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Centripetal force involved with a pulley system

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data
    If you have a motor pulley and a drum pulley connected by a belt then how do you explain why the belt can sometimes slip if the washing machine is rotating too fast?

    2. Relevant equations

    3. The attempt at a solution
    I know that the answer involves the tension of the belt being unable to provide the centripetal force but I can't see where there is a centre seeking force in this system. The pulley grips the belt due to friction I assume but wouldn't that be a sideways force, I can't see where the centripetal force is?
     
  2. jcsd
  3. Nov 15, 2014 #2

    mfb

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    Tension will certainly give a force inwards.
     
  4. Nov 15, 2014 #3
    Please could you explain how? Isn't tension a pulling force so I thought this would act along the length of the belt? What does this question free to as the circular motion, is it the belt? Would the belt also not be circular as they are more oblong shaped?
     
  5. Nov 15, 2014 #4

    mfb

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    The direction of tension changes along the wheel. To balance forces, there has to be an additional force from the wheel.
    Take a rubber band and see if your finger feels a force if you extend it.
     
  6. Nov 15, 2014 #5
    Thanks. So is this referring to the centripetal force of the belt? What do you mean by balance as isn't centripetal force a net force? How do you figure out the direction of the tension as the belt goes around the wheel?
     
  7. Nov 15, 2014 #6

    gneill

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    The tension acts along the length of the belt. But consider a small segment of the belt wrapped around the drum. The tension at the ends of the segment act tangentially to the drum surface since the belt follows the curve of the drum. What is the direction of the sum of the two end tension forces? So then what's the direction of the net force due to tension on that small belt segment?
     
  8. Nov 15, 2014 #7
    Thanks that was very helpful. If you take that segment as you say then you would have a tangential component either side which if you add them tip to tail you would get a force towards the centre? Is that right? Would the part of the segment that is directly in line with the centre of the pulley have a force already pointing towards the centre? So, are you saying that the belt provides its own centripetal force?

    Why does it slip if the tension force isn't big enough then? I'm still a bit confused with that.
     
  9. Nov 15, 2014 #8

    gneill

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    Sketch it out and prove it to yourself :)
    What force might that be?
    Draw the sketch as suggested and see.
    What force determines the force due to friction?
     
  10. Nov 15, 2014 #9
    Thanks again. I have tried to think hard about this and have drawn a diagram (attached to this thread).

    What has friction got to do with the tension not being enough when the belt slips? My thinking - when the belt is pulled tight around the pulley, the pulley pushes on the belt (normal force) which causes friction. The friction produced is proportional to the normal force. The normal force is therefore proportional to the tension. So if the tension is not strong enough the belt will not grip the pulley and they will move relative to each other. I just don't see how you can argue this case in terms of circular motion and centripetal force? Does the circular motion only apply to the belt when it is wrapped around the pulley? So the bits I have marked in green do not have (or need) any centripetal force?
     

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  11. Nov 15, 2014 #10

    gneill

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    The centripetal force, which keeps the belt pressed against the drum, is supplied by tension. The belt is not massless. So the faster it moves the more centripetal force is required to keep it moving in a circular path. So what happens as the drum spins faster and faster and the belt moves faster and faster? What happens to the normal force?
     
  12. Nov 15, 2014 #11
    Thanks for the help. Ok, so as the belt spins faster and faster it needs a bigger and bigger centripetal force. The centripetal force is provided by the tension which itself is provided by the normal force (I think). So if the tension of the belt cannot cope with the speed of the drum it will slip and as its does so will move relative to the drum and the normal force will become zero during the slipping? Is that correct? Am I correct in saying that the belt provides its own centripetal force since we are talking about the belt undergoing circular motion and the centripetal force being provided by its own tension? I always thought centripetal forces were provided by external forces.
     
  13. Nov 15, 2014 #12

    gneill

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    No, the normal force is the reaction of the drum pushing back against the belt. The belt pushes on the drum thanks to the tension being "resolved" into an inward force.
    Maybe not zero, it just has to drop below the threshold where static friction fails to hold the belt from slipping.
    The tension is provided by the pulley. Usually the pulley is spring-loaded to supply the tension. The tension results in a centripetal force around the drum. The centripetal force that exceeds that required for circular motion of the belt pushes the belt against the drum. The drum then provides the normal force that leads to friction between the belt and drum.
     
  14. Nov 15, 2014 #13
    Thanks, I think I get it now. So, if the tension force was only big enough to provide only the centripetal force for the belt then the belt would not press on the drum? The pressing of the belt on the drum only happens if the net inward force is bigger than the required centripetal force of the belt? So, in effect, it is the centripetal force of the belt that is exerted on the drum which in turn exerts an equal and opposite force (normal force) which causes friction (as frictional force is proportional to normal force). When slipping occurs is there still a centripetal force which is enough for the belt to move around the drum but not enough to cause friction between the drum and belt?
     
    Last edited: Nov 15, 2014
  15. Nov 15, 2014 #14

    gneill

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    The tension in a rope is due to some external forces acting on different parts of the rope. Such as two people pulling on each end, or two weights hanging from the ends while the rope passes over a pulley.

    If the rope is a continuous loop (or a belt) it can also be imparted by pulling apart two pulleys that the rope passes around. For the pulley and drum setup of a clothes dryer there's usually a tensioning spring pulling the belt tight, sometimes pulling the motor pulley, sometimes as a separate pulley.

    Fig1.gif
     
  16. Nov 15, 2014 #15
    Thank you for the picture, that makes sense now. Are those points I made in my previous thread correct (as I edited them just a second ago).
     
  17. Nov 15, 2014 #16

    gneill

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    Yes, obviously, right?
    Any pressure by the belt on the drum is due to the portion of the centripetal force exceeding that required to keep the belt's mass moving in circular motion. If the centripetal force is less than that required to maintain the circular path then the belt loses contact with the drum (maybe not everywhere at once around the circuit of the drum). But definitely the static friction force will gradually decrease as the speed increases.
    It could happen. But more likely static friction fails and dynamic friction takes over before that happens. Then you get the belt slipping and heating up, likely leading to failure of the belt accompanied by a strong burnt-rubber smell :D
     
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