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I am trying to understand how to size an electric motor to pull a person up a sloped incline.

I know that I need to take into account the speed desired, the weight of the person and the friction/resistance counteracting (including the efficiency of the motor).

I'd be interested to know how you would begin to calculate the requirements

  1. From a purely hypothetical perspective
  2. If we could measure it in the real world with a force guage
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The speed is not a factor but the acceleration of the person is.

The equations are

  • m= mass of the person

  • $ \mu=friction\ index$

  • $\alpha= acceleration$

  • g=9.8ms^2

  • $\theta= angle\ of\ the\ ramp$

$$F_{net}=mgsin \theta-\mu mcos\theta=ma $$

Dividing both sides by m we get

$$a= g sin\theta-\mu cos\theta$$

This is the acceleration downslope if the person is let to slide-down free.

So we need a force to be at least greater than this to stop the downslope motion and push the person up. If you know your desired acceleration upslope the total force will be the sum of the

$$F_{final}=F_{net}+m\alpha_{up}$$

So if your motor is pulling the man up by a rope winding around a pully with radius, R, the torque required is:

$$\tau=F_{final}R$$

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  • $\begingroup$ Thanks Kamran. is 𝜇 the mass of the person? $\endgroup$ Jan 4 at 15:57
  • $\begingroup$ So a person weighing 100kg on a slope of 45 degrees would be (9.8^2)(sin(45)) - (100)(cos(45)) = -2.8 $\endgroup$ Jan 4 at 15:57
  • $\begingroup$ no, $\mu$ is the friction index. mas is m. $\endgroup$
    – kamran
    Jan 4 at 15:59
  • $\begingroup$ Beware these types of calculations. In the real world the frictional forces will be significant, maybe even dominant. A more practical method would be to find a machine which does a similar job and copy the motor specs for that. $\endgroup$
    – Drew
    Jan 5 at 5:03

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