# A Torque applied on a lighting column

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1. Feb 20, 2017

### aon3t

Greetings!

I am currently making a program which is about lighting columns. And I am currently at a halt because I need some basic physics done before I can complete it.

I have a lighting column which is
L = 8 m
D = 0.087 m
M = 69 kg

I have a force, applied from the wind load, at
F = 1235 N

Now I need to find the torque applied on the whole column. I have used this formula earlier
T = ½ ⋅L⋅M⋅g (where i've used F = mg)

This have given me T = 4944 Nm.

I just need a validation if these calculations are correct.

Because my real problems starts now:

I want to add an arm to the top of the lighting column with these data

L = 3m
M = 36 kgs
D = 0.120m

And I wonder what the combined torque applied to the column is now?

The arm will be at a 90° angle outwards from the column like this → Γ

Later on I want to add a lighting armature to the arm and do the same calculations (but that won't be a problem when I understand how the arm affects the column).

English is not my first language, so bear with me if I am unclear, or not using the correct terms.

2. Feb 20, 2017

### BvU

Hello aon,

Just for clarification: is the column in a vertical position ? If so, there is no torque ½ ⋅L⋅M⋅g (action line of mg goes through base of column).

With the arm (if it's horizontal) you can add T = ½ ⋅Larm⋅Marm⋅g to the wind torque T = ½ ⋅Lcolumn⋅Fwind

Not that at the top of the column the arm exerts this same torque T = ½ ⋅Larm⋅Marm⋅g

3. Feb 20, 2017

### aon3t

Thank you very much. This is currently how I've done it (I used the winload "Fwind" instead of Mg on the column just as you wrote). And yes, the column was in a vertical position.

So if I understand it right: I find the wind torque, and add this to the torque applied by the arm (which, as you said, applies the same torque to the top of the column)?

And one more question:

If I now want to add a lighting armature to this arm. Which have the following data:

L = 0.758m
M = 12 kg

Which is the best way to find the new Tarm?

a) Just add the mass and length to the arm (seeing it as a bit wider and heavier arm than before) and use the same formula as earlier?
b) Find the torque seperately and add them? (this gives me a lower torque than the suggestion above)
c) Find the Farmature=mg of the armature and just add this to the torque formula? (Idea is that then the whole force of the armature is applied to the tip of the arm)

I have the 3 examples (g ≈ 9.81):

a) T = ½ ⋅(Larm+Larmature)⋅(Marm+Marmature)⋅g = ½⋅(3+0.758)⋅(19+12)⋅9.81 = 571.42 Nm

b) Tarmature = ½ ⋅L⋅M⋅g = ½⋅0.758⋅12⋅9.81 = 44.61 Nm
Tarm = ½ ⋅L⋅M⋅g = ½⋅3⋅19⋅9.81 = 279.6 Nm
T = Tarmature + Tarm = 44.61 + 279.6 = 324.21 Nm

c) T = ½⋅Larm⋅(Marm+Farmature)⋅g = 456 Nm

4. Feb 20, 2017

### BvU

Picture is worth ...

So far I have (at O, the base)
Fwind * 4 m ,
marm g * 1.5 m
marmature g * (3 + 1/2 * 0.758) m

At the top, only the latter two​

5. Feb 20, 2017

### aon3t

Thank you! Now I have all I need, I shall write a thanks to you when I hand in my paper for the program.

I had my eyes on how you solved the Tarmature, but I disregarded it because I found the T to be a bit smaller than expected, but that was because I forgot to add it to the Tarm.

Thanks a lot!

6. Feb 20, 2017

### BvU

You're welcome. Good luck with the program.