Connected particles - Force on pulley by a string : ExamSolutions
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- čas přidán 29. 10. 2008
- Tutorial on connected particles when theres is a force on a pulley by a string.
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still saving lives in 2020
if it wasn't for you thousands would be failling right now. all the luv
Spreading the word etc is all I can ask. Thanks.
Cool - Hope you can get that A this time
You are an absolute legend. Definitely boosted my grade, no doubt about it.
I wish you every success.
YOU SIR, ARE THE BEST! ❤️
Thanks mate, this clears up a lot as I was confused about the tension too :) Ah man, I really am grateful for your help and support - it must take up a LOT of time.
Heyy mate how are you doing now !!!!!
You saved my life!!!!! My M1 official exam is in 3 days and finally i know how to do this!!! Thank You!
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Can't believe that, this video was uploaded before 12 years ago
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thank you :) i'm revising for my M1 exam soon, and that 3rd example came up almost exactly in a past paper, quite worryingly i'd never seen a question like it :S but thanks to this video i completely understand it now :)
you may have just saved me a grade :'D
really helpful with my work. Thanks
thank you no one could explain this to me and i understand now completely
That's what I like to hear!
u are great man
helpful to no end, thank you so much.
Pleased to hear it helped. Thanks for the comment.
Thank you for your videos, Great for going over things Im a little unsure of, and thanks to your C1 videos I was one mark of an A in my january module.
thank you so much! I have a mechanics exam tomorrow and my teacher didn't prove any of that! this was super helpful :)
Thanks sir for your explanation,it has really build so much skills in me.
Thank you sooo much for this🙏🙏🙏
@TheSimiarora It seems you are quite confused. The first tutorial was about forces acting on the particles and in this we are looking at the forces acting on a pulley. The tensions are the same but in opposite directions due to Newtons 3rd law: To every action there is an equal and opposite reaction.
thank you Sir...you should be my maths teacher...like ur type of teaching
Thank you it helped me✌️👍👍
Thank you very much! .You answered all my doubts :D
Thanks
Regarding your point at 7:50, when marking in the angles, you do it so, it goes from the direction of the resultant force, (into the plane) to the actual string. Would it be incorrect if it was just marked inside the triangle?
Thank you very much!
thanks man! u the best!
THanks broo, like our teacher doesn't tell us all these subtle exam tricks and expects us to get full marks
I hope you can now get those marks if it comes up. Good luck
Nice explanation!!!👍
You are very welcome
Yes
@1Little1Wayne007 Pleasure - Thanks for your support
im doing mei exam in June, your vid really helps, im hoping getting a good grade.
but why the cosine acting upwards? is that alright if i use -Tsin45?
thanks please reply...:)
Thank you :D I've got a M1 exam today :D
@ExamSolutions I love you for that comment, and for all of your help!
Hi just wondering why T is equal on both sides is it because they are in equilibrium thanks
thank you very much
thank you
Thank you so much =)
I have a question regarding part 5:00 of the video. At this point you explain that the resultant force=2Tcos(45). I do not understand this part as by C=A/H, i would think this would make the resultant force=2T/cos(45), as T is the adjacent and we are finding the hypotenuse, solving for H=A/cos, so R=2T/cos(45), please help if you can.
Check this out www.examsolutions.net/maths-revision/mechanics/dynamics/connected-particles/force-on-pulley/tutorial-1.php
Hope it helps
ExamSolutions
Hi there, thanks for your help :). Realised i was placing the right angle in the wrong area therefore changing the sides. Thankyou so much for these videos.
Thanks so much for this video
You're welcome
Good luck with the M1
That too would be correct as you have parallel lines
@xenomorph321 I hope so. Cheers.
Thanks for all your help, unfortunately I don't have an awful lot of money to donate, but I'd be happy to support you in any other way if thats possible.
No probs.
Lifesaver. These questions usually aren't worth more than two marks though, thankfully!
ur Awsome.....
Why does it only take in to account the tensions, and not the weights of the particles too? Surely the weights affect the resultant force?
tension is found out using weight
@ninjaturtle205 Thanks for the complement
thank you so so so much
No problem. Best wishes.
how did you bisect the lines to five Tcos35????
Love you
how would i substitute for the tension I don't get it?
Next week is my M1 exam thanks for uploading this
Me too
there’s another example when there’s 2 particles ; one being on an inclined plane and the other being on a horizontal table. R = 2T(180-a/2) where a is the angle that the fixed plane is inclined to
I really appreciate it if you answer its really really confusing me.There seems to be a contradiction on the way you answered 2 "forces on pulley exerted by a string" questions. On paper 2009 Q7 Part C, you drew the resultant arrow in the same direction as the tensions, however here you drew it on the opposite direction . Really confused! Really appreciate if you answer.
+Mohammed D I understand your point. In fact the question you are referring to in the paper, says the magnitude as opposed to direction so from that point of view the direction does not matter. I drew it in that direction though as the tensions in the strings were pulling the pulley in towards the plane and this was the force on the pulley. However, for the pulley to stay in equilibrium the pulley would need to exert an equal but opposite force outwards as in my video tutorials. I hope that makes sense.
@MrATJones Good luck.
any hints as to what's on the exam??? i'm bricking it atm! :P
Can any explain why T is acting downward instead of upward thanks.
The tension forces in the string are equal and opposite so when you look at the forces on the particles they act upwards but on the pulley they act downwards.
It was 2Tcos 35
I love you
Please could you clarify why it is R=2Tcos45 and not R=2T/cos45. Many Thanks
the right angle is placed between the dotted line and the line connecting the dotted line to the Tension force which is downwards or to the left on the diagram, that means that the dotted line is the adjacent and the Tension force is the hypotenuse to get Tcos45
What if the angle of the inclined plane was alpha or beta, would the same rules apply?
Yes
Hey,
It says examsolutions.co.uk but it's .net You may want to add a little note or something into the video.
Thanks for the great vids
luke filer These were early videos when I first started under the domain examsolutions.co.uk. I then went on to change it to the now examsolutions.net. If you try my old domain name it will forward you to the new one. To change these old video covers would take too long but they will gradually be faded out.
Hey,
Ah okay didn't realise that it re-directed you. That's fine then :) Thanks again for the great uploads.
Doesn't tension oppose weight, if so why is tension facing downwards when it should face upwards?
Thanks for the video though !
Late reply, but he is talking about the tension acting upon the pulley. Obviously the tension on the particles is opposing the weight, but here we're just looking at the forces between the pulley and the string.
Your voice is like steve banyard commentator for barclays premier league
@XDAdzXD I sure hope you get that A then!
Why the T tension is acting downward?
OH MY GOODNESS. *subscribes*
Pleased to be of some service.
@udsamassj4 Cool
MASHALLAH.....SUBHANALLAH
But what about the other kind of pulleys.
What kind of pulleys?
ExamSolutions you showed three examples here, but what to do for others?
There are more like (inclined plane and horizontable table in one), then theres obtuse angled pulleys and so on..
Is this A-Level stuff?
Yes
Ah ok brilliant thank you. I am at GCSE at the moment and have been considering taking A-Level Physics for a while now, would you say it is worth it? It certainly looks very interesting
@@CamHarrisVlogs Yes, physics is very interesting but I am sure others would disagree. Surely it should depend on what you may want to do in the future that may make you lean towards a subject. If you enjoy maths then it will help support physics especially the mechanics side. Good luck.