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A confusing torque phenomena

  1. Aug 16, 2008 #1
    to be able to understand my point of view please check this picture which I attched>>

    the question is in the picture please check it

    thanks
     

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  2. jcsd
  3. Aug 16, 2008 #2

    Doc Al

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    I'm looking at the diagram but I don't understand the setup. Can you explain it a bit?

    Is that rod supposed to be horizontal?
     
  4. Aug 16, 2008 #3
    how F2 is in the opposite direction of the motion of the disk is it posssible ??

    in classthe proffesor called F2 as the tangential force that cause the resistive torque

    can we say that f2 is not the only force that is acting on disk2 ?

    and can we say as well that f2 must be accompagned by another force that is equal to it but in the opposite direction to balance the troques in disk2 so it can rotate with constant angular velocity ?
    if yes where from does this second force come ? - is it from disk 1 ??

    please i need a very detailed explanation on how can we rotate the disk2 with constant angular velocity with only having F2 i THINK it is not possible there must be another forces that are acting on disk2 in order to balance the torques and to get constatn angualr velocity , what is the origin of the second force that is acting on disk2
     
  5. Aug 16, 2008 #4
    the setup is :
    2 gears in a shaft where there is an imput gear that transmit power through the shaft to the other gear .

    the poblem describe F1 AS the acting tangential force on disk 1 , and the problem is giing u a given that is constituted by : constant angular velocity , 100 kw of power , and F1 tangential of 100 kN,

    the requirement is : F2 THE FORCE THAT is causing the resristive torque on disk2 .

    my question is:

    if we take disk2 alone , we obviousely see that it is rotating with constant angular velocity so the angular accelration is 0 , but the confusion is , how can we have an angular accelration = to 0 if we have 1 tangential force on disk2 , isnt it impossible cause we know sum of torques = 0 , and if we have 1 force that casue 1 torque then the sum of torques would be not eqaul to zero but in the same time we see that the disk2 is rotating with constant angular velocity , it is impossible we got to have another force on disk2 to make an equilibrium of torques.
     
  6. Aug 16, 2008 #5

    Doc Al

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    These two disks are connected by rigid rod (I presume). I further presume the rod is held in position by frictionless bearings, such that it is free to rotate. So you can view the entire system--both disks plus the rod--as a single rigid body. If the only torque-producing force exerted on the disks is F1, then the system will have an angular acceleration. If F1 is balanced by F2, such that F2 produces a counter torque, then you can have no net torque on the system.

    Realize that the connecting rod allows a force exerted on one disk to affect the other.
     
  7. Aug 16, 2008 #6

    Doc Al

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    If you view the disks as separate bodies, then you are right: F1 is not the only force on disk 1; the rod itself is exerting a torque on the disk. Same goes for disk 2.
     
  8. Aug 16, 2008 #7
    first >. ur assumptions are `100 % right

    so you are saying that that F1 will affect f2 and f2 will affect f1 in reverse porcess so we would have equilibriumed torques and sum of torques will be equal to zero ???
     
  9. Aug 16, 2008 #8
    my second question is :
    how does the rod perform a second force or an another way an inverse torque on diske 2 ?

    is it due to the transmission of F1 to disk2 number 2 ?

    secondly : the resistive torque is does it exist on both disks or only on disk 2

    if it exist on both disks does that mean that the resistive torque on disk 1 is caused by disk2 and the rotation of disk2 is a responisbility of F1 ?

    and if we have 2 forceson eachdisks why can we say that F1 = power x velocity , why we chose f1 not the resistive torque on disk1 caused by disk2 is it cause the resistive torque on disk1 cause by F2 is eqaul to power x velocit in the spot of action of this resistive torque ?
    p.s : it is complicated and my english is bad , but I am very greatfull for ur patience .
    i AM STUDYING mechanical engineering , I know how to calculate and to sole all such problems it is piece of cake but I would love to know the systematic of acting forces..

    thx
     
  10. Aug 16, 2008 #9

    Doc Al

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    If the torque exerted on the system about the given axis by F1 is equal an opposite to the torque exerted by F2, then the net torque on the system is zero. The connecting rod links the two disks. (You don't really have to worry about the forces exerted by the rod on the disks as those forces are internal to the system and thus in themselves exert no net torque on the system. But they certainly do act!)
     
  11. Aug 16, 2008 #10
    Now I got it we have to consider the shaft as the body and 2 forces that are acting on it the first is F1 and the second is F2 and both F1 and F2 are opposite sides in order to make 2 torques of diefferent direction and same amplitude so the shaft can turn with constant angular velocity , what maight ocnfuse is taking every disk by itself , but when u study the body as a whole it is easy , thanks for u ur help .
     
  12. Aug 16, 2008 #11

    Doc Al

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    The force on one disk creates torsion in the rod that transmits a force to the other disk. Example: Hold a fork by the handle and give it a twist. How come the end of the fork turns when you only touched the handle? Same idea. Another example: Poke something with a stick. Clearly the force of your hand on one end of the stick is transmitted to the object you are poking.

    Looking at the system as a whole, I'd say that it makes more sense to say that the resistive torque acts on disk 2. (If I understand your problem correctly. But I'm no mechanical engineer.) But since they are connected, a resistive force on one disk affects the entire system.

    I suspect that they want you to calculate the power you have to apply to overcome the resistive force. F1 is the applied force; F2 is the resistive force (the "load") that you must overcome to make the system rotate. Since the net torque is zero, the applied power just equals the "reverse" power due to the resistive force.

    Let me know if this is making sense.
     
  13. Aug 16, 2008 #12
    it is true power in = power load which is power out ...

    and as u said looking at the system as a whole can make ur life easiyer and less complicated , cause if we look at the body as a whole we can consider that the toque on disk 1 - the torque on disk 2 =0(where disk 1 and disk2 are a part of the whole shaft that is rotating so if we have the sum of torques = 0 then my delima is answered , sum of torques = 0 then the angular accelration = 0 and the angular velocity = cte

    about power , we can calcuate power in 2 methods , the first is taking F1 and multiplying it with the linear velocity or taking F2 and mulitplying it with the linear elocity on disk2 , it would be the same , cause we have an equilibrium that is established ...

    I really thank you for your help u gave me the hint of stop thinking about studying the disks alone cause it would complicate my understanding, though if we go deeper inthis study we would understand how the microscopic reactions occur ...
    by the way my real name is daniel and I am a 4th year mechanical engineering , internal combustion engine as a major --- I hope u dont think that I am stupid , I know how solves such problems but i like as well to dive deep inside the problem , though we stoped taking these kinds of probelms now , this problem is at my second year of engineering.
    Again I really thanks u , u were very helpfull.
     
  14. Aug 16, 2008 #13

    Doc Al

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    For many purposes, you don't have to worry about the details of the torsion created in the rod. But sometimes you do: The engineer who must guarantee that the rod is strong/rigid enough will have to worry about such details. That's a bit more advanced.
    Welcome to PF, Daniel. You are asking good questions; many do not. Good on you.

    And you are welcome. :wink:
     
  15. Aug 16, 2008 #14
    well about the torsion on the shaft this point concern the tortional stress on the shaft which is calculated by T.R /J

    J = 2 I , the unit of torsoniall stress is MPA... the course that details these stuff is called : machine design and mechanics of material .

    in thic course we take >>> bending stress , tortional stress , shear tress ,"drect-shear , secondary shear<<speciallly on bolts and welding >> "
     
  16. Aug 16, 2008 #15

    Doc Al

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    You'll soon be an expert on such things.
     
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