Sum of torque

[dieter]

Looks good, as far as I understand it.


BTW. I'm off for a break


BR

[EOW]

Ok. I drawn anothers views at different positions. The torque from F7/F8 is not constant it depends of the position, so forth image shows the system lost energy but it's possible to change the position of the friction and have a clockwise torque to the support3 (fifth image). The sum of torque from F3 and F4 to support1 and support2 is 0 like I show before.

with

angular velocities (lab ref):
Support1 = 0
Support2 = 0
Support3 = +2w
Pulley1 = +4w
Pulley2 = +12w


The motor need -12FRwt
The friction gives +2FRwt
The Support3 gives +12FRwt
The Support2 need 0
The axes of pulleys gives +4FRwt

The sum is at +6FRwt

[EOW]

With one motor and one brake on a support. The motor drives the brakes with a crossed belt. The motor turns clockwise at +10w around itself. The brake turns counterclockwise around itself at -10w. The support turns clockwise at +w. So, the motor turns at +11w in lab ref and the brake turns at -9w in lab ref. The motor needs -11FRwt, the brake gives +11FRwt (the brake turns at -10w and the support turns at +w, the difference is 11w), torque F1/F2 gives +2FRwt to the support, torque F3/F5 needs -FRwt to the support, the sum is at +FRwt. All angular velocities are constant. I guess no mass for pulleys and belt.

[EOW]

I can replace the friction with an electric generator for example. I guess the efficiency of the motor, generator, belt, etc is 1 (no losses). The motor gives F1 and F2 to the Pulley1. The motor gives F3 and F4 to the support. The belt gives the force F5 to the Pulley2. The belt gives the force F6 to the Pulley1. F5 gives F7 to the axis of the Pulley2. F6 gives F8 to the axis of the Pulley1.The generator gives F9 and F10 to the support.

I noted R the radius of the Pulley1 and the Pulley2.

With |2F1|=|2F2|=|2F3|=|2F4|=|F5|=|F6|=|F7|=|F8|=|2F9|=|2F10|=|F|

The motor need to give -10FRwt
The support reveices -FRwt from F2 and F3
The support receives -FRwt from F9 and F10
The support receives +2FRwt from F7 and F8
The generator receives +11FRwt because the difference between the rotor and the stator is 11w

The sum of energy is +FRwt

All angular velocities are constant. The sum of torque on the support is 0, so its angular velocity is constant. The energy lost from the motor is -10FRwt and the generator can recover +11FRwt.

[EOW]

The stator of the motor is fixed on the support so it turns clockwise at +w. The stator of the generator is fixed on the support so it turns clockwise at +w.

The rotor of the motor turns clockwise at +10w relatively from the stator, so the rotor turns clockwise at +11w (lab ref), the power needed for the motor is -10FRw and the support needs -FRw. The rotor of the generator turns counterclockwise at -11w but the stator turns clockwise at +w so the difference is 12w, but the support receives a torque -FR and need the power FRw.

Forces on axes of the pulleys give a power 2FRw to the support and with -2FR from 2 startors, the sum of torque on the support is 0.

So, the motor needs -10FRw and the generator gives 12FRw.