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## Homework Statement

An 8m and 200kg telephone post is laying on the ground. What work one needs to do to reposition the post vertically?

## Homework Equations

## The Attempt at a Solution

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- #1

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An 8m and 200kg telephone post is laying on the ground. What work one needs to do to reposition the post vertically?

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Umm, a telephone-post maintenance worker's job?

Are you asking about the amount of work to be done?

Are you asking about the amount of work to be done?

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Oh sorry, I meant work :uhh:

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thanks!

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

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dx

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Any book which explains torgue in detail? I imagine all on classical mechanics must have an explanation, just tell me

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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.

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.

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Yep, we actually already did that. Read the previous postsWhy 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.

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

thanks

- #12

dx

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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

Any book which explains torgue in detail? I imagine all on classical mechanics must have an explanation, just tell me

[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.

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How can it shown to be [tex](xF_{y} - yF_{x})\Delta \theta[/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].

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?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.

thanks

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dx

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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].How can it shown to be [tex](xF_{y} - yF_{x})\Delta \theta[/tex] ?

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

thanks

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