Does 60 kg-cm in motors mean it can lift 60 kg?
60 kg-cm refers to the torque available at the shaft of the motor.If one were to mount a pulley of radius R and make it lift a weight W, then depending on the radius of the pulley, the amount of weight that can be lifted would vary, but the product of the radius of the pulley R and the weight W that can be lifted would be constant and equal to the torque developed by the motor. In our case we consider two examples, one with a pulley of radius 1 cm and the other with a radius of 2 cm. Since the torque available is 60 kg cm, the weights that can be lifted are 60 kg and 30 kg respectively.
Which DC motor is used in lift meachanism?
DC compound motors are used in elevators. A DC compound motor can be called as a superimposition of DC series and DC shunt motor. So, it has high starting torque like a DC series motor and has better control of speed like a shunt motor. Both these features make it ideal to be used in lifts where the lift needs to break high stall torque and also needs efficient speed control over its operation - be it travelling one floor or 50 floors at a time.The compound motor responds better to heavy load changes than a shunt motor because of the increased current through the series field coils. This boosts the field strength, providing added torque and speeds.Thus, DC compound motor satisfies the load and operational requirements to be used in an elevator and is the most sought for.
What DC motor should be used to lift 10kg?
Well I think that you should use a parallel DC motor, combined with gear reduction, to reduce speed and incrase torque. Also there are some stepper and servo motors that can lift a lot. Depending on what are you using it for you can pick an appropriate motor. If you are making something bigger, a parallel DC motor is probably a way to go, but for smaller applications steppers and servos are better and simpler.
How do I build a basic robotic hand using servo motors and Arduino in a short amount of time?
Here's what you can do-Step 1. Collect the following hardware -a. 5 Servo Motors. 1 for each finger. The size and torque should not matter unless you are planning to make the hand fit into a smaller area (need smaller sized servos for that), or are planning to lift heavy objects with it (need servos with higher kg-cm values for that). A servo with <2kg-cm should fit your needs.b. Arduino Compatible board. Something that can control the positions of your 5 servos. Usually some Motor Driver shield is needed for the purpose if you are using an Uno/Mega etc., but you can explore other options such as -Mini Driver, Magician Controller or Micro Magician boards by DAGU, which will allow you to do the same thing without buying an add-on shield.c. The Humanoid Hand from 4M Kidz Labs. This has the complete hand, finger and strings.Step 2. Mess around with the 4M Humanoid Hand and at the end of the strings, in place of the rings that it has, attach the horn of your servo motors. Modify the body of the Hand too if you want to.Step3. Code the servo position control for your servos.This step should not take more than a few hours if you are good at Arduino programming.Voila! You're done!If you are in India, the parts I mentioned can be bought from http://RoboRium.comEdit (14th November 2014): I made this project for myself, just to see if I could.Instead of 5 servos I used just 3. During the build I realized that 3 was enough, but if you want to extend it, it can go up to 4 or 5. P.S. there won't be enough space for 5 servos if you are putting them where I have done in the video. https://www.youtube.com/watch?v=...
How much power is required to lift a 10 kg weight by a DC motor?
It depends upon speed of lifting.Equation of the power is:P = f*v (P is power in Watt, f is force in Newton and v is velocity in meters /sec)Force required for lifting 10 kg weight is (10*9.81=) 98.1 Newton. If you assume lifting speed to be 0.1 meter / sec, then power required is (98.1 * 0.1 =) 9.81 Watt. This much power is required without considering losses in lifting mechanism.Generally rated motor speed is 1500 rpm.Consider a simple lifting arrangement consisting of a rope and pulley. Rope tied to the weight is wound on the pulley which is rotated by the motor. Even if you use a very small pulley of say 25 mm dia directly mounted on motor shaft, it will result in lifting speed of (25*3.14*1500) mm/minute = 117750 mm / minute = 1963 mm / sec = 1.963 meter / sec, which is too high speed. Hence you should use a reduction gear, to match your speed requirement.If you use a speed reduction ratio of 19.63:1, you will get lifting speed of 0.1 meter / sec and power requirement (ignoring losses) is 9.81 Watt. Considering losses in the speed reduction gear and pulley, efficiency may be 50% and motor output power has to be 9.81 / (50/100) = 19.62 Watts.If you use seed reduction of 39.26, speed of lifting as well as power requirement will be reduced to half of above example.Hope, I have answered the question. If you need any more information related to this answer, please comment.
Why is lifting a metal rod from the end harder than lifting it from the center of the rod?
Lifting a rod at the middle requires only a force opposite the center of gravity equal to the weight (plus a little force to accelerate the rod up).When lifting the rod at the end, not only is the upward pure force required, but an additional moment is required, equal to one-half the rod length times the rod weight. This moment is difficult to generate with only a twist from a hand.
What type of motor will work best, If i have to lift a load of 500kgs with very precise & quick start / stop and variable lifting speed control?
You can use induction motor if load is heavy and if you want perfect position or motion you can use servo or stepper but for the variable speed and torque control you need to use VFD (variable frequency drive).
How do I calculate the electric power required to lift 100 kg of material?
For this question to be complete, you need two ask about two more parameters, for what hight ? and in how much time ?.Anyway, here is the answer.F = mg = 100*9.81 = 981 NSuppose we will lift it for 5 meters.W = F * L = 981* 5 = 4905 N.m = 4905 JSuppose we will lift it in two minutesP = 4905/120 = 41 watt