Swear. My professor spent half the class one time trying to explain centroids and he couldn't figure out why his numbers weren't adding up. He confused me even more that day, it was discouraging.
i legit watch like 20 videos, waste my time dont understand annything but then I come and watch a 10 min video of yours and i understand everything tysm please upload more videos.
@@ScienceShorts the only question I have is where does the perpendicular distance to the line of action come in? Since you only drew it on but never used it.
Mate, thanks so much. I just spent 40 mins trying to understand moments using a textbook and my notes from school and I couldn't understand a thing. Your video just solved all my questions in 5 mins and 50 seconds
Thank you so much, you are such a life saver, again like many of the viewers, I too came here after my teacher spent the entire class confusing the hell out of all of us.
Awesome video. Really helped clarify for me. You have a great teaching ability, very clean and easy to understand. I was hoping that you would give an example of the 'toppling' style question. Any chance you could do this in another video, or send me a photo of a page of working of how to do them? Thanks
Thank you so much, Science Shorts! You have really helped me through my entire sixthform education! Im about to go for the exam, thank you again! I must have watched this video 100 times in the past year haha!
A weight on a teetertotter causes it to rotate around the pivot like hands on a clock. seconds minutes "moments" - moments of inertia! thanks, I always wondered why they called it that.
For those studying A level Physics, are you using John Bird's electrical and mechenical text books? If so, are they that helpful in terms of aiming for top grades?
@@hasanmohamed4213 that’s because he made point T1 the pivot and the total distance from T2 to T1(which happens to be the length of the beam) is 2m. If the centre of the beam was the pivot, then the distance for T2 will be 1m.
Did not get it 5:50 How does the T2 x 2 put in the right side of the equal sign can someone explain to me the formula or FBD formula of this??? thank you!!!
Based on: sum of clockwise moments= sum of anticlockwise moments The moments going clockwise are the light and the weight of the beam, while the anticlockwise moment is only T2 so T2 x 2 goes on the right side of the equation
I’m like 4 years late but it’s because the distance between T1 (the pivot) and T2 is 2m , and since the formula is moment=force*distance , it is going to be T2*2 , how is life going tho
For the forces at angles section, I’m a bit confused on the line of action bit. I normally would resolve the tension, to find its vertical component, times that by the length, and then equal it to the clockwise moment. That would result in Tsinθ not Tcosθ though.
@@josh-kd4gbthe reason why it’s Tcosθ is because he’s moving “through” the angle. Look at his resolving vectors video, it helped me a lot. Plus, if you do plenty of mechanics questions, you will get used to it! Took me ages to get it right lol
The rod, or beam or whatever the object is. Its weight acts in the centre, assuming it is uniform, hence the distance from one side to the middle is 1m, that means the middle to the other side is also 1m. 1+1=2, hence 2m distance when taking pivot at T1
@@mathiedits534 moment = force × distance. T1 is 0 because the distance is 0 since we are taking the moment about T1. T2 is 2 meters away from where we are taking the moment (T1) So it is the force of T2 × distance = T2 × 2 meters
At 12:47 tou choose a specific pivot while in the previous example you said we could choose the pivot point where ever we want. But, apparently, that's nit the case. Could you elaborate on that?
At 10:30 I don't understand why the vertical component is Tcos0. Every time I have tried to work it out using triangles I seem to end up with a different answer. Could anyone help, would be much appreciated! :)
Well, in this scenario, tension is acting up, but at an angle. We cant just use "T" if it is at an angle. The tension T has a vertical component and a horizontal component. Think of it like two forces, one is acting horizontally and one is acting vertically, but they have different magnitudes. The resultant force of these 2 forces is T. So using trig (SohCahToa) we can find the magnitude of the V component. We want the component of T acting up from that point on the beam (or whatever it is), so we draw a line (light dotted line so it's less confusing) going up from that point. That makes an angle (theta) between T and the like (the vertical component of T). Using SohCahToa, T is the hypotenuse (h), and the side of the triangle we want is adjacent (a) to the angle, therefore to get this vertical component, we multiply T by cos(angle). Angle is always in degrees in mechanics. Useful tip is if you also need the other (here, the horizontal) component for something else, it will always be the other trig function (in this case, to get the H component, we ×T by sin(angle)). It is rare that we have to use tan. Even if the option to use tan was there, I would still use either cos or sin (meaning I would have to use a different angle).
I looked at your other comment. the other guy who said they would end up with Tsin(theta) is right, but the angle would be different. They are drawing a line downwards from the top of the line T, rather than what was done in the video, a line upwards from the bottom of T. The reason why it's sin and not cos is because Tcos(theta) = Tsin(90-theta). So using sin, drawing a line down from the top of the line T, we would get a new angle of 90 - theta. Do some past paper questions from 2017-2019 and check your workings. Or, try easier questions from your textbook so you can understand it a bit better before going to the big complicated questions.
Take moments about the bottom of the ladder. The centre of mass pulls, say, anticlockwise, but the reaction force from the wall pushes clockwise. The distances are going to be along the ground to the centre of mass's line of action, and vertical height the top of the ladder is above the bottom.
In the roundabout section of this video, If F1 and F2 have the same magnitude, are they a couple? Because if the roundabout is thought about like a clock, F1 could be considered (about) 7:30 and I would have thought that the opposite force to make a couple would have been (about) 1:30.
You are actually counting the gravitational force in the downward force (weight) as gravity would act upon mass thus it would create a downward force pulling the mass down. So actually when we take weight in to account, we automatically know that gravitational force is present
i'm a bit confused at 10:42 u say the equation is TcosO x L/2 = W1L but isnt for the string, shouldn't the length be perpendicular to the line of action, so the length of the red line you drew not the half of L? ( O = theta) please answer if possible :)
@@aurnaurrrcleaur The red line was just to show the perpendicular distance from T to the pivot. However, it is difficult to find the length of the red line, so you resolve it to TcosO and use the perpendicular distance L/2 from TcosO to the pivot.
9:32 Wait so is the moment for the top part of the system equal to T multiply the length of the red line with the 90 degree angle that is perpendicular to the forces line of action? Or is it still Tcos theta?
always come here to get a subject properly explained after my teacher mucks it up
Phillip Banks not wrong there chief, same here
Hello uncle
FIRST THINGS FIRST RIP UNCLE PHIL
Swear. My professor spent half the class one time trying to explain centroids and he couldn't figure out why his numbers weren't adding up. He confused me even more that day, it was discouraging.
Same I've learned half of my As phy from here
Perfectly balanced, as all things should be
lmaaoo
🤣
I see that avengers reference
Bro is not thanos😂🙏🙏
okay thanos
i legit watch like 20 videos, waste my time dont understand annything but then I come and watch a 10 min video of yours and i understand everything tysm please upload more videos.
Thank you!
@@ScienceShorts What a guy
@@ScienceShorts the only question I have is where does the perpendicular distance to the line of action come in? Since you only drew it on but never used it.
@@AbdullahiIdri2001 DO U EAT PORK？LOL
i agree omfg
I seriously can not believe that after 3 years I finally get moments. I honestly feel like crying right now 😂 Thank you! Thank you so much!!!
These videos got me an A for my AS Physics. Thanks a lot Sir
nice bro
Congrats👏🏽
What you got in a level?
@@Awai_quotes they might not have done A-Level
Mate, thanks so much. I just spent 40 mins trying to understand moments using a textbook and my notes from school and I couldn't understand a thing. Your video just solved all my questions in 5 mins and 50 seconds
Just discovered this channel by UA-cam algorithm and it's been helping me a lot with ENEM down here in Brazil. Cheers
I'm so thankful for this channel, without it, I would be failing my physics 😩
this guy is better than any teacher you can have. had a 2 hour lesson on this and learnt more in the first 3 minutes
Bro ur a legend I have a physics test tomorrow an I can actually do well now. God bless
Thank you so much, you are such a life saver, again like many of the viewers, I too came here after my teacher spent the entire class confusing the hell out of all of us.
Keep up with the great vids! It helps a lot cause I have a test next Monday!!!
What did you get?
Got mine this Monday!
Gilbert Virgo how did the test go
Anyone doing physics for A level this man is teaching me better than my physics teacher
Awesome video. Really helped clarify for me. You have a great teaching ability, very clean and easy to understand.
I was hoping that you would give an example of the 'toppling' style question. Any chance you could do this in another video, or send me a photo of a page of working of how to do them?
Thanks
I have a physics test coming up next week and this video is saving my life!!!!!!!! THUMBS UP FOR ALL YOUR PHYSICS VIDEO!!!
I saw a few videos before this but they weren’t as helpful as this one.Thank you so very much!!
Thank you so much, Science Shorts! You have really helped me through my entire sixthform education! Im about to go for the exam, thank you again! I must have watched this video 100 times in the past year haha!
+Mustard Good luck! :)
What grade did you get?
What grade did you get??
what grade did you get?
What grade did you get
Phew! Thanks man!
Great video! I had watched the video once and
I forgot this channel name!
Glad I had liked the video and found it again!
you actually make the stuff easy to understand. thank you.
These videos are an actual lifesaver! Absolutely perfect explanation and examples, keep up the fantastic work :)
Sir your teaching in its way are SO unique. I understood every word and now starting to like physocs
can u plz come to my school and teach us physics your literally an amazing teacher!!!!!!!!!!!!
I got what i wanted in the first 10 seconds 😳😳🤣
Thank so much
Second semester of uni and already passed Mechanics 1,and just I understood moments from your video!
my G helping me out with content i didnt even know i needed to know 😂
This is a great video, and very practical. Thank you for the resource!
You are a blessing from the heavens.
Thanks this helped me study for my physics test liked and subscribed :)
I don't understand how the distance of the clockwise moment is 2. at 6:12
This video explained it so much better than my physics teacher. Thank you so much.
A weight on a teetertotter causes it to rotate around the pivot like hands on a clock. seconds minutes "moments" - moments of inertia! thanks, I always wondered why they called it that.
For those studying A level Physics, are you using John Bird's electrical and mechenical text books? If so, are they that helpful in terms of aiming for top grades?
eugenechelseafc89 try books by I.E.Irodov
I use Hodder, CGP, edexcel revision guide and Edexcel student book.
hey man im just winging it youknow
@@cal6738 ayy my guy
@@justaracoonchillinginatoilet69 I got a D 🤣🤣
I really appreciate his efforts for helping students
The fact that we did 2 months of work on this and I get it in under 15 minutes here. My teachers are really lacking
Concisely explained and clearly demonstrated, thank you so much.
This helped immensely, thank you so much!
This person is a Godsend. Thank you so much!
PLS , more as level stuff
Thank you so much, this was extremely helpful!
Making my academic life easier lot of appreciation ❤
Still very confused, but that’s not your fault, it’s a good video, I’m just a little bit thick
No, you're not MJ. Which part confuses you?
lmao same
@@beingprospera584 why does he do t2 X 2m and not t2 X 1m as is it not meant to be from the axis
@@hasanmohamed4213 that’s because he made point T1 the pivot and the total distance from T2 to T1(which happens to be the length of the beam) is 2m. If the centre of the beam was the pivot, then the distance for T2 will be 1m.
Same I’m very confused at forces and angles
good video but invest in a ruler
Can't afford one.
@@ScienceShorts What's the price of a ruler in UK? Let me see if I can donate you some for buying one
haha hunter bailey you just made an idiot of yourself coz science shorts guy is a genius
@@mynahification you don't understand jokes do you
Not necessary
Helped so much with understanding moments🙏🙏
Thanks a lot! Plz keep it up. You are helping so many!
absolute life saver
Did not get it 5:50
How does the T2 x 2 put in the right side of the equal sign
can someone explain to me the formula or FBD formula of this???
thank you!!!
Based on: sum of clockwise moments= sum of anticlockwise moments
The moments going clockwise are the light and the weight of the beam, while the anticlockwise moment is only T2 so T2 x 2 goes on the right side of the equation
I’m like 4 years late but it’s because the distance between T1 (the pivot) and T2 is 2m , and since the formula is moment=force*distance , it is going to be T2*2 , how is life going tho
very helpful for my as levels
thank you 'science shorts'
For the forces at angles section, I’m a bit confused on the line of action bit. I normally would resolve the tension, to find its vertical component, times that by the length, and then equal it to the clockwise moment. That would result in Tsinθ not Tcosθ though.
same with me lol
@@josh-kd4gbthe reason why it’s Tcosθ is because he’s moving “through” the angle. Look at his resolving vectors video, it helped me a lot. Plus, if you do plenty of mechanics questions, you will get used to it! Took me ages to get it right lol
you are amazing.
Thank you for this.
It will work for generations 😂😂
delightful
When you use the trig to calculate the upwards force, are you calculating the Nm?
is the weight of the plank in the force at an angle not counted?
Thank you for content!!!!! It was so helpful, please do cover more topics. Thank you :)
thank you for this!!!
thank you for existing
superb explanation,you have tremendously helped me.
why we multiplied t2 with 2 ?
Thank you man.
Sir can you please do videos on the new specification
Don't give this man a medal
Because he will probably throw it at a wall to find the momentum
Nah, I'd flog it.
Been watching these videos for past 2 days Exam is tomorrow I’m kinda sick of ur voice even tho it is quite amazing thanks for the videos!
Empire joshvir262 edexcel? Me too. Good luck brother
8:36 is it possible to hold everything on the stick without any string attached to the stick?
5:47
Im not sure where he got the x2 from, could someone explain?
It’s the distance from pivot
The rod, or beam or whatever the object is. Its weight acts in the centre, assuming it is uniform, hence the distance from one side to the middle is 1m, that means the middle to the other side is also 1m. 1+1=2, hence 2m distance when taking pivot at T1
could,not understand
@@mathiedits534 moment = force × distance.
T1 is 0 because the distance is 0 since we are taking the moment about T1.
T2 is 2 meters away from where we are taking the moment (T1)
So it is the force of T2 × distance = T2 × 2 meters
Well done😊
delightful, thanks
by any chance are you going to do more A-level videos because i find them quite useful
Could you ignore the perpendicular rule if you use components?
For angled moments when do we use sin and when do we use cos
At 12:47 tou choose a specific pivot while in the previous example you said we could choose the pivot point where ever we want. But, apparently, that's nit the case. Could you elaborate on that?
perfect...thanks a lot
Thank you sir.
thanks man
At 10:30 I don't understand why the vertical component is Tcos0. Every time I have tried to work it out using triangles I seem to end up with a different answer. Could anyone help, would be much appreciated! :)
Well, in this scenario, tension is acting up, but at an angle. We cant just use "T" if it is at an angle. The tension T has a vertical component and a horizontal component. Think of it like two forces, one is acting horizontally and one is acting vertically, but they have different magnitudes. The resultant force of these 2 forces is T.
So using trig (SohCahToa) we can find the magnitude of the V component. We want the component of T acting up from that point on the beam (or whatever it is), so we draw a line (light dotted line so it's less confusing) going up from that point. That makes an angle (theta) between T and the like (the vertical component of T).
Using SohCahToa, T is the hypotenuse (h), and the side of the triangle we want is adjacent (a) to the angle, therefore to get this vertical component, we multiply T by cos(angle).
Angle is always in degrees in mechanics.
Useful tip is if you also need the other (here, the horizontal) component for something else, it will always be the other trig function (in this case, to get the H component, we ×T by sin(angle)). It is rare that we have to use tan. Even if the option to use tan was there, I would still use either cos or sin (meaning I would have to use a different angle).
I looked at your other comment. the other guy who said they would end up with Tsin(theta) is right, but the angle would be different. They are drawing a line downwards from the top of the line T, rather than what was done in the video, a line upwards from the bottom of T. The reason why it's sin and not cos is because Tcos(theta) = Tsin(90-theta). So using sin, drawing a line down from the top of the line T, we would get a new angle of 90 - theta.
Do some past paper questions from 2017-2019 and check your workings. Or, try easier questions from your textbook so you can understand it a bit better before going to the big complicated questions.
Hi, I find your videos really helpful! Thank you so much for posting such helpful videos, I'm actually watching them in preparation for a test! :)
Fantastic video, really useful
Given to calculate the reaction between for ex a ladder and a rough wall, where do you make the reaction point to?? Thankyou
Take moments about the bottom of the ladder. The centre of mass pulls, say, anticlockwise, but the reaction force from the wall pushes clockwise. The distances are going to be along the ground to the centre of mass's line of action, and vertical height the top of the ladder is above the bottom.
A-level paper 1 in 5 hours, good luck everyone :)
hi can you please make a video on kinematics and dt vt graphs etc. btw LOVE your videos such dense and full of actual syllabus content.
10:34.....why is it cos.....why not sin
In the roundabout section of this video, If F1 and F2 have the same magnitude, are they a couple? Because if the roundabout is thought about like a clock, F1 could be considered (about) 7:30 and I would have thought that the opposite force to make a couple would have been (about) 1:30.
Great question - the answer is no, as there would be a resultant force - couples must consist of forces that add up to zero i.e. no acceleration.
so interesting!!
thank u so so muchhh
Thank you Thank you Thank you .......
one question, should we count gravitational force as one of the forces that cause moment?
You are actually counting the gravitational force in the downward force (weight) as gravity would act upon mass thus it would create a downward force pulling the mass down. So actually when we take weight in to account, we automatically know that gravitational force is present
Do you have Cambridge igcse questions regarding moments? Complicated crane problems
5:07 how is the 200n and 80n going clockwise, shouldnt it be the other way round
Thanks for teach me.
THANK U :)
i'm a bit confused at 10:42 u say the equation is TcosO x L/2 = W1L but isnt for the string, shouldn't the length be perpendicular to the line of action, so the length of the red line you drew not the half of L? ( O = theta) please answer if possible :)
TcosO is the vertical component of T, hence the length perpendicular from the pivot to TcosO is L/2 and not the red line
@@Milosaur15 yh ok but then what was the purpose of the red line?
Like how is it related to L/2?
@@aurnaurrrcleaur The red line was just to show the perpendicular distance from T to the pivot. However, it is difficult to find the length of the red line, so you resolve it to TcosO and use the perpendicular distance L/2 from TcosO to the pivot.
9:32 Wait so is the moment for the top part of the system equal to T multiply the length of the red line with the 90 degree angle that is perpendicular to the forces line of action? Or is it still Tcos theta?
atu bro bnu atu pendingi, god bless.
@ 5:55 Why is T1 ignored when finding the moments?
moment= force x perpendicular distance... the moment for T1= (T1) x (0) which is just 0.
Where can I get some good moment questions to practice from ?
thanks a lot ... for retrieving me from getting muddled up ..with these topics .
Why did we x2 at 5:58
That's the length of the beam.
At 10:48 isn't it TSin0 because it's vertical?
it's tcos0 because it goes through the angle
quick question: why did you multiply T2 by 2?
nice sir
Legit video