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Please I need some help. If I have a driving shaft with 10 Nm torque and 1000 RPM, and I add two 3-inch pulleys on the shaft to transmit power to another shaft with a pair of 6-inch pulleys mounted on it and connected to the initial 3" via a belt, will I have transmitted a total torque of 40 Nm to this new bar at 500 RPM?

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    $\begingroup$ torque times speed is constant. You halve the speed and double the torque, so 20Nm $\endgroup$ – agentp Aug 7 '16 at 20:26
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The pulley shaft on the driving shaft is only required to determine the rotational speed of the second pulley on the free shaft. Since the motor can only produce 10Nm of torque, you know the torque and the pulley size is irrelevant.

What agentp was referring to is conservation of energy, meaning in this case the power of your driving shaft is transmitted to the free shaft, assuming no losses.

The power equation is as follows:

Power = Torque x Rotational Velocity

So you can solve for the power of the drive shaft and the rotational velocity. The rules for pulley or gear diameter to speed ratio is:

D1/D2 = N2/N1

Which can be determined by inspection in this case as the pulley with diameter twice as big will logically rotate half as fast.

Then substitute those values into the equation and solve....which results in 20Nm

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