Basic torque problem

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long_vacation
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#1
Report Thread starter 9 months ago
#1
So I saw two very similar problems on Youtube videos:

https://ibb.co/YjNQSGr
https://ibb.co/NTTmFSF

I'm a bit confused now. For the first image, the person said that only the vertical component of the tension provides moments so that
L/2 *Tcosθ =W1*L.

However, for the second image, the other guy said that the the whole tension provides moments so that T*Lsinθ = mg* L +Mg*L/2

So for a question like this, is the vertical component of the tension the only part that provides moments? Or it's actually the whole tension.
In my opinion, I think only the vertical component of the tension contributes to the equilibrium because the horizontal component of the tension acts directly on the pivot, which means no turning effect would be caused by it. Hence, only the vertical component provides moments

Thanks
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Stonebridge
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#2
Report 9 months ago
#2
(Original post by long_vacation)
So I saw two very similar problems on Youtube videos:

https://ibb.co/YjNQSGr
https://ibb.co/NTTmFSF

I'm a bit confused now. For the first image, the person said that only the vertical component of the tension provides moments so that
L/2 *Tcosθ =W1*L.

However, for the second image, the other guy said that the the whole tension provides moments so that T*Lsinθ = mg* L +Mg*L/2

So for a question like this, is the vertical component of the tension the only part that provides moments? Or it's actually the whole tension.
In my opinion, I think only the vertical component of the tension contributes to the equilibrium because the horizontal component of the tension acts directly on the pivot, which means no turning effect would be caused by it. Hence, only the vertical component provides moments

Thanks
They are both correct.
The tension in the cable privides an anticlockwise moment about the pivot. That is clear.
To calculate it you either:
Take the vertical component of the tension and multiply by the whole length of the beam. (If the cable is attached to the end)
or
Take the whole tension and multiply it by the perpendicular distance of the cable from the pivot. (d3 in the diagram he has there)
Last edited by Stonebridge; 9 months ago
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long_vacation
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#3
Report Thread starter 9 months ago
#3
(Original post by Stonebridge)
They are both correct.
The tension in the cable privides an anticlockwise moment about the pivot. That is clear.
To calculate it you either:
Take the vertical component of the tension and multiply by the whole length of the beam. (If the cable is attached to the end)
or
Take the whole tension and multiply it by the perpendicular distance of the cable from the pivot. (d3 in the diagram he has there)
Thank you. So these two methods will give us the same moment right?
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Stonebridge
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#4
Report 9 months ago
#4
(Original post by long_vacation)
Thank you. So these two methods will give us the same moment right?
Yes, so long as you get the correct angle and formula for the vertical component, and the correct geometry for the perpendicular distance.
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long_vacation
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#5
Report Thread starter 9 months ago
#5
(Original post by Stonebridge)
Yes, so long as you get the correct angle and formula for the vertical component, and the correct geometry for the perpendicular distance.
Thank you so much
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