The pulley is only an intermediary. There is no net torque on the pulley. The torque exerted by the motor should be exactly canceled by the other forces acting on the pulley. With the magnitudes you claim, that won't happen.V711 said:2 pulleys are connected with the belt. If I want to turn the big pulley (with a small mass), I need to give a torque, what torque I need to give ?
I'm agree. No net torque on the big pulley.haruspex said:There is no net torque on the pulley.
For me 6*R*F1 = 3*R*F5 so there is no torque on the big pulley. In my last image F1=0.5*F5haruspex said:With the magnitudes you claim, that won't happen.
Yes, that all looks ok now.V711 said:I'm agree. No net torque on the big pulley.For me 6*R*F1 = 3*R*F5 so there is no torque on the big pulley. In my last image F1=0.5*F5
I said ok but for me forces are :
|2F1|= |2F2|= |2F3|= |2F4|= |F6|= |F7|= |F8|= |F9| = |F5|, it's possible, no ?
Yes, I said so in post #25.V711 said:ok, great ! I would like to know if the forces F6/F8 gives a torque to the support ?
Let's look at all the torques.V711 said:ok, and at a time, the motor needs the energy -3FRwt (the motor turn at 2w but it is on a support that turn at w), the support receives -3FRwt (from F3/F4). The small pulley receives +6FRwt and the support receives +2FRwt, the sum is not at 0 so I suppose the small pulley don't receive 6FRwt but 4FRwt, no ?
Like I drawn my forces in my last image, I supposed the small pulley is not brake, the small pulley has a mass and it is accelerate more and more.haruspex said:Support : small pulley = brake torque (B)
haruspex said:Motor:large pulley = T
Axle+belt:large pulley = F7*3R
Motor : support = T
Two axles : support = F6*2R
Belt+axle : small pulley = F7*R.
Sorry, I don't understand your question.haruspex said:What equations do those give you?
Ah, yes. But that means there's an unbalanced torque on the support, so that will accelerate the other way. And nothing will make sense unless you assign a moment of inertia to it.V711 said:Like I drawn my forces in my last image, I supposed the small pulley is not brake, the small pulley has a mass and it is accelerate more and more.
haruspex said:Ah, yes. But that means there's an unbalanced torque on the support, so that will accelerate the other way. And nothing will make sense unless you assign a moment of inertia to it.
It depends where the brake is mounted. I would imagine it would be mounted on the support, so the torque of the brake would be transferred to the suppor. If the pulleys are not accelerating then there should be no net torque on the support. Alternatively, the brake may be mounted on something outside entirely.V711 said:Ok, I understood. So with a brake on the small pulley, the support don't receive a torque, correct ?
With a brake, the friction gives the energy 5FRwt (at a time) ?
yes, and like the angular velocity of the small pulley is 6w, on the support that angular velocity is 6w-w=5w, correct ? So I thought the work from friction is 5FRwt.haruspex said:I would imagine it would be mounted on the support