# Calculating optimal angle on a rolling pulley wheel, which holds weight, on a horizontal cable

The 40 kg load on the rope accelerates from 0 to 50km / h in 6 seconds.

The rope is tightened with a force of 500kg while the whole assembly is on it. The middle pulley is driven by a dc motor.

I want to calculate the optimum angle between the middle (drive) pulley and the end pulley (pulley 2 on the picture) for driving without slipping or without too much load on the motor.

An angle between two pulleys that is too small would cause slippage on the rope, while too big angle would cause a higher load on the motor, and a higher power consumption.

The pulleys are made of aluminum and the rope has a coefficient of friction of 0.1

The current angle between the two pulleys is 15 degrees, as seen in the picture.

Motor spec are: 24V, 3560 oz/in stall torque, 4.3 HP

If you need more information, please ask.

• Who downvoted the question ? Nov 21 '19 at 7:05
• I see no reason for downvoting this question Nov 21 '19 at 15:39
• Are the load and pulleys 1 and 2 constrained vertically?
– Drew
Nov 22 '19 at 7:16
• Oh, nvm, I think I see what you did. The drive pulley moves along with the load and pulleys 1 and 2 doesn't it?
– Drew
Nov 22 '19 at 7:21
• No, I'm sorry if I wasn't clear, I'm sorry for that. The other two pulleys are for support only, and the whole assembly moves along the rope over the drive pulley in the middle. All three pulleys are fixed to the housing together with the load and move along the rope. Pulleys 1 and 2 are fixed and the drive pulley can be moved vertically. Nov 22 '19 at 19:53

## 1 Answer

Sorry if my explanation is simple (anything wrong please point out).
With my understanding of your situation the right way should be something on this line.

Your objective velocity is 50Km/h (13,88m/s)
To obtain this velocity in 6s you'd need calculate aceleration
$$a=v/t$$ $$a = \frac{13,88[m/s]}{6[s]}$$ $$a = 2,315[m/s²]$$
The force needed to obtain our aceleration is $$F=m*a$$ $$F = 40[Kg] * 2,315[m/s²]$$ $$F = 92,6N$$

To acelerate without slipping you need estatic friction force higher than your aceleration force so $$F\mu_{est} > F$$

$$F\mu_{est} = F_{motor} * \mu$$ ( $$F_{motor}$$ for traction force on wheel)
$$92,6N = F_{motor} * 0,1$$ $$F_{motor} = 926N$$

Now, for the angle you want we use trigonometry.
Considering the cable pulling force is 500Kg total, I'll consider 2.500N traction in each side.

$$sin\theta = 930/2500$$ $$\theta = 21,84º$$