a pulley system...

  1. Aug 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Two blocks ##A## and ##B## having masses ##m_1= 1 kg, m_2 = 4 kg## are arranged as shown in the figure, The pulleys ##P## and ##Q## are light and frictionless. All the blocks are resting on a horizontal floor and the pulleys are held such that strings remain just taut.

    At moment ##t=0## a force ##F = 30 t (N)## starts acting on the pulley P along vertically upward direction as shown in the figure. Calculate,

    (i) the time when the blocks A and B loose contact with ground.
    (ii) the velocity of A when B looses contact with ground
    (iii) the height raised by A upto this instant.
    (iv) the, work done by the force F upto this instant.
    scan0001.jpg

    2. Relevant equations
    $$F=30t \\ m_1=1 kg \\ m_2=4kg \\ F=ma$$

    3. The attempt at a solution
    (i)
    FBD of pulley ##P##
    geogebra-export.png
    From the above image it is clear that $$T=10t$$

    FBDs of block ##A##, ##B## and the pulley ##Q##

    adha.png

    From the FBD of pulley ##Q##, we get
    $$T'=2T=20t$$

    From the FBD of block ##A##, we get
    $$N_1+T=m_1g$$
    As the block ##A## is about to loose contact with the ground the normal reaction form the ground vanishes, i.e. ##N_1=0##
    So, $$T=m_1g\implies 10t=1\times 10\implies t=1\text{ sec}$$

    From the FBD of block ##B##, we get
    $$T'+N_2=m_2g\implies 2T+N_2=m_2g$$
    Same as for block ##A##, block ##B##, when about to lose contact with ground ##N_2=0##
    $$\therefore 2T=m_2g\implies 20t=4\times 10\implies t=2\text{ sec}$$

    (ii)
    When ##B## loses contact with the ground, ##A## would have traveled with the acceleration ##a=10t## for ##t=1\text{ sec}##, i.e. from ##t=1 sec## to ##t=2 sec## so the velocity of the block ##A## as ##B## loses contact is found as follows:-

    $$V=\int_{0}^{1}{10(t)}dt=5\text{ m/s}$$

    (iii)When ##B## loses contact with the ground, ##A## would have traveled with the velocity ##a=10t## for ##t=1\text{ sec}##, i.e. from ##t=1 sec## to ##t=2 sec## so the displacement of the block ##A## from its initial position as ##B## loses contact is found as follows:-

    $$X=\int{dx}=\int_{0}^{1}{vdt}=\int_{0}^{1}{(5t^2)dt}=\left(\dfrac{5}{3}t^3\right)^{1}_{0}=\dfrac{5}{3}m$$

    (iv) Finally, the question where I am having trouble.

    As the force acting on the system is ##F=30t##, but the force that pulls the block ##A## through a distance is the tension, so the work done should be $$\int_{0}^{1}{T.dx}=\int_{0}^{1}{(10t).(5t^2)dt}=\int_{0}^{1}{50t^3dt}=\dfrac{25}{2}J$$.

    In my calculation of the work done by the force ##F##, I found it odd that I was using ##T## instead of ##F##, but according to what I know is that that the work done by a force acting on a body is given by the dot product of the force with the displacement vector, so, here the only massive body being displaced is the block ##A##, and the force acting on it is the tension, so I considered tension as the force that acts directly on the block. So, why is my answer different from that of the book's.
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2016 #2
    You are being asked to calculate work done by 'F' so just do that . 'F' and 'T' are different forces acting on different objects doing different work . Displacement of massive body as a result of force doing any work is irrelevant .

    Do you believe the topmost pulley P on which 'F' acts does not move during the time when block A has been lifted till the time block B leaves the ground ?
     
    Last edited: Aug 30, 2016
  4. Aug 30, 2016 #3
    That's what I thought at first that before the block loses contact with the ground the topmost pulley had been moving, so whether to consider that in the displacement that had occurred due to the force ##F##.or not, or just the displacement of the block ##A## was to be considered. After, this what I thought would seem to be very weird but do clear it for me, so what I thought was that as the force does work on the system the system stores energy. This energy now has to be used so, only massive bodies can have energy, so that's why I ruled out the case where the displacement of the pulley had to be considered. The above line of thought might seem very weird but do clear it for me. Also, if all of this, that I thought, was for naught then does this mean that only the displacement of the topmost pulley has to be considered.
     
    Last edited: Aug 30, 2016
  5. Aug 30, 2016 #4
    Yes , only the displacement of the top pulley is taken into account for calculating work done by 'F' .

    According to Work Energy theorem ,the work done by a non conservative force ( F in this case) is equal to the change in the mechanical energy of the system under consideration . Now , you have to be clear about what your system is . If you treat the upper pulley as your system , since it is mass-less , net work done by all the forces will be zero . Work done by force 'F will be non -zero .

    If you treat the entire set up ( two pulleys +two blocks ) as your system , then external forces acting on the system are force 'F acting upwards and force due to gravity acting on the CM (Center of Mass)of the system . If W represents the work done by force F , then ##W =ΔPE + ΔKE ## . Potential Energy PE and Kinetic Energy KE refer to those of the CM of the system .

    Now when you talk about system storing energy , understand that only PE is stored in the system and that too only when conservative force is present . Just like gravity is present in this set up . So , yes , PE is stored in the system , but this PE refers to the work (negative work) done by gravity on the CM of the system. Second , when you say "energy has to be used " , it doesn't make sense .

    The energies refer to that of the Center of Mass of the system under consideration , not that of only massive bodies .
     
    Last edited: Aug 30, 2016
  6. Aug 30, 2016 #5
    Your explanation was as clear as crystal, the only thing that bugged me was did you consider the pulleys approximately mass less or they have no mass(or exactly 0 mass). Cause if its the latter one then the energy of the bodies with mass is only present and the pulleys have no energy, so the energy of the centre of mass is the same as that of the bodies with mass' energies combined. Am I right to conclude what I did o_O
     
  7. Aug 30, 2016 #6
    They are equivalent ( as far as maths is concerned ) . In reality nothing is massless . But when something is considered massless (or approximately massless ) it means its mass is taken to be zero .

    You are partially correct . The massless objects do not contribute as far as possessing mechanical energy is considered .

    The PE of the CM is equal to the sum of the PE of the objects comprising the system ( i.e sum of the PE's of objects having mass ) .

    But , The KE of the CM is not equal to the sum of the KE of the objects comprising the system .

    While applying Work Energy theorem , we are concerned with the KE of the CM , not with the KE of the bodies having mass .

    Suppose you consider only pulley P as your system , then since it is massless , it neither has PE nor KE . So when you apply Work Energy theorem , W =ΔPE + ΔKE , W = 0 . This means total work done by forces acting on the pulley is always zero between any two instants . The work done by force 'F' would be equal to the total (net ) work done by the three tension forces acting downwards .

    I guess you still haven't covered chapters on Work Energy and Center of Mass . When you study them , then you will be able to appreciate these things more .In the meantime enjoy applying Newton's Laws :smile: .
     
  8. Aug 30, 2016 #7
    According to what you said the work done by the force ##F## will be equal to the summation of the work done by the tension in the string, so the work done by the tension force is

    Work done by tension on ##A##:-
    $$\int_{0}^{1}{T.d{x}_{A}}=\int_{0}^{1}{10(t+1).{5t^2}dt}=\dfrac{175}{6}J$$

    Work done by tension on ##B##:-
    $$\int_{0}^{1}{T.d{x}_{B}}=\int_{0}^{1}{20(t+1).(0t)dt}=\int_{0}^{1}{0dt}=0$$

    So, work done by force ##f=30t## is ##\dfrac{175}{6}J##, am I right?

    I found it very strange, that the book has the same answer but its penultimate step in the hint given is ##\displaystyle \int_{1}^{2} {(30t).10\left[\dfrac{t^2}{2}-t+\dfrac{1}{2}\right]dt}##, which is equal to ##\dfrac{175}{2}##.

    Thanks for all your time and effort, it was a great help:wink:
     
    Last edited: Aug 30, 2016
  9. Aug 30, 2016 #8
    Good work !

    ##\dfrac{175}{2}## is the correct answer :smile: . The final book answer is incorrect . There should be a 2 instead of 6 :wideeyed: . ##\dfrac{175}{6}## is the magnitude of work done by tension ( only one ) on pulley P . The total work done by all three tension forces on the pulley is ##\dfrac{175}{2}## . Actually ##\dfrac{-175}{2}## J :wink: .

    I said work done by tension on the pulley , not on A and B . Remember our system is Pulley P :cool: .

    Your calculated values are correct as far as work done by tension on blocks A and B are concerned , but work done by a single tension (any one of three ) on Pulley P is ##\dfrac{-175}{6}## and total work done by all the tension forces is ##\dfrac{-175}{2}## .
     
  10. Aug 30, 2016 #9
    Okay...so that clears things up for me. Thanks again.
     
  11. Aug 31, 2016 #10
    On second thoughts , I think the correct answer should be ##\dfrac{175}{4}## i.e work done by force F on pulley P should be ##\dfrac{175}{4}## J , whereas work done by each tension on pulley should be ##\dfrac{-175}{12}## J.

    Why ? Because distance moved by the pulley is half the distance moved by block A (assuming no slipping of the string on the pulley ) .

    I would request the experts @haruspex , @TSny to kindly confirm the result i.e should ##\dfrac{175}{4}## J be the correct answer ?
     
  12. Aug 31, 2016 #11

    haruspex

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    Your answer is correct. The method used in the hint appears to be ##\int Fv.dt##. That is valid, but they used the wrong v. What matters is the velocity of the force. The force moves at only a third the speed of m1.
     
  13. Aug 31, 2016 #12
    haruspex ,

    Are you saying 175/6 J is the work done by force F ?
     
  14. Aug 31, 2016 #13

    haruspex

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    Yes.
     
  15. Aug 31, 2016 #14
    Ok . Thanks .

    And would you agree that real work done by tensions on pulley P would be ##\dfrac{+175}{18}## J (left string) , ##\dfrac{-175}{18}## J (middle) , ##\dfrac{-175}{6}## J (right) ?
     
  16. Aug 31, 2016 #15

    haruspex

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    A tension, or compression, is not a force. It results in a force at each end, so it is more like a pair of equal and opposite forces (though that gets trickier when you have an accelerating rope with mass, since the tension will vary along it). It follows that in the massless case, the two ends will do equal and opposite work, so the rope as a whole does none.
    In the present case, the rope clearly does no work on mass 2, while all the 175/6 J is done on mass 1. Since it does positive work on mass 1, that section of rope must do negative work on the pulley. So if you wish to think in terms of the work done on the pulley by three tensions they would be,left to right, 0, +175/6, -175/6.
     
  17. Aug 31, 2016 #16
    @TSny ,

    What do you think about post 14 ?
     
  18. Aug 31, 2016 #17

    TSny

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    I'm not confident thinking about the work done on pulley P "by the left string" or "by the right string". I'm more comfortable thinking about the work done by the contact force of the string on the pulley along the rim of the upper half of the pulley. Each infinitesimal section of string that is in contact with the rim exerts some force on the pulley. Integrating the vertical component of the force over all the sections that make contact with the pulley will give the net downward force on the pulley due to the string that passes over the top of the pulley. It is easy to show that this net downward force on the pulley is 2T, where T is the tension in the string. In addition, the section of string that is attached to the center of the pulley exerts a downward contact force of T on the pulley. So, the net downward force on the pulley due to the string is 3T, as perhaps was expected. The net work done by the string on the pulley is ∫(-3T)dyp, where dyp is the vertical displacement of the pulley.

    It seems to me that "technically", the "left" and "right" strings are not making contact with the pulley, and so are not doing any work on the pulley. Only the portion of the string that passes over the top of the pulley is making contact with the pulley (and also the portion of the string attached to the center of the pulley makes contact with the pulley).

    Anyway, that's how I Iook at it.
     
  19. Aug 31, 2016 #18
    Thank you . Your reply has cleared up few things that were bothering me :smile:
     
    Last edited: Aug 31, 2016
  20. Aug 31, 2016 #19
    Could you please show how you get this result .

    Thanks
     
    Last edited: Aug 31, 2016
  21. Aug 31, 2016 #20

    haruspex

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    I can understand that, but one can instead treat the section of rope in contact with the pulley at a given instant as being part of the pulley. The force exerted on that by the left-hand string does not move, so does no work.
     
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