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Braking band calculations

  1. Oct 13, 2015 #1

    I'm trying to analyze the following system to get some idea about the forces present.


    The red line is supposed to represent the belt wrapped around the two drums.

    There is an extra part to the belt, however. Tangentially from the drum, a part of the string is attached to an unmovable object. A string force (T3) is acting on the string here, and this force keeps the whole belt from moving around as the larger drum turns with an angular velocity of ω.

    So when the system is running, the drive-belt is static and unmoving. The larger drum is turning around with ω and therefore there is obviously a friction force acting on the string.

    I have been trying to analyze this with respect to the capstan equation, and deriving some form of formula, but I'm at a loss as how to take into account the force T3.

    What I would like to end up with, is some form of equation describing the size of T3.

    I hope someone can perhaps give some pointers or know references to somewhere where a similar system is analyzed?

    Thanks in advance,
    Last edited: Oct 13, 2015
  2. jcsd
  3. Oct 13, 2015 #2
    Alright, I've given it a try, but I have a nagging feeling that I'm doing something wrong...
    If anyone have the mathematical insight to check the calculations I would be extremely grateful!

    Here follows:

    Part 1: The larger drum

    Okay, so the force F1 is due to the belt tension. So we know the belt tension at that point on the drum. Using that we want to derive an expression for the tension at point D.

    What is used here is the capstan equation: ## \frac{F_D}{F_1} = e^{\mu\theta_1}##
    We are interested in knowing the tension at the point D, which means that we will have ##\theta_1 = (270-\phi) ##
    Another thing to remember is that ##\mu## in this case is not the static friction coefficient, but the dynamic friction coefficient.

    Checking the equation gives us the correct idea that at point D the tension is higher than at F1 (since the friction force is being added from all the way around).

    Looking closely at the point D, we have the following:

    Since the tension at at either side of the small strip of the braking band (brown) must be the same, we have that
    $$F_3 = F_D - F_{hold}$$
    Next, we still need to analyze the small part of the belt going from point D and up until the spot where the belt "lets go" of the drum (where the tension/force is F2).

    This is simply the same as before (the capstan equation), and we arrive at: ## \frac{F_2}{F_3} = e^{\mu\theta_2} ##
    And here the angle is: ## \theta_2 = (90 - \phi) ##

    We can insert this into the equation relating the three forces:
    $$ F_3 = F_D - F_{hold} $$
    $$ F_2 e^{-\mu\theta_2} = F_D - F_{hold} $$
    $$ F_2 e^{-\mu\theta_2} = F_1 e^{\mu\theta_1} - F_{hold} $$
    Next we turn our attention to the small drum (where the tightening mechanism is found in the final system).

    Part 2: The smaller drum

    At the smaller drum, nothing is moving. The drum is not moving or turning, since the belt is not moving (it is held by the holding force).
    What I arrive at is that the two forces must be equal to each other: ## F_1 = F_2 ##

    Is this a mistake? I'm unsure...

    Part 3: Putting it all together
    We now know that ## F_1 = F_2 ## and we have an equation relating those two forces as well.
    $$ F_2 e^{-\mu\theta_2} = F_1 e^{\mu\theta_1} - F_{hold} $$
    Using the equality:
    $$ F_1 e^{-\mu\theta_2} = F_1 e^{\mu\theta_1} - F_{hold} $$
    This allows us to get an expression for ## F_{hold} ## from ## F_1 ##.
    $$ F_{hold} = F_1 \cdot \left( e^{\mu\theta_1} - e^{-\mu\theta_2} \right) $$

    Of course, ##F_1## can be found as a function of the tightening force ##F_{tighten}##, but that is of lesser interest.

    I'm not entirely confident of this calculation, so if anyone can spot my mistakes it would be cool!

    Thanks community!
    Last edited: Oct 13, 2015
  4. Oct 13, 2015 #3


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    The only initial simple issue I see is that since the F tighten on the small drum is what is creating the friction force on the large drum then for the capstan θ1 = 180 + 2 * Φ.
    Also T1 = T2 is only true if the problem statement includes either " the small drum is free to rotate" or " the small drum is frictionless" but this may only be a bit of nitpicking on my part
    There may be other issues because I am still struggling with this problem as well; because, it appears to me that F hold might also act to increase to the drum friction force.
  5. Oct 13, 2015 #4
    Thanks for you feedback JBA!

    • I'm unsure about your first remark. As I see it, the ##F_{tighten}## is in a way responsible for creating the normal force of the onto the drum.But I don't know if that is what you're getting at.. Can you clarify?

    • Your second remark regarding the T1 = T2 is a good observation! And indeed the statement "The small drum is free to rotate" is applicable in this system.

    • Your last remark regarding the ##F_{hold}## - This force is not an 'active' force. There is no mechanism pulling the string. In case the large drum is not rotating, the ##F_{hold}## would be zero, since there is not friction force "trying to rotate the belt". Could it still somehow increase the friction force? As I see it, it has to counter the friction on the belt (since the belt is not accelerating but remaining static).
      My first instinct was the ##F_{hold}## would simply be equal to the sum of all the tangential friction applied to the belt...
  6. Oct 13, 2015 #5


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    The capstan formula is based upon the entire angle of contact of the wrap on the drum. which in the figure = 180 + 2 x Φ

    On further consideration, I agree with your last statement on F hold.
  7. Oct 14, 2015 #6
    Hmm, I see what you mean. However, what I specifically did was to calculate the string tension using the capstan formula, at Point D. Therefore, the wrapping angle is not the whole angle (and I think you mean ## \theta_{wrap} = 360 - 2\phi ##, since the angles are measured from the "x-axis" of the circle?).


    What I did in the calculation of the tension at point D, was to look at a small segment of the belt, as illustrated to the right in the picture above. Since the part of the belt is not accelerating, the forces must sum to zero, which means that
    $$(T+dT) cos(\frac{d\theta}{2}) = T cos(\frac{d\theta}{2}) + \mu\cdot F_N$$
    Because the angles are differential, the cosines end up being equal to 1.
    $$T+dT = T + \mu\cdot F_N$$
    This equation easily simplifies to ## dT = \mu F_N ##, which is intuitive because the change in tension is equal to the friction force applied to the small piece of belt.

    Next, the normal force ##F_N## is given by
    $$ F_N = (T+dT)sin(d\theta/2) + T sin(d\theta/2)$$
    Here I use the small angle approximation of the sines and end up with
    $$F_N = Td\theta$$
    This result is used in the equation for dT giving
    $$ \frac{1}{T}dT = \mu d\theta $$
    Finally I integrate from ##\theta = 0## to ##\theta = 270 - \phi##, since that is until point D.
    On the left hand side I integrate from ##T = T_1## to ##T = T_D##, and assume that I know ##T_1##. (I have changed notation from F to T in this reply).
    $$\int\limits_{T_1}^{T_D}\frac{1}{T}dT = \int\limits_{0}^{270-\phi}\mu d\theta$$
    All of this is simply the capstan equation derivation and I end up with the tension at the point D.
    $$T_D = T_1 \cdot e^{\mu\theta_1}$$
    And here we set ##\theta_1 = 270 - \phi##

    The next steps are simply as I mention in my calculations in the first reply.

    1) I look at the tension at the point D and assume that the tension point to the right is a sum of the ##F_{hold}## and the tension of the part of the belt still following the drum around.

    2) I calculate the tension in the belt from the point D and until the point where the belt "lets go" of the drum and goes toward the smaller drum.
  8. Oct 14, 2015 #7


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    Your solution looks perfect to me. Sorry about the mistake on the angle reference, I think my age is creeping up on me. Maybe I need to see an optoneurologist because my brain is clearly having problems comprehending what my eyes are seeing.
  9. Oct 14, 2015 #8
    How is the other end of the rope attached (if at all)?

  10. Oct 15, 2015 #9
    Hello Chet!

    Using the first image i posted as a reference, the part of the belt with ##T_3## is attached to a force sensor in one end. It is also sewed to the belt wrapped around the drums.
  11. Oct 15, 2015 #10
    Are you saying that the opposite end of the rope is sewn to the drum?
  12. Oct 15, 2015 #11
    Oh no, not at all. I'll try to explain in another way.

    The belt is wrapped around the larger and the smaller drum. Also attached to that belt is a string. This string is sewn to the belt and is used to hold the belt when the larger drum rotates.


    So there are no attachment points other than the extra string. And the extra string is hooked up to a force sensor.
  13. Oct 15, 2015 #12
    I got the same result as you.

    Last edited: Oct 15, 2015
  14. Oct 16, 2015 #13
    Thanks a lot Chet and JBA for your interest!
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