Sum of torque

[EOW]

Hi, Dieter,

In the post #130, the Pulley2 is not fixed. The Pulley2 turn of 360°, one round, because its angular velocity is 4w (lab reference), the Pulley1 turns at 2w (half round). I will draw another example.

For understand, put the Pulley1 and the Pulley2 in the same support, if the Pulley1 is the motor, the motor need to give -3FRwt and the support lost -3FRwt, the Pulley2 receives 4FRwt and the support receive 2FRwt the sum is at 0. Now, with 2 supports, the sum of the torque on Support1 and Support2 is 0. If the motor is the Pulley1, the support lack 2FRwt. If the motor is the Pulley2, the support don't lost -2FRwt.

But the gears will force the supports to rotate at the same speed, in the same direction?

Yes, it's not necessary to add gears for understand. But like the Support1 receives a torque T and the Support2 receives the torque -T, the sum is at 0, if I use gears +T-T = 0, like that supports turn always at w. For one support the torque change its sign in a round.

So it cannot rotate at all?

Support1 and Support2 turn at w. Pulley1 turns at 2w (lab reference), Pulley2 turns at 4w (lab reference).

The pulleys will give F and -F to the supports.

correct

I added an image with an angle of 45°, for look at A and B. The support turns of 45°(w), so A turns of 90° (2w), B turns of 180° (4w).

[dieter]

Hi EOW,


ok, that makes sense now.


BR

[EOW]

Are you ok with a torque at 0 for Support1 and Support2 if I use gears ?

[EOW]

Hi Dieter,

I think d1=d2, always. The force F on the Support1 gives a clockwise torque T but the force -F gives a counterclockwise torque -T. |T|=|-T|. I drawn forces with the Pulley1 a motor.

Have a good day

[dieter]

Hi EOW,


But the torque of the pulleys will affect the supports only when there is, example given, friction...?



Have a nice day too! 


BR