The 40 kg weight accelerates and runs on a rope from 0 to 50 km/h in 10 seconds.

I want to calculate the minimum required rolling friction that I need (acceleration without slipping) to achieve acceleration in a time of 10 seconds from 0-50km / h.

Friction adjustment (later for different speeds and accelerations) I would change by increasing /decreasing the angle between the drive pulley and the pulley 2 as shown in the figure. Currently, the angle is 15 '.

My question is: How to calculate the minimum required rolling friction (or better say, minimum required angle which is 15' now) between the drive pulley and the rope of 40kg of weight moving along the rope, as shown in the drawing?

enter image description here

All pulleys are covered with rubber, and the rope is synthetic, Dyneema. What aditional information do I have to apply for the correct calculation?


1 Answer 1


let's find out acceleration first.

$$v= at = 50000M/h * 1h/3600s *1000M/kM =13.888M/s \\ a= \frac{v}{t} =13.88/10 =1.388 m/s^2$$

the tributary weight on the two pullies is

weight on the right pully is $$40 *(1147.5-154.6)/1147.5= 34.6kg \ and \ on \ the \ left =40-34.6 = 5.4 kg $$

the angle of rope to left is $15 * 154.6/1147.5 = 2.02$ degrees.

say negligible, we assume the entire friction from the right wheel.

The friction force is

$ F_f = \mu*40*9.8/ sin(15) \quad = 40*1.388$

  • $\begingroup$ So, we got the friction force of the existing setup. How to calculate the minimum required friction force for the specified acceleration? As the friction is less, the motor consumes less current. I do not want to have more friction than needed, because the battery life is shortened and the system is exposed to more current and the stresses in the whole system are much larger. $\endgroup$
    – littlerock
    Commented Mar 9, 2019 at 21:17
  • $\begingroup$ Maybe I wasn't clear , sorry for that. What is the lowest angle of rope on right side, needed for that acceleration without slipping? I want to bring friction force lowest possible. $\endgroup$
    – littlerock
    Commented Apr 4, 2019 at 12:06

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