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Pipe climber

  1. Feb 25, 2014 #1
    Hi I got a physics question and would like you to help me abit urgent. I have a device that has to climb a pipe of 105mm in diameter. I have the basic concept of my device I just need to know a few things, the device has 4 wheels acting on the inside wall of the pipe it has to climb up 2.5 meters then come down pulling a chain of 0.2 kg/m the tube is set 0.5 meter above the ground, so the chain will need to be carried up all the way to the top. So 3 meters in total minus the height of the device which is 20cm. The chain is attached to the bottom. Wheel is 9mm

    1. What force should the wheals act on the inner
    2. what motor is right to be used to pull the device and the chain, the device weight roughly 370 g without the motors. Thanks you

    Attempt at solution
    Total weight of chain and device is 0.2kg x 2.8m = 0.56kg
    Adding the device weight of 370g to 560g we total to 0.93kg

    Well with two motors on two wheels, I can calculate the force acting down on the device which is force= mass x gravity
    F = 0.93kg x 9.81
    F= 9.12 N
    So there fore I need two motors producing more force than 9.12N to create an upward force, therefore
    If I have a motor attached to a wheel of 9mm in diameter. And when I buy I motor I am given the angular velocity (w), and the torque (t)
    So if torque = force x radius of wheel x sin90
    And I have torque and radius then rearranging the equation to get force upwards is
    Force = torque/ (radius x sin90)
    So now I have force acting upwards and I can get a good enough motor to have a higher force acting upwards than downwards.
    The motor I buy will have the info of the torque and angular velocity
    So to get a tangent velocity on a position on a wheel
    Tangent velocity = radius of circle x angular velocity

    Having the tangent velocity in rad/sec can I work out the time it takes for the wheel to travel up to the top, and is there any thing a posing the velocity to decrease it if so how do can I calculate it and what is it. Thank you
     
  2. jcsd
  3. Feb 25, 2014 #2

    haruspex

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    Not sure how you plan to do this with only four wheels. What is the geometrical relationship between them when in the pipe? Are their centres all in the same horizontal plane or all in the same vertical plane?
    Other than that, what you have posted looks fine. The main loss will be in rolling resistance (losses in deformation of tyres). Since you have to maintain enough lateral force to prevent slipping, that could be significant.
     
  4. Feb 25, 2014 #3

    BvU

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    Hello Done, and welcome to PF. Nice "Introductory Physics Homework" you are posting here. Apparently you missed the instructions obliging you to use the template, for your sake and for the sake of potential helpers.

    Well, it is clear you are in a big hurry. So much that it has become very difficult for someone a bit further away to understand what this is all about. (Meaning: if you can find someone close by to help you, it might go faster; a teacher or his/her assistant).

    I am going to spend some time trying to decode your story. But I am not going all the way, because I am almost certain I get lost very quickly.

    So clearly the chain is attached to the device after it has climbed up these 2.5 m. Correct ? Probably not. Does it jhave to go up and down with this chain attached ? How pointless, but who knows ...

    Just to wake you up: "Wheel is 9mm" can be interpreted in several ways. Do you really want to waste your time by not telling what this means ?


    Please go through your story once again and make sure it is crystal clear. We want to understand you as best we can; don't make it too difficult. And if you have some drawing it will save you 1000 words.
     
  5. Feb 25, 2014 #4

    haruspex

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    No, I believe it is to go up and come down, both with the chain attached.
     
  6. Feb 25, 2014 #5

    BvU

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    I would rather hear that from Done, so we all save time.
     
  7. Feb 25, 2014 #6

    haruspex

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    It's there in the OP, twice:
     
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