As easy as it can get

Umm, a telephone-post maintenance worker's job?
Are you asking about the amount of work to be done?

 

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?

 

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.

 

You can integrate [tex] - \tau d\theta [/tex], where [tex] \tau [/tex] is the torque on the rod due to its weight.

 

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.

 

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.

 

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.