Calculating the linear force from the motors torque

Hi guys

I need a motor for my project, but I´m not quite sure if my calculations are correct.

I need a motor to drive 10 kg on a linear slide, where the friction constant is 0.1. The motor will drive a belt the the object is fixed to the belt.

To calculate the force required to move the object:

F = m * g *u , F = force, m = mass of object, g = gravity constant, u = friction. F = 10 kg * 9,82m/s2 * 0,1 F = 9,82 N to push the object. That must be right.

Calculating the motors torque:

T = F * d, where T = torque, F = force, d = distance

T = 9,82 N * 0,01 m (distance from motor shaft to belt) T = 0,0982 Nm

Is that right, that I dont need a larger motor, or am I missing something? Is the distance when using torque meassured from the center point of the shaft or on the shafts edge?

Kind regards Kasper

You calculations look absolutely ok assuming you belt drive wheel has a pitch diameter of 20mm. The length of a torque lever is the distance between the turning point (center of shaft) and the point where the force vector makes a 90 degree with the lever (in this case: pitch diam/2) But observe 2 points:

  • The force calculated is the minimum force needed to overcome friction, nothing else. as calculated that force will give you zero acceleration
  • The calculation assumes that the whole system is absolutely horizontal

Given these two points a motor with a torque of greater than 0,3Nm should perform fine. Then you will have 20N for acceleration which gives you 2m/s/s

Hi Nilton

Thank you for your quick reply and explanation :) You are ofcause right about the to points - I´ll remember that.

I´m wondering what would happen to a motor like the Nema 23: http://www.ebay.com/itm/Nema-23-Stepper-24V-2A-Motor-Mill-Robot-Lathe-RepRap-Pulley-44-Timing-Belt-/181108472334 if I programmed it to run at a certain speed and amp, and I then intentionally put more resistance in and turned it backwards. Would the motor break down, or what would happen?

Kind regards Kasper

Always allow a generous factor for friction losses in the rest of the system and to overcome static friction (greater than dynamic friction). So 0.2Nm would be a reasonable first estimate.

If your belt is stiff and tight you'll need more torque just to move that, so try to use flexible belting (rubber+glass fibre, not polyurethane+steel cores)

A NEMA23 stepper motor is total overkill, they can produce much more torque than you need and are expensive to drive. You'll also may struggle to find a belt pulley 20mm in diameter for an 8mm shaft (semi standard for NEMA23).

NEMA17 has 5mm shaft, so 20mm pulleys no issue, and torque ranges are upto about 0.5Nm typically.

Thanks for your reply Mark.

The size of the belt pulley is not that important, 20 mm was just an example - so a larger one is also ok. It is actually not just a stationary load of 10 kg which have to be moved. I´m building a device which have to extend and flex a cadaver knee. Since the knee is pretty stiff, I must add some more torque, and I would be afraid of only having 0.5N.

But yes, a NEMA 23 might be to much. I´ll remember to look for a rubber belt :)

Since it is just a pilot study - the voltage is not a problem.

Robin2 has written a excellent text on stepper motors

Not the usual load then!

Sometimes its better to measure torque than trying to estimate it and get it wrong, measuring torque just requires some string, a drum to wind it round and a spring-balance or weights / weight-pan.

No I guess not Mark ;) I´ll definently try measuring the force and torque needed with some weight - that was a good idea.

Thanks Nilton, the guide looks very informative for newcomers. I´ll have a look at that one tommorow - its getting late in Denmark!