A-Level Maths: S1-25 Non-Parallel Forces: An Introduction to Ladders
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- Опубліковано 14 жов 2024
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You're an absolute legend mate. Honestly.
Thanks!
usual excellent tutorial . Well done Jack.
These questions stink so this video was a big help. you're the man jack 👍
Great video as always! Could you please explain as to why we cant use F=Mu x R. Didn't quite understand what 'limiting equilibrium' means. Thanks
So imagine you're pushing a sofa across a carpet. You can apply a little bit of force to the sofa so that it won't budge - in this case you're not beating friction - in this case Fr < mu*R. So the force you're applying is equal and opposite to the friction (no movement).
If you increase the force you apply, up the point at which the sofa is just about to start to move, then we say that the sofa is in limiting equilibrium. It isn't moving but is on the point of moving - in this case Fr = mu*R. So the force you're applying is still equal and opposite to the friction and hence there is still no movement.
Now you push the sofa a little bit harder - you are now beating friction and Fr = mu*R. Friction increases only up to a certain point, and then you overcome it. So now you're applying more force than the friction, and so the sofa now moves.
@@TLMaths Thank you for the reply. So, in the case of the ladder we can't use the equation as there is no external force being applied to it? As in it is not at the point of moving?
Quite often you'll be dealing with someone climbing the ladder and working out how high they can go before the ladder slips - so the ladder will be on the point of slipping (limiting equilibrium) so that you can say Fr = mu*R
Sir I have a question.Here the value of friction is 28.9 N and the normal force at bottom is 100 N. So the resultant of this two force is 104.09 N something . But this force is acting 73.8 degree with vertical between friction and normal.That means the ladder does not apply force along it's incline.if it gives force along it's incline then the friction and normal are different in this case.
Sir please solve my doubt...Love from India.
What if the top of the ladder overhangs over the vertical surface it’s leaning against? Would the normal reaction at that top point be perpendicular to the ladder instead of the wall?
It would be perpendicular to the ladder
Do we need to know about ladders for the aqa spec? Can't seem to find it in my textbook.
No you don't - AQA and OCR MEI don't do ladders (much easier!!)
The video content is so excellent, congratulations
Thank you very much
Did you remove the ladders section on your website? Why?
It's still there: sites.google.com/site/tlmaths314/home/a-level-maths-2017/full-a-level/s-moments/01-moments. I removed them from the Teaching Order if that's what you mean as they're not on AQA.
when do we know if reaction force is angled compared to when it is perpendicular???
A reaction force is perpendicular to the surface the object is in contact with.
Very helpful, thank you!
Does ocr need to know how to do hinges problems?
Are you OCR A or OCR B MEI?
@@TLMaths ocr a!
Yes, 2020 paper had a hinge problem
you’re awesome keep it up
Hi, I've just seen a past paper question where the ladder is resting on a smooth vertical wall, and the reaction force between the wall and the ladder seems to be perpendicular to the ladder (jn the mark scheme) rather than horizontal, as I would usually draw it. Could you explain why this may be? Thank you
Was the ladder resting against the very top of the wall (and overhanging it)?
YOU ARE A GOD. Excellent tutorial.
Thank you very much 🌹🌷
You solved a very big problem for me 💙❤
Your the best