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AC Cable resistance

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What size AC motor do I need to turn a 6" diameter pulley that will wind up a cable with 50kg resistance?
I sucked at physics in undergrad and I suck at physics even more now. Thanks.

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/s²) on earth, a = 9.81 m/s², 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.
Catenary | Read more
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How to calculate the impedance of ug cable?
Am doing a project with UG 3 core 400sqmm XLPE cable i have got the values of the reactance resistance(ac), and capacitance for per kilo meter for per core , is there these value for all the three cores together or any formula to find it ....i need to have value for 3 core i.e, the net impedace for a kilometer.. pls help me

Resistance Between Terminals AC and AD?
What is the resistance between terminals AC? What is the resistance between AD? An underground telephone cable, consisting of a pair of wires, has suffered a short somewhere along its length (at point P in the Figure). The telephone cable is L long, and to determine where the short is, a technician first measures the resistance between terminals AB to be RAB; then he measures the resistance across the terminals CD to be RCD. http://capa-new.colorado.edu/teacher/capalibrary/Graphics/Gtype58/prob08.gif So I know that to find resistance we do R=pl/A but I dont really know how I would find it in this situation since it is not like finding AB or CD like I have done before. Please help

Since R = ρL/A the only variable here is L Let x be the distance from A to point P For RAB we have RAB is proportional to x and RCD is proportional to (L - x) So RAB/RCD = x/(L-x) so x = RAB/RCD*(L -x) Therefore x( 1 + RAB/RCD) = (RAB/RCD)*L So x = (RAB/RCD)*L/(1 + RAB/RCD) Now RAB = 2*ρx/A => ρ/A = RAB/2x = RAB*(1 + RAB/RCD)/(2*(RAB/RCD)*L) so RAC = ρL/A = RAB*(1 + RAB/RCD)/(2*(RAB/RCD)) If the cables are the same size and material then RAC = RAD
How to work Lifestyler System 50 cross trainer-electrical resistance?
Our neighbors are moving and gave us one of those cross trainer workout machines. It has a digital display on a stand that plugs into the workout machine with what looks like a telephone type cable, but there is no battery compartment or AC plug in, so we cant change the resistance level. I was just curious if anyone else had one of these machines and knew how to give it power. It is on probably the least amount of resistance now and basically is useless. Thank You!

Let a man look at it he will figure it out (especially with no directions) lol. but there has to be a power source to make the digital display work. Check to see if there are any compartments that require a screw driver. If that does not work check this site out you can look up the manual for the products Good Luck http://www.iconservice.com/manuals.html
tcarrw | Read more
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