I am currently designing a motorized zipline trolley that will carry a person on a horizontal cable, so I am trying to calculate the friction on the wheels to determine what kind of motors I need. I initially approached the problem using a static friction model using the coefficient of static friction between two steels. However, my professor told me to replace my model with a rolling friction model, so I took the coefficient of rolling friction between two steels. This greatly reduced the calculated value of friction; however, I feel like there should be more friction because the weight of the person will cause deflection in the cable, so a portion of the wheels' circumference will be in contact with the cable. I want to assume no slipping for the wheels, so I don't know how to account for the static friction while keeping the rolling friction model. How can I calculate the additional frictional force that I am not modeling?

  • $\begingroup$ Can you draw a diagram of what you're trying to do? I'm imagining something like a ski lift system, where the cable is driven by the pulley. In that case, there is no (or should be no) relative motion between the cable and pulley. For an idler pulley in the same system, again there shouldn't be any relative motion between the cable and pulley (or else the pulley isn't doing its job!). I think you're confused by the same thing that I am - how do you have a rolling friction model between two objects that are moving in unison? Rolling implies relative motion, but there isn't any. Again: diagrams. $\endgroup$
    – Chuck
    Sep 23, 2016 at 14:01
  • $\begingroup$ I can draw you a diagram when I have chance but for now, literally think of a zipline. The wheels DO have a relative velocity to the cable because the wheels roll on the cable. The only difference is that this device is propelled by a motor across a horizontal cable, as opposed to the potential energy due to the elevation difference. $\endgroup$
    – Skipher
    Sep 23, 2016 at 15:16
  • $\begingroup$ when a wheel rolls on a surface, or in this case a cable, there is no relative velocity at the contact point. Rolling friction is what you use. Note there is no chance your cable will be perfectly straight. The force required to climb uphill due to slight cable sag is quite likely larger than the friction you are worrying about. $\endgroup$
    – agentp
    Sep 28, 2016 at 0:50

1 Answer 1


There are a two items I think you are wondering about, the effect of static friction and the extended length of the contact area between the rope and the pulley.

Start up torque on a motor can be greater than the design torque of the motor, so your professor may be suggesting to ignore the rolling friction because of that. I would say it's a valid assumption to assume the motor can overcome the static friction if it's overcoming the dynamic friction. Check out motor-torque curves on engineering toolbox for example. I can't remember the complete explanation and terms for this, but when you start a motor there is not a strong magnetic field produced by the rotor, because it is not moving yet, and so the magnetic field that is being produced by the AC current in the stator gets stronger and produces more torque than when the rotor is spinning at its nominal rating. That's why you have high inrush current when starting a motor, no matter what (assuming you have no starter controls on it).

Secondly, the contact area of the rope on the pulley is irrelevant. You have assumed no slip for starters, so the total frictional force between the rope and the pulley that would be altered by contact length is not going to change. Draw a free body diagram of the pulley. You will have to have tension in the rope between the two pulleys, and you can theoretically make that tension to whatever you want. The tension in the rope (due to the tension you set as well as the mass of the trolley load) will act as the torque on the pulley of the motor. That is the torque the motor has to overcome.

There will be some deflection in the rope due to the load, which you can solve for. Then it's just the added tension in the rope from the load that the motor has to account for since the rope tension is always perpendicular to the pulley.

Break that out into components to and balance then solve for the new tension. I'm on my phone right now and can't enter the equation. But I will try once I have access.

  • $\begingroup$ I see. Overcoming the tension in the cable helps me understand the model much better than attempting to quantify the additional torque required due to the deflection in the cable. $\endgroup$
    – Skipher
    Sep 23, 2016 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.