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Force on a massless inextensible string passing around a pulley

  1. Aug 20, 2013 #1
    oft14.gif
    I just can't get why for a massless inextensible string passing around a frictionless pulley we easily take T1=T2 (in fig.).My profs. never did explain this & jumped on further & when i ask them they won't answer saying it was very basic and other students say it's obvious. pls. help
    also pls tell me when can the tensions be unequal i.e if friction in pulley would do so etc.
     
  2. jcsd
  3. Aug 20, 2013 #2
    If there is no friction (this is the important thing !!!!) then the tension in the string is the same everywhere.
     
  4. Aug 20, 2013 #3
    yes, tension will be same but i don't get why so?
     
  5. Aug 21, 2013 #4
    Have you learned how to do a moment balance yet? If so, take moments about the axis of the pulley with T1 on one side and T2 on the other side. If there is no friction at the bearing of the pulley and the pulley is in equilibrium, what does that tell you about the relationship between T1 and T2?
     
  6. Aug 21, 2013 #5
    by 'equilibrium' do you mean that the pulley is not rotating about its axis.
    and how can we conclude that
     
  7. Aug 21, 2013 #6

    PhanthomJay

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    The pulley may be rotating, however, the pulley is also assumed massless and bearing frictionless if the massless inextensible string is to have equal tensions on both sides. If the pulley has mass, or friction in its bearings, or both, the tensions generally will not be equal for the rotating pulley case.
     
  8. Aug 21, 2013 #7

    arildno

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    PhantomJay:
    It is irrelevant whether the pulley has mass or not, or whether the pulley is rotating; if there is no friction between the massless rope and the pulley, then the tension in the rope is necessarily constant.
    Why?
    Let the tensile force acting on a rope segment (covering the angular interval ([/itex](\theta,\theta+d\theta)[/itex]) at angle [itex]\theta[/itex] be [itex]\vec{T}(\theta)=-T(\theta)\vec{i}_{\theta}[/itex], whereas the tensile force at [itex]\theta+d\theta[/itex] be [itex]\vec{T}(\theta+d\theta)=T(\theta+d\theta)\vec{i}_{\theta+d\theta}={T}(\theta+d\theta)(\vec{i}_{\theta}\cos\theta-\sin(d\theta)\vec{i}_{r(\theta)}\approx({T}(\theta)\vec{i}_{\theta}+d\theta(\frac{dT}{d\theta}\vec{i}_{\theta}-T(\theta)\vec{i}_{r(\theta)}[/itex] regarding [itex]d\theta[/itex] as small.

    The sum of the tensile forces equal therefore, to order [itex]d\theta[/itex]: [tex](\frac{dT}{d\theta}\vec{i}_{\theta}-T(\theta)\vec{i}_{r(\theta)}[/tex]

    Since no frictional forces act upon the rope segment, we have, at arbitrary angle [itex]\theta[/itex], the equation:
    [tex]\frac{dT}{d\theta}=0[/tex]
    that is, T equals some constant, irrespective of pulley mass.
     
  9. Aug 21, 2013 #8

    PhanthomJay

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    A frictionless pulley is usually understood to be pulley with no friction in its bearings or axle, not friction free between rope and pulley grooves. If the pulley is accelerating angularly and has mass, then since net Torque= I(alpha), then T1 and T2 cannot be equal. If the pulley is massless, then I = 0, net torque is thus 0, and therefore T1= T2 for the massless case, or for the equilibrium case where alpha is 0.
     
  10. Aug 22, 2013 #9


    i can't understand parts of your comment & therefore can not really get what you are saying.
    So can you please include an image or something.I'd appreciate the help.
     
  11. Aug 23, 2013 #10

    PhanthomJay

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    Getting back to your original post, I don't think it is a 'very basic' or 'obvious' question at all.

    First off, there is confusion bewtween what is meant by a frictionless pulley. There are at least 2 types of friction to consider: one where there is axle friction at the axle bearings if the pulley rotates. The other is traction friction between the rope and pulley surface, a necessity for pulley rotation, without which the pulley cannot rotate, and instead the rope would just slip around it. Most problems do not clearly identify this distinction.

    When m1 and m2 in a simple Atwood machine are equal and at rest, it doesn't matter about whether or not there is axle friction or traction friction, or whether the pulley is massless or not. In all cases, the tensions are the same on both sides of the pulley, based on equilibrium consisderations and Newton's first law applied to a free body daigram of either mass.

    When m2 is greater than m1, here is where it can be confusing. If the pulley/rope interface is smooth and completely frictionless, the pulley will not rotate because there is no means of transferring the tangential force from the cord to the pulley to cause a torque on it. In which case, the rope will slip and the cord tensions will be the same on both sides of the pulley, as arildno has noted, independent of the pulley mass, and independent of axle friction also, since there is no rotation.

    But if there is friction between pulley and rope (static 'traction' friction), but no axle friction, the pulley will rotate, and if the pulley is massless, tensions on both sides will still be the same, but if the pulley has mass, tensions will not be the same because a net torque is required to rotate and accelerate the pulley. The friction force does no work, hence, it need not be considered in calculations for acceleration or speed using Newton 2 or energy methods.

    Then there's the case of axle friction, which will slows the rotation and must be considered as applying a counteracting torque, and unequal tensions in the cords.

    Most problems when talking about frictionless pulleys imply no axle friction. Most also assume that the pulley rotates without slipping and thus imply that there is traction friction. When a pulley has mass, the problem will so state its size (radius) and mass, so that its rotational inertia can be calculated.

    Obvious and basic?? I don't think so.
     
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