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I've been trying to find a way to calculate the contact pressure between a rope under tension and a pulley (free to move, or fixed). I'm expecting to find an equation that considers the wrap angle - similar to the capstan equation - but I'm specifically looking for an expression for the contact angle. Not concerned about the mass of the rope, in my problems the force applied is far greater than the weight of the rope.

EDIT: I need this to consider friction between the rope and the pulley - just like the capstan equation. Consider a fixed pulley with a very large radius, and a very high contact friction. In this case, you would expect the region close to the centre of the contact area to have a lower contact pressure.

Any suggestions?

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  • $\begingroup$ Is the center of mass of the pulley constrained? $\endgroup$ – Bob D Jan 29 '20 at 11:42
  • $\begingroup$ Pressure is defined as force over area. In this case you can find force over length (or angle), and it will vary from point to point around the pulley. $\endgroup$ – R.W. Bird Jan 29 '20 at 15:55
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In a pulley we can assume that the friction can be neglected, so that the tension in the rope can be assumed to be constant. Thus, consider the following diagram, showing the forces acting on the rope due to a turning angle $\mathrm{d}\theta$:

enter image description here

From the symmetry of the problem (any part of the rope that is in contact with the pulley will feel the same pressure, since the curvature and the tension are constant), you conclude that the force per unit length in the normal direction is given by:

$p = \frac{\mathrm{d}F_N}{\mathrm{d}\theta R}$

since the length that supports the force $\mathrm{d}F_N$ is $\mathrm{d}\theta R$.

But since, from the diagram, $\mathrm{d}F_N = 2 T \sin{(\mathrm{d}\theta/2)} \sim T \mathrm{d}\theta$ when the angle tends to zero, the above equation becomes:

$p = \frac{T}{R}$

Therefore, the pressure does not depend on the total angle bent, but rather on the curvature at which you are bending it (or one over the radius of the pulley). Obviously, the total force on the pulley will depend on the total angle, which can be calculated easily by integrating the projection of $p$ over the angle bisector along the total arc along which there is contact with the rope.

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  • $\begingroup$ If you are going to use a (tiny) dƟ on one side of an equation then you need a dF on the other side. An integration needs to keep on mind that the dF is a vector. $\endgroup$ – R.W. Bird Jan 29 '20 at 16:09
  • $\begingroup$ I changed the name of the normal force to reflect your comment. About the fact that it is a vector, I specified that one should integrate the projection along the required direction. $\endgroup$ – Guillermo BCN Jan 29 '20 at 16:38
  • $\begingroup$ Thanks for your response, however I neglected to add that it needs to consider friction between the rope and the pulley. Similar to the capstan equation, I'm expecting a contact pressure that varies along the length of the rope. $\endgroup$ – user1473810 Jan 30 '20 at 6:55
  • $\begingroup$ Can you not use the Capstan solution then, with a wrapping angle is $\pi$? $\endgroup$ – Guillermo BCN Jan 30 '20 at 8:14
  • $\begingroup$ The Capstan equation alone doesn't consider the normal force (and therefore pressure). So that equation alone doesn't help. However, I think the answer lies in the derivation of the equation - see my answer response below. $\endgroup$ – user1473810 Jan 31 '20 at 3:44
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I found this video on Youtube which explains the derivation of the Capstan equation.

https://www.youtube.com/watch?v=H0T7m537YT0

From what I can gather, the normal force ($\Delta N$) is a function of $\Delta T$, $T$ and $\Delta\Theta/2$. If $\Delta T$ goes to $0$, that is, the pulley is in equilibrium, then $\Delta N$ is then only a function of $T$ and $\Delta\Theta/2$ - therefore the sum of forces in the X direction goes to $0$, and $\Delta F_{fr} = 0$, but since $\Delta N \ne 0$, then $\mu _s$ must $=0$.

This neglects weight of the cable (or in the case of this example - a flat belt), but for my case that's OK. The example also uses a flat belt rather than a round cable - I'm assuming this is so they can ignore any geometric changes as a round cable compresses and the contact area changes.

Hope this makes sense.

Edit: After reviewing Guillermo's response once more, I now understand that his answer seems to agree with my explanation above - Thanks.

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