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As easy as it can get

  1. Jun 16, 2008 #1
    An 8m and 200kg telephone post is laying on the ground. What work one needs to do to reposition the post vertically?

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Jun 16, 2008
  2. jcsd
  3. Jun 16, 2008 #2
    Umm, a telephone-post maintenance worker's job?
    Are you asking about the amount of work to be done?
  4. Jun 16, 2008 #3
    Oh sorry, I meant work :uhh:
  5. Jun 16, 2008 #4
    Okay any thoughts about how you want to go about solving this problem? To simplify, assume that the post's mass is uniformly distributed. Where is its centre of mass?
  6. Jun 16, 2008 #5
    If the mass is uniformly distributed then the center of the mass is in the center. Thus I need to lift the center by half of the lenght of the pole [tex] A = mgl/2 = 8 kJ [/tex]

    Now what about a numerical way to solve this? I was wondering about integrating somehow using the moment of force or something of the kind
  7. Jun 16, 2008 #6
    What do you mean numerical method? You can find the centre of mass if required by integration, but other than that I don't see what else you should do.
  8. Jun 16, 2008 #7
    Well if there would be a force acting on one end of the pole then I thought I could write and integral of some sorts and find out what work is done by the force to place the pole vertically
  9. Jun 16, 2008 #8


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    You can integrate [tex] - \tau d\theta [/tex], where [tex] \tau [/tex] is the torque on the rod due to its weight.
  10. Jun 16, 2008 #9
    Thanks dx

    Any book which explains torgue in detail? I imagine all on classical mechanics must have an explanation, just tell me
  11. Jun 16, 2008 #10
    Why even bother with torque?

    If you lift this log the 4 requisite meters and put in a magical frictionless bearing in the center, every orientation of the log has equal potential energy. So isn't the solution the gravitational potential of the center of mass at 4 meters?

    Any solution involving torque would include the inertia of the pole about some hinge, which, although not unnecessarily complicated, is far more complicated than the gravitational potential formula.
    Last edited: Jun 16, 2008
  12. Jun 17, 2008 #11
    Yep, we actually already did that. Read the previous posts

    I wanted to involve torgue that's why I asked how to do it

  13. Jun 17, 2008 #12


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    Torque is basically the rotational equivalent of force. For example, you know that the work done by a force [tex] F [/tex] in moving an object a distance [tex] \Delta x [/tex] is

    [tex] F\Delta{x} [/tex].

    Now consider some particle on the plane with a force [tex] (F_x, F_y) [/tex] on it. It you now rotate the particle about the origin by and angle [tex] \Delta \theta [/tex], the work done on it can be shown to be

    [tex](xF_{y} - yF_{x})\Delta \theta[/tex].

    So this strange combination [tex] xF_{y} - yF_{x} [/tex] is the torque, which is also the vector cross product of (x,y) and
    [tex] (F_x, F_y) [/tex]. This can also be written [tex] |r||F|sin\theta [/tex] where [tex] \theta [/tex] is the angle between
    r and F.

    In the case of your problem, the torque is [tex] - \frac{l}{2} mg cos\theta [/tex] where [tex] \theta [/tex] is the angle between the ground and the rod. If you integrate minus this from 0 to pi/2, you get the answer.
    Last edited: Jun 17, 2008
  14. Jun 17, 2008 #13
    How can it shown to be [tex](xF_{y} - yF_{x})\Delta \theta[/tex] ?

    why is it l/2 if the force is acting at the end of the pole?Is it because of the mass distribution?If it wouldn't be symmetric, then what?

  15. Jun 17, 2008 #14


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    Displace the particle by a small angle [tex] \Delta \theta [/tex], and the find the small displacements in the x and y coordinates in terms of this. If these are [tex] \Delta x [/tex] and [tex] \Delta y [/tex], then the work will be [tex] F_{x} \Delta{x} + F_{y} \Delta{y} [/tex].

    No, the force acts effectively at the center of mass, which is the middle of the rod if it is uniform.

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