I sucked at physics in undergrad and I suck at physics even more now. Thanks.
NO! no longer even on the top of the worldwide! Revelation 7:a million (Infallible King James version!) And after those issues I observed 4 angels status on the 4 CORNERS of the earth, keeping the 4 winds of the earth, that the wind shouldn't blow on earth, nor on the sea, nor on any tree. plenty for Christian technological understanding…!
Looks like you need to solve for torque. 6 is a distance - your radius is 3 inches. 3 inches times 2.54 cm per inch = 7.62 cm, or 0.0762 metres. 50 kg is a mass. To convert mass to force in Newtons, multiply by the acceleration due to gravity of 9.8 m/s^2 = 490 Newtons. To get Joules (energy), multiply force by distance: 490 N * 0.0762 m = 37.338 Joules. We're missing a time value. How fast do you want the motor to turn? You might also be able to solve this by answering how far and how fast do you need to lift the mass? Divide the work in joules by the time in seconds to get the wattage of the motor. Power = work / time. Divide the power by the AC rms voltage to get the current rating.
We'll need more information. Technically 50 kg is a mass and you're trying to overcome a force. If you put the object on a scale and the scale read 50 kg, the scale was converting the 50 kg mass into Newtons on the earth's surface and expressing the result (wrongly) in kg. Remember F = ma? F in Newtons = mass (in kg) x acceleration (in m/s?) on earth, a = 9.81 m/s?, so a 50 kg mass exerts a force of 9.81 x 50 = 490.5 Newtons Now we'll need to know how rapidly you want to lift this mass. The formula for motor horsepower is Power (in HP) = 0.7457 (Horsepower per Kilowatt) x Torque (in Newton-meters) x 2π x rotational speed (in rpm) / 60 (seconds per minute) / 1000 (watts / kilowatt) This gives you shaft power output. Because of the motor's ineffiecencies, you will need to apply between 25% and 33% more input power than you expect to extract. If you attach the cable directly to a 6 diameter drum and the cable is lifting a 50 kg mass, the torque on the shaft is 490N x 3 inches (the pully radius) x 0.0254 (meters per inch) = 37.33 Newton-meters The shaft output power for a typical 1760 rpm electric motor will be HP = 0.7457 x 37.33 x 2π x 1760 / 60,000 = 7.01 You'll need a motor rated at about 10 HP to do this job. You will also want to consider stepping down the speed, which will decrease the efficiency of the system considerably but will also reduce the speed the cable is wound onto the pully (in this example it's at 46 feet per second or about 30 miles per hour). If you use a step-down gear to reduce the take-up speed to 1 foot per second, the rotational speed drops to about 38 rpm and the shaft horsepower drops to 0.15 HP. Of course the gearing introduces a great deal more inefficiency and the resulting rated horsepower of the motor might have to be as much as 3 times the calculated output. You would be looking at a 1/2 or 3/4 HP motor.
