You can approximately calculate some kind of upper limit for the "power" a motor can deliver for an infinite duration, due to thermal design limits at lets say 'standard conditions' (normal room temperature, normal ventilation, ...):

A Motor has one or multiple phases (coils that will induce a magnetic field), which have:

• I_max - a maximum design current (in the unit: Ampere or milli-Ampere),
• R- an electrical resistance (given in the unit Ohms)
• U_max - a maximum design voltage (given in Volts)

At least two of the above must be known. The relation between them is: R = U_max / I_max, so you can calculate the third one. You can measure the current for a given voltage or the restance of the coil, if you need to. But, make sure the motor doesn't move while measuring (e.g. by fixing the axle).

If you have I_max and U_max you can calculate P_max = I_max * U_max [Unit: Watts]. This will be the maximum power the Motor will -consume- without destruction under normal conditions! It can stand more power, if you improve cooling. And you can overload a motor for a limited time, until it gets too hot and finally breaks...

Each motor has an efficiency which depends on the motor itself and how fast it is spinning (see datasheet). This means the ratio of energy in form of motion or acceleration that is produced relative to its consumed energy. At optimum speed this can be as low as 50% or even higher than 90%, this depends on the motor.

Each good motor datasheet has some kind of curve that gives you an value for efficiency (%), torque (typically in: Ncm, Nm, mN*m) and motor current. At speed=0 the motor current is the highest (I_max) and the torque (=holding torque) is also at its highest value. The torque is the value for the rotational force. The holding torque is achieved when applying the nominal/design voltage of your motor at stand-still (=very short period until it moves faster) and/or blocked motion (=until not blocked anymore, good for measurements). If you use a higher voltage (which is valid but might damage the motor), the holding torque will also be higher. You can measure the holding torque, by mounting a little lever to the motor (e.g. r=10 cm=0.1m long) which the motor pushes onto e.g. a kittchen scale. The mass m (in Kilograms, e.g. m=34g=0.034kg) measured by the scale can then be used to calculate the force, which is F = m * g, where gis the earth gravity constant (approx. 9.81 m/s²). Then you can multiply the length of your lever by the measured force, you'll get the torque: M = F * r = m * g * r = 0.034kg * 9.81 m/s² * 0.1m = 0,033 Nm ... where "Nm" means "Newton-Meters" or if you use the lever length in cm it is: M = 0.034kg * 9.81 m/s² * 10cm = 3,33 Ncm (Newton-Centi-Meters).

Mechanical systems usually need to overcome the stiction first, which usually depends on the mass of it (which pushes it onto a surface), the materials of the material that has contact to the surface and the material of the surface itself.

What is Stiction? Here is a little experiment: You might have noticed that your smartphone will not slide downward an slightly inclined surface... the gravitational force is not strong enough, yet. But, if you push it a bit, it will overcome stiction ("sticking to the surface") and will slide on its own. As said, how strong the stiction is depends on the material of the thing... e.g. rubber on wood has an higher stiction than flat, polished plastic on polished glass. So also the roughness of the materials have an influence. The force to overcome Stictions can be calculated by the formula: Fs = mu * Fg, were mu is the constant for the materials and Fg is the force that pushes the thing to the surface. In a simple case (e.g. a smartphone laying on a flat and horizontal surface, no inclination) this can be Fg = m * g, where m is the mass (e.g. of the phone) to be moved and g = 9.81 m/s² (earth gravity constant) again.

The needed torque depends on the mechanical system. In a simple case you just mount some wheels on the motor. Then the radius (half of the diameter) of your wheels are r. And your needed torque to overcome stiction will then be Ms = Fs * r. If you use larger wheels you need a higher torque. If you use gears you can also influence the needed torque (in both directions), but you'll decrease the efficiency of the system and - at the same time - introduce a now stiction torque of the gears (which might or might be significant).

As soon we've overcome stiction, you will have still friction that might stop things from moving... that means deaccelerating it until it finally stops again. But, theses forces are usually much smaller (for slow speeds only!). And remember you can - for a short period of time - drive your motor with a higher current/voltage than specified!

All those friction, which depends on a lot of other factors, limits how fast it can move (top speed)... and finally how fast it can accelerate to that speed (maximum acceleration). This all falls under the topic "dynamics" and there is lots of it to learn... but, the basics can be learned quite fast.

So your implicit questions actually are:

Please note that this is ' SDwarfs's answer from (https://robotics.stackexchange.com/questions/14943/how-much-weight-can-dc-motor-carry), not mine.

The shaft torque of your motor can be calculated as

Tshaft = 9.55*Output Power/N;

where N is revolution per seconds;

then, the load torque can be calculated as,

Tload = F*r;

where is the weight of the block you want to lift & r is the radious of the pulley connected to motor shaft.

Tshaft should be equal to or greater than the Tload

What is your motor's power output?

Should you have it, you can compute a few essential parameters.

Follow these steps: https://pages.mtu.edu/~wjendres/ProductRealization1Course/DC_Motor_Calculations.pdf

Sulaymon Eshkabilov Do you think for the AC motor calculation steps are the same?

What is your motor's power output?

Steps are pretty much likewise, but with the ac motors how many phases are present (and unit conversions) should be taken care of carefully.