# Understanding required torque for a motor lifting a weight

This is a continuation of me trying to understand torque and stepper motors in my other question. I'm trying to understand the torque a motor would be required to generate to lift a small weight, and the formulas involved.

The first part of my question is to verify if I am calculating this correctly:

Let's say I have a 450 g mass (roughly one pound) then the force of gravity pulling it down is:

\begin{align} F &= ma \\ &= 0.450 \:\mathrm{kg} * 9.8 \:\mathrm{m}/\mathrm{s}^2 \\ &= 4.41 \:\mathrm{N} \\ \end{align}

If I have a stepper motor with a spindle for my string that pulls up my motor with a radius of 5 cm. I think my torque needed would be:

\begin{align} T &= Fr \\ &= F * 0.05 \\ &= 0.22 \:\mathrm{Nm} \\ \end{align}

So now if I want to move that mass I need to find a stepper motor that can output more than 0.22 Nm of torque, right?

The follow-on to my question is that if I want to see how fast I can move it then I need to look at a Torque speed curve, right?

My confusion is this: do I have to ensure that I'm moving slow enough to get the torque I need, or does that curve say if you need this torque you won't be able to go above this speed because the motor won't let you?

• A nice, clear answer, explaining in stages how you need extra 'power' at stage X to overcome Y, and that Z will require more. So many sites show you an algebra formula that leaves you more puzzled than when you started. (Personally, trying to work out the motor power needed to move a pivot in scissor lift arrangement forward 25cm when there is 10kg fore and aft of another pivot point !!) Feb 14, 2021 at 8:45
• This is a very good thread, and helps a lot. However there's one thing that I don't understand what is the radius that you use? Sorry, maybe this is clear for everyone, but I'm coming from an IT background, and these things are not that clear to me. I would like to build an adjustable standing desk for myself, and would like to know how powerful motors I need. Aug 27, 2021 at 17:50
• In my question above I used a radius of 5cm, but in my math I'm off by one decimal place. I'll edit it. Aug 27, 2021 at 19:13
• @sztech The radius of the pulley or the axile of the rotor, whichever produces the force.
– r13
Aug 27, 2021 at 19:39
• Thank you! I think that this is enough info to start. I'm still not sure if I need to add any other mechanism to hold the weight of the tabletop/monitors/printer, or if a stepper motor could hold the weight. Aug 29, 2021 at 10:39

You have the right concept, but slipped a decimal point. 5 cm = 0.05 m. The gravitational force on your 450 g mass is 4.4 N as you say, so the torque just to keep up with gravity is (4.4 N)(0.05 m) = 0.22 Nm.

However, that is the absolute minimum torque just to keep the system in steady state. It leaves nothing for actually accellerating the mass and for overcoming the inevitable friction.

To get the real torque required, you have to specify how fast you want to be able to accellerate this mass upwards. For example, let's say you need at least 3 m/s². Solving Newton's law of F = ma:

(0.450 kg)(3 m/s²) = 1.35 N

That, in addition to the 4.4 N just to balance gravity means you need 5.8 N upwards force. At 0.05 m radius, that comes out to a torque of 0.28 Nm. There will be some friction and you want some margin, so in this example a 0.5 Nm motor would do it.

Note also that torque isn't the only criterion for a motor. Power is another important one. For that you have to decide what's the fastest speed you want to be able to pull the mass upwards at. Let's say 2 m/s for sake of example. From above, we know the highest upwards force is 5.8 N.

(5.8 N)(2 m/s) = 11.6 Nm/s = 11.6 W

After accounting for some losses due to friction and leaving a little margin, the motor should be rated for about 15 W minimum.

• I have a comment sir. Your explanation was really good and clear but I have a question: let's say I want to spin the motor very fast so I can pull up the weight fast. Let's say I want to pull it at 10m/s with constant speed. If I provide the initial 10 m/s speed with a push from my hand so the motor reaches that speed, will the motor only have to provide 0.22Nm plus a little more to overcome gravity and friction in order to keep the 10 m/s speed? The same applies to 100 m/s? If I, somehow, give the initial push to the motor achieve 100 m/s, will the motor only need to provide 0.22Nm? Apr 19, 2016 at 15:13
• @Samul: Yes, you seem to have the right concept regarding torque. Note that the power required to sustain the upward speed still has to be there. Apr 19, 2016 at 18:35
• @Samul: At any steady speed, you only need enough torque to counter gravity and friction. However, there are two gotchas with this. First some part of the friction will be proportional to speed (viscous friction). Air drag is even worse, proportional to the square of the speed. Second, the power required is the speed times the torque. In a ideal system with no friction, the torque to maintain 10 m/s and 1000 m/s is the same, but the power required to produce that torque at the higher speed is 100 times more. If you needed 15 W at 10 m/s, then you'll need 1.5 kW at 1000 m/s. Apr 19, 2016 at 21:46
• @inf3rno you can use multiple motors to distribute the load (e.g. one per leg). That's a heavy desk to lift with a single gear-less motor. Jul 28, 2020 at 23:02
• Unfortunately not. Aug 29, 2021 at 23:24