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Homework Help: I want to know better

  1. May 12, 2012 #1
    I want to know what is the best motor can carry 1000 kg
    And the weight is not large and also small size
    And working with electricity, not fuel

    Can you help me
  2. jcsd
  3. May 12, 2012 #2
    How do you want to carry 1000 kg? To lift it vertically upward, or horizontally?
  4. May 12, 2012 #3
    I want to rise vertically, but not continue to rise, such as the helicopter
    To reach the high fixed
  5. May 12, 2012 #4
    How fast do you need it lifted?
  6. May 12, 2012 #5
    Sorry, I'm not an expert so these things can explain more
  7. May 12, 2012 #6
    any help plz Dickfore
  8. May 12, 2012 #7
    Which things?
  9. May 12, 2012 #8
    I want to lift the body off the ground for a distance meter and a 1000 km grams all of this on the air like a helicopter and then be fixed
  10. May 12, 2012 #9
    I'm sorry I don't know what you mean by this. What helicopter? What does "distance meter and a 1000 km grams" mean?
  11. May 12, 2012 #10
    To not have a man and weighed 1000 km sitting on a chair and I put motor bottom, bringing the chair to a distance of meters and holds that person
  12. May 12, 2012 #11
    Do you want it to work on this principle (only bigger):

  13. May 12, 2012 #12
    Something like this
  14. May 12, 2012 #13
    If the mass of the load is [itex]m[/itex], and it is pulled by a constant speed [itex]v[/itex], then, the power of the motor needs to be at least:
    P = m \, g \, v
    where [itex]g = 9.81 \, \mathrm{m} \, \mathrm{s}^{-2}[/itex] is the acceleration of free fall.

    For example, if you want to raise a ton (1000 kg) for 15 m (approximately a five story building) in half a minute (30 seconds), you would need a motor with a power:
    1000 \, \mathrm{kg} \times 9.81 \mathrm{m} \, \mathrm{s}^{-2} \times \frac{15 \, \mathrm{m}}{30 \, \mathrm{s}} = 4905 \, \mathrm{W} \approx 5 \, \mathrm{kW}
    Last edited: May 12, 2012
  15. May 12, 2012 #14
    Thank you
    i will see it
  16. May 12, 2012 #15
    Then, you need to adjust the diameter [itex]d[/itex] of the head of the motor to the torque it provides:

    \tau = \frac{m \, g \, d}{2}

    The angular velocity (in rpm - rotations per minute) is:
    n = \frac{60 \, v}{\pi \, d}
  17. May 15, 2012 #16
    wouldn't this kind of hard to achieve?
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