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So, my company works with electric motors, and my boss wanted me to calculate how much weight could be lifted by a motor rated at 750W.

I know that I can't just calculate the work done by it without any further specification of characteristics, but what I what to know is a couple of formulas that could get me there using measurable values, for example, i can measure current consumed when lifting a 10kg weight.

Also, I'm open to suggestions, if you think there is an easier way to get where I want to, I am open to new ideas.

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  • $\begingroup$ A motor can't lift anything. A motor applies torque to a rotating shaft. The thing that you are asking about is called a winch. That is, a rotating drum that pulls on a rope that is wound around it. The amount of weight that a winch can lift is proportional to the torque applied to the drum, and inversely proportional to the radius of the drum. Besides the drum and the motor, an electric winch usually also has a gear train, so the gear ratio is also going to appear in your calculation. $\endgroup$ – Solomon Slow Nov 9 '15 at 14:16
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There are two separate issues here. The maximum weight your motor will be able to lift depends on its torque. The rate at which your motor can lift the weight depends on its power.

The torque of the motor can be increased or decreased by running it through a gearbox, so in principle you could lift as big a weight as you want as long as you use a very low gear (and therefore lift the weight slowly).

The rate of lifting is easier to be precise about. If you lift a mass $m$ through a height $h$ in a time $t$ the power required is:

$$ W = \frac{mgh}{t} \tag{1} $$

or putting it another way, if you lift the mass at a velocity $v$ the power is:

$$ W = mgv \tag{2} $$

(because v = h/t).

So to take your example of a 10kg weight, to lift this at a speed of 1 m/s would require a power:

$$ W = 10 \times 9.81 \times 1 = 98.1W $$

which is well within the power of your 750W motor. Alternatively you could flip this around and ask how fast you could lift the 10kg weight if you ran the motor at full power. Rearranging equation (2) gives:

$$ v = \frac{W}{mg} = \frac{750}{10 \times 9.81} \approx 7.6 \text{m/s} $$

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  • $\begingroup$ I know you know this, but I would only add for the OP that the 750W is rated input power which won't equal the mechanical output power when it comes to lifting a mass. The power available for torque will be less due to the efficiency of the motor (maybe 80% or less) adding gears also reduces efficiency and available torque. $\endgroup$ – user6972 Jun 19 '14 at 18:53
  • $\begingroup$ Thank you, i will probably apply the 80% efficiency, which seems fair for electric motors, thank you for the answer, that's exactly what i was looking for :) $\endgroup$ – Antero Duarte Jun 20 '14 at 11:15
  • $\begingroup$ If the motor is rated at 750 W, then that is the output power, not the input power. $\endgroup$ – Eric Jun 23 '14 at 12:26
  • $\begingroup$ Seconding @Brad. Generators are rated for electric output, motors for mechanic output. There could be some losses in the pulley system but that won't be the matter in such a simple exercise. $\endgroup$ – WalyKu Mar 9 '15 at 13:30

protected by Community Nov 9 '15 at 13:10

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