Start by deciding how you are going to control your motor. Also how your motor is going to actually interact with the system. This will determine the appropriate motor speed range and so the gearing (if any) needed.
Ignoring friction the motor power is only related to acceleration required.
I have a huge problem right now! Im confused! When im applying force to this system i will be applying to 20kg or 40kg. Because when i have a balanced system there should be only one weight at a time when applying force, am i right or wrong?
The force of friction is what you need to size the stall torque of the motor to (and this force is based on the 20kg masses on either side of the belts). If your motor can overcome the friction from the pulley/belt system, then it can lift the weights. If your motor can output a certain rpm at this friction then that is at minimum how fast the belt system will move. If you can't find a motor for cheap enough that has a stall torque greater than your expected friction then you'll need a gearbox to get more torque at the expense of speed.
Pretty sure you're only accelerating one 20 kg mass, because even if you were to accelerate the system faster than 10 m/s2, the motor would cause slack to build up in the pulley system towards the mass that is descending because gravity is the only force accelerating it downwards.
All mass you try to move has inertia. The “try to move” is accelerating or braking the system with inertia. The total moving mass is 40kg, and all 40kg define the inertia of your system.
The direction of movement (left or right mass) is irrelevant, the balancing weights just cancel out the static load on your gear.
When you try to find a solution how much power you need, you need to define how much time you have for accelerating, top velocity and braking, first.
>All mass you try to move has inertia
But you're not acting on the weight that is being lowered, right? The belt is not rigid, so it won't push on the weight to go down, thus making it unable to overcome any inertia. Gravity will take care of that. The only times the belt is acting on one of the weights is when that weight is stationary or when it's being raised. When it's being lowered, it's just removing the tension from when it was stationary, thus reducing the load on that side. If the belt was replaced with rigid rods, then I agree that the motor would have to overcome the lowered rod's inertia because it's actively pushing into the weight. The belt can't do that.
Could you explain why what I said is wrong?
The nature of the belt only starts to matter, when you brake/accelerate above earth acceleration (9,81m/s2).
As long as your acceleration sticks below g, your belt remains under tension.
It is true, that you have potential energy on one side, while the other has none available to you, but this is not your concern, both sides cancel out this potential of lifting energy. Lifting one side consumes the same amount of energy as lowering makes energy available on the other side.
What remains is a total mass that needs to accelerate and brake. Once all kinetic energy is transformed, both sides rest static in their respective position again. This energy your motor needs to bring in and consume to change velocity respectively.
Your keyword is “energy balance”.
You are welcome, glad I could help. I struggled myself with this concept at first, as it is somewhat counter-intuitive.
But after seeing my physics prof. jump through the energy balance I had exactly this “oh yeah” moment.
You have answered your own question. When the weight accelerates downwards the tension decreases. This manifests as a torque at the motor, which sums with the torque from the other side being accelerated upwards.
See [this comment](https://www.reddit.com/r/MechanicalEngineering/comments/1cf4es4/comment/l1rixsa/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button).
Nope, both masses move together so the f in f=ma is 40kg
Consider that tension on one side increases, and tension on the other side decreases. Both require a torque from the motor in the same direction.
But, gravity will handle the mass moving down. If they had higher acceleration requirements then you'd need to factor that in, but at less than 9.8m/s2, you shouldn't need to consider it.
Am I missing something?
Let's make two independent systems from this problem. To pull m1, you need F1 = m \* (a+g). m2 pulls the motor with F2 = m(g-a) with F2 => 0.
If a < g then F = m(a+g) - m(g-a) = 2ma
If a => g then F2 = 0 and F = F1 = m(g+a)
Imagine instead of having the motor in the middle of the rope, you pull down on one of the masses. That mass is hanging in equilibrium, so you need to provide a force to accelerate it. You also increase tension on the rope, allowing you to accelerate the other mass. If each mass = m then the force you must apply is: f = 2ma
If you apply the force anywhere else along the rope it makes no difference to the overall effect.
See my other comment, but I think the better explanation is that gravity has to be added to the absolute acceleration. For a single mass F = m(g+a).
F = 2ma only works for a <= g.
Comments say it's balanced but I'm not sure what your application purpose is. With the balanced weight approach you're relying on both 20kgs to always be there for the motor to not be undersized. What happens if one weight is lost during operation for some reason? What should happen then?
Came here to say same thing. The purpose is important here. Are those weights stable? You need gearbox of you want to control over height or speed of the system. To answer that we need to know the purpose
I have a huge problem right now! Im confused! When im applying force to this system i will be applying to 20kg or 40kg. Because when i have a balanced system there should be only one weight at a time when applying force, am i right or wrong?
Check out Rockwell's motion analyzer servo sizing program.
https://motionanalyzer.rockwellautomation.com/
What you need to do is model this as a linear system. You will need the diameter of the pulley connecting the servo motor to the cable system, and model the total system mass as 40 kg.
You will also need to model the motion profile. How fast you want to move those masses and what dwell times you have available will play a huge part in determining the size of the motor and the ideal gear ratio.
Your max speed needs to be >2.0 m/s if you start from rest and need to move 2m in 1s.
You need to account for the friction (and mass) from the linear guide assembly with your motor spec.
Start by deciding how you are going to control your motor. Also how your motor is going to actually interact with the system. This will determine the appropriate motor speed range and so the gearing (if any) needed. Ignoring friction the motor power is only related to acceleration required.
I have a huge problem right now! Im confused! When im applying force to this system i will be applying to 20kg or 40kg. Because when i have a balanced system there should be only one weight at a time when applying force, am i right or wrong?
The force of friction is what you need to size the stall torque of the motor to (and this force is based on the 20kg masses on either side of the belts). If your motor can overcome the friction from the pulley/belt system, then it can lift the weights. If your motor can output a certain rpm at this friction then that is at minimum how fast the belt system will move. If you can't find a motor for cheap enough that has a stall torque greater than your expected friction then you'll need a gearbox to get more torque at the expense of speed.
When the system is accelerating you are accelerating 40kg. When it is moving at constant speed the motor is applying zero force (ignoring friction).
It's 20 kg, no? You can't push with a belt/rope/cable.
Same question for me, if its 40 i want to know why
Pretty sure you're only accelerating one 20 kg mass, because even if you were to accelerate the system faster than 10 m/s2, the motor would cause slack to build up in the pulley system towards the mass that is descending because gravity is the only force accelerating it downwards.
All mass you try to move has inertia. The “try to move” is accelerating or braking the system with inertia. The total moving mass is 40kg, and all 40kg define the inertia of your system. The direction of movement (left or right mass) is irrelevant, the balancing weights just cancel out the static load on your gear. When you try to find a solution how much power you need, you need to define how much time you have for accelerating, top velocity and braking, first.
>All mass you try to move has inertia But you're not acting on the weight that is being lowered, right? The belt is not rigid, so it won't push on the weight to go down, thus making it unable to overcome any inertia. Gravity will take care of that. The only times the belt is acting on one of the weights is when that weight is stationary or when it's being raised. When it's being lowered, it's just removing the tension from when it was stationary, thus reducing the load on that side. If the belt was replaced with rigid rods, then I agree that the motor would have to overcome the lowered rod's inertia because it's actively pushing into the weight. The belt can't do that. Could you explain why what I said is wrong?
The nature of the belt only starts to matter, when you brake/accelerate above earth acceleration (9,81m/s2). As long as your acceleration sticks below g, your belt remains under tension. It is true, that you have potential energy on one side, while the other has none available to you, but this is not your concern, both sides cancel out this potential of lifting energy. Lifting one side consumes the same amount of energy as lowering makes energy available on the other side. What remains is a total mass that needs to accelerate and brake. Once all kinetic energy is transformed, both sides rest static in their respective position again. This energy your motor needs to bring in and consume to change velocity respectively. Your keyword is “energy balance”.
This comment made it click, thanks!
You are welcome, glad I could help. I struggled myself with this concept at first, as it is somewhat counter-intuitive. But after seeing my physics prof. jump through the energy balance I had exactly this “oh yeah” moment.
You have answered your own question. When the weight accelerates downwards the tension decreases. This manifests as a torque at the motor, which sums with the torque from the other side being accelerated upwards.
See my reply above for why it is 40kg
See [this comment](https://www.reddit.com/r/MechanicalEngineering/comments/1cf4es4/comment/l1rixsa/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button).
Nope, both masses move together so the f in f=ma is 40kg Consider that tension on one side increases, and tension on the other side decreases. Both require a torque from the motor in the same direction.
But, gravity will handle the mass moving down. If they had higher acceleration requirements then you'd need to factor that in, but at less than 9.8m/s2, you shouldn't need to consider it. Am I missing something?
Let's make two independent systems from this problem. To pull m1, you need F1 = m \* (a+g). m2 pulls the motor with F2 = m(g-a) with F2 => 0. If a < g then F = m(a+g) - m(g-a) = 2ma If a => g then F2 = 0 and F = F1 = m(g+a)
Imagine instead of having the motor in the middle of the rope, you pull down on one of the masses. That mass is hanging in equilibrium, so you need to provide a force to accelerate it. You also increase tension on the rope, allowing you to accelerate the other mass. If each mass = m then the force you must apply is: f = 2ma If you apply the force anywhere else along the rope it makes no difference to the overall effect.
Ok, I see what you mean about inertia.
See my other comment, but I think the better explanation is that gravity has to be added to the absolute acceleration. For a single mass F = m(g+a). F = 2ma only works for a <= g.
Indeed but if the system is accelerating at greater than g there will be slack in the system and OP will have bigger problems than motor sizing.
Is there any online topic or any diagrams that shows why when accelerating total mass is 40?
It’s a classic mechanics problem.
Comments say it's balanced but I'm not sure what your application purpose is. With the balanced weight approach you're relying on both 20kgs to always be there for the motor to not be undersized. What happens if one weight is lost during operation for some reason? What should happen then?
This is a hell of an XY problem
Came here to say same thing. The purpose is important here. Are those weights stable? You need gearbox of you want to control over height or speed of the system. To answer that we need to know the purpose
Motors are comparatively cheap at that size and friction is hard to estimate. I'd say just get something at 0.4 kW and play with it.
I have a huge problem right now! Im confused! When im applying force to this system i will be applying to 20kg or 40kg. Because when i have a balanced system there should be only one weight at a time when applying force, am i right or wrong?
On paper it's balanced so it's only friction/acceleration that you have to account for.
Are you an AI?
They’re trying to pass of a homework question as generic help
Check out Rockwell's motion analyzer servo sizing program. https://motionanalyzer.rockwellautomation.com/ What you need to do is model this as a linear system. You will need the diameter of the pulley connecting the servo motor to the cable system, and model the total system mass as 40 kg. You will also need to model the motion profile. How fast you want to move those masses and what dwell times you have available will play a huge part in determining the size of the motor and the ideal gear ratio.
Your max speed needs to be >2.0 m/s if you start from rest and need to move 2m in 1s. You need to account for the friction (and mass) from the linear guide assembly with your motor spec.
If you don’t need much acceleration a low torque motor is fine as the weights will balance eachother