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How to understand tension force existing in slack side of belt drive?

  1. Mar 25, 2013 #1
    Hi,

    For a long time, it is always unclear for me to understand that there is a tension force acting in the slack side of a belt drive. Tension acting in the tight side is quite straight forward and hence no doubt!
    When I see a belt drive in motion, I always see the slack side kind of deformed and wobbling. This leaves me an impression that there is no force on that portion of belt. But in all our text books it is said that there is a tension acting. Can you explain the easiest way to understand this? May be with some examples or videos? Thanks.
     
  2. jcsd
  3. Mar 25, 2013 #2

    Simon Bridge

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    The "slack" side is not slack - even if the pulleys are not turning, there is a tension there. Hook one roller/pulley/end/whatever to a spring and you'd see the spring extend under the force. The tension is provided by gravity. The belt looks saggy for the same reason a power line does.

    When the pulleys turn, you'll see the "slack" side tighten up a bit compared with rest - showing you the acceleration adds some extra tension there. Clearly the tension in the top has to pull on the bottom just as hard, otherwise the belt would stretch.
     
  4. Mar 25, 2013 #3
    Thanks, Simon. Is it something like, the belt is wound / installed with a pre-tension around the pulleys already and this is the slack side tension?
     
  5. Mar 26, 2013 #4

    Simon Bridge

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    The tension wheel supplies an extra component to the tension - useful for stopping the belts slipping.
     
  6. Mar 26, 2013 #5
    Thanks, Simon. I think I have to think a little more about what you said to understand clearly. Thanks again! :)
     
  7. Mar 26, 2013 #6
    Hi Simson! I exactly caught what you meant! If there is no tension in the slack side, then there will be no friction existing between pulley and belt and hence no motion transfer. Thank you so much!
     
  8. Mar 29, 2013 #7
    what causes slack side and tight side is the difference in tension induced due to friction between the pulley and belt. If pulleys are rotating in clockwise direction, relative motion is anticlockwise for belt. hence friction will be in clockwise direction where ever is considered on belt. so belt above pulleys will have higher tension, hence belt below pulleys form slack side and they sag due to self weight.
     
  9. Mar 29, 2013 #8

    jim hardy

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    Think of the belt's two sides as two springs.
    At rest they are both stretched equally.
    When rotating their tension is unequal by the amount of load being transferred.
    The loose side has a lower natural frequency because it's not stretched so tight - so it'll flop around more visibly, like a guitar string..

    If the slack side becomes very loose the belt will slip at high load.
    This used to happen a lot back when cars used vee-belts.
    The alternator belt would become loose through stretch or wear, alternator could no longer keep up with headlights so you'd run the battery down on a long drive after dark. Yet its voltage would check fine in daytime.


    A lot of folks bought new battery, alternator & regulator when they just needed the belt tightened.
    This also tended to show up at start of summer when airconditioner gets switched on.

    hope this helps.
     
  10. Jun 5, 2013 #9
    Figure belt tension for tight and slack side (T1 & T2)

    force = power/velocity

    Assuming you have an motor of 2 hp (1492 N-m/s) and the belt is moving at 12 mph (5.3645 m/s). Metric units are better here so ...

    F = (1492 / 5.3645) = 278 newton (62 pound force - lbf)

    So now

    T1 = 278 * (1 + (1/(2.7182 * 0.25 * 0.5934))) = 967.4072 newton (217 lbf)
    T2 = 278 * (1/(2.7182 * 0.25 * 0.5934)) = 689.4072 newton (155 lbf)
    P = (967.4072-689.4072) * 5.3645 = 1491.331 N-m/s

    It appears the proof works because 1491.331 N-m/s = 1.999 hp (close enough)

    u = coefficient of friction is a tough one, but 0.25 for a dry running belt is a good base
    a = find out your pulley groove angle, convert to radians (i'm using 34 deg = 0.5934 radians)

    Assuming I didn't screw things up, it appears the slack side, in this case, is 71% of the tight side. So it's fairly tight even on the slack side.

    1. Just figure what horsepower your motor has (convert hp to N-m/s)
    2. How fast the belt is moving (mph to m/s)
     
  11. Jun 5, 2013 #10
    Thank you all. All the replies helped me to understand this by different ways. Thanks again!
     
  12. Jul 5, 2013 #11
    jryer, in your formula, what is the 2.7182? Do you have a reference to the original formula that you could supply? Thank you!
     
  13. Jul 6, 2013 #12
    F = Moving force in the pulley driver
    T1 = tension on the tight side
    T2 = tension on the slack side
    e = base of the Neperian or natural logarithms (2.7182)
    μ = coefficient of friction
    α = angle subtended by contact surface at the center of the pulley

    T1 = F * (1 + (1/(e^ua - 1)))
    T2 = F * (1/(e^ua - 1))
     
  14. Jul 6, 2013 #13

    Simon Bridge

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    @kem0022: welcome to PF;
    ... jryer (basically) just started with a free-body diagram for the pulleys and followed the math from there.
    ... you should be moving away from memorizing equations to understanding the processes required to get those equations in the first place, but you know that right?
     
  15. Jul 8, 2013 #14
    Compression of slack side

    I understand the tightening of "slack side" of the belt when the pulley is being driven. But if the tension on the slack side(bottom as u have referred) was not the same and if it were lesser than the top then wouldn't the slack side compress? how does it stretch? Please share your knowledge Simon.
     
  16. Jul 8, 2013 #15

    Simon Bridge

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    Welcome to PF;
    Looks like you should make a pulley system using a very stretchy rubber band and turn the wheels and watch.

    The tight and slack side go at the same speed so there is no compression.
    Try doing the free-body diagram to see how the tensions balance out.
     
  17. Jul 9, 2013 #16
    Thanks Mr. Simon. I did happen to study the free body diagram.

    I found out that initially when the pulley is stationary, there is a tension force which is nothing but pre tension which is required to generate friction to make the belt and thereby the pulley to move.

    Now, when the driving pulley starts rotates there is an increase in tension on one side and decrease in tension(depends on the pre tension). The side with the increased tension is the tight side and the side with lesser tension happens to be the slack side. This tension is also a function of coefficient of friction. Higher the coeff friction, higher is the power transmission capability. Hence we have difference in tension which acts as the driving force of the belt. Thus the belt rotates at almost the velocity of pulley(there will be belt slip). This causes the driven pulley to rotate and hence power is transmitted.

    As i pointed out in my earlier reply, there will be contraction and expansion in the belt Mr. Simon. They call it the phenomenon of elastic creep. It is nothing but, in the driving pulley, the area of belt under the tight side of pulley will undergo elongation and the area of belt above the slack side of the pulley will undergo contraction.

    So yeah. Thats it. Thanks Mr. Simon.
     
  18. Jul 9, 2013 #17

    Simon Bridge

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