Moments of forces | Ladder leaning against a smooth wall and floor (1/2)

Here I look at the moments of forces for a ladder resting in equilibrium against a smooth wall and ground and held by a tie to prevent it slipping.
Check out my video on a ladder leaning against rough surfaces. • Moments of forces | La...
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[@JohnSmith-xo9me]

Life saver. 8 years later.

[@tyresereid1275]

I can relate

[@amattullahshajida8113]

I can relate so much alevels saviour

[@floatingoncloud91]

thank you so much sir! i finally understand mechanics after being confused for an entire year during A2. i was wholeheartedly relying on statistics pulling up my grade for applied but now I think I can do much much better

[@ExamSolutions_Maths]

I am pleased to hear that you have gained some confidence back. Best wishes for every success.

[@kyliek.9481]

ExamSolutions thank you for making these videos!

[@Captain_Rhodes]

i have started M2 and have never learned that thing about when the angle is 'not contained' we use sin. where should I learn about that? I dont know why its not mentioned anywhere leading up to these problems!!

[@mahrinahmed4957]

9 years later, and still relevant. THANK YOU.

[@mlfacts7973]

wow, how ?

[@issaabdalla281]

@@mlfacts7973 the specification has changed but the concepts are still the exact same

[@nathanpeck2817]

theres now way you just said that

[@ExamSolutions_Maths]

Great to have a quick response. Thank you.

[@ExamSolutions_Maths]

Thanks - I intend to if I keep getting support for this project

[@christinewen7119]

Great series! Very clear format for the paper-graders to follow-- thus insuring max marks. The advantage of Daniel's choice of origin is that you can immediately write down the answer AND need not decompose any forces AND you can see that the length of the ladder does not matter. I used the point halfway down the wall, which similarly balances the torque of gravity versus the torque of the rope.

[@christinewen7119]

Three more comments: 1) Above I meant to use the term “force couple” [not "torque of the rope"] for the moment counter-balancing the moment due to the ladder’s weight-- . Forget it! 2) More importantly, it is NOT a matter of the unusual position chosen to compute moments; choosing point A is equally the best! What happened is that the solution presented by the author breaks forces into perpendicular components, rather than seeking the lever arm of the ladder’s Mg force and the lever arm of the rope force (T). From which we can simply write T = Mg/(2tan70) which follows from: b*Mg = (2b*(tan70)) *T which balances the Torques. [2b = base of triangle that the ladder, ground and wall define.] 3) The use of the (USA dynamics?) term Torque leads one to question the stability of the equilibrium-that is, how easily is the Torque balance upset by a perturbation? Anyhow, it gets me thinking about what a tough job retro-fitting a home to withstand earthquakes must be …etc…

[@ReMoBoY97]

I didn't understand the point Daniel is mentioning, can any one clear it for me or better mark it on a picture and upload it,please. thanks in advance.

[@mensamoo]

Thanks for the video. There was a massive presumption in this question that the string holding the ladder in equilibrium was along the floor. It could have been at any angle as it states in the question 'attached to a point on the wall' . This could have been at any point of the wall below the top of the ladder, and therefore fixed at various angles to the bottom of the ladder. Lets hope it wasn't an exam question.......

[@dhanushivan]

Wrong, the question explicity states that the string is "horizontal" and since it is attached to the base of the ladder, through common sense you can identify that the point of the wall where the other end of the string connects to is the base of the wall.

[@ryanathwal8819]

He doesn’t explain how he got r1sin70 at 9:58 I’m so confused

[@mathew66]

U don’t need it because the moments are around point A. Because no moments would occur.

[@allyphil826]

Thank you so so much

[@3469Josh]

Thanks a lot, really helped me

[@DanielWalvin]

Why did you not take moments about the point above R_2 and to the right of R_1, i.e. the point where R_2 and R_1 do not exert moments? You can get the answer that way, I assume it's quicker.

[@DanielWalvin]

A__P ..\..| ...\.| ....\| ....B At P.

[@ExamSolutions_Maths]

Daniel Walvin You can, it is up to you. That position tends to be unusual though.

[@ReMoBoY97]

I didn't understand the point Daniel is mentioning, can any one clear it for me or better mark it on a picture and upload it,please. thanks in advance.

[@ihaashyhd5957]

if it Contains the angle then it's a Cos Contains Cos

[@mahrinahmed4957]

Minor point, but you apparently have to give it to 2 significant figures as gravity is used

[@ExamSolutions_Maths]

I am sorry but you are wrong and I am correct. You missed out g=9.8 in your calculation.

[@jesssoph3658]

@bumbread that’s what you use in a level maths for g

[@emily-sz4jf]

wow thank you so much!!

[@ExamSolutions_Maths]

That's okay. Pleased you found it useful. Best wishes

[@Anudari11]

love you and your channel !!!!

[@lucky_em9039]

at the T, why is it that you use 70 degrees instead of 20?

[@caitlinhannigan9294]

because he's using sin. if he were using cos he would use 20.

[@joelbeeby866]

If you took moments about B would you solve by R1 = T

[@YoussefMedhatAboutaleb]

Thank you for this, and as you have suggested experimenting really gives you a good grip of the concept. Please continue this project.

[@yoyo9winner]

bless ur heart and soul man

[@Michael-cq5qk]

hello, i tried taking moments at the centre but i got the wrong answer, i got double of the correct answer. Is there a reason for this @examsolutions

[@katherinekeogh8213]

This made so much sense THANK YOU

[@ExamSolutions_Maths]

Good to hear. Thanks for watching.

[@raindrop9314]

Hello sir but at 19:00 when you made t the subject of the formula. why cant you do tan^-1(70)x40*9.81/4, because i assumed if your taking tan to the other side it would be tan^-1

[@ExamSolutions_Maths]

That would be incorrect. If solving an equation for an angle, say tan(angle)=something then the angle = tan^-1(something).This equation is not of that form.

[@raindrop9314]

@@ExamSolutions_Maths i also took moments around the middle point and i seemed to get the same answer however when i took it from the point at the bottom i got 71.4 i did 20x9.81xcos(70)/sin(70) . What did i do wrong ? Thanks in advance.

[@ExamSolutions_Maths]

@@raindrop9314 You forgot the lengths of 2 and 4 metres in your calculation. It should be 2x20x9.81xcos(70)/4sin(70) to give 35.7

[@josheee3e]

this may be a silly question, but when you were working out the vertical components, why didn't you do moments? (like from A,2X20g-4XR=0) it may be really obvious, but i have been revising a while and i think my brain might be fried.... :D

[@johnsergie3906]

Hi, Is this AS or A2?

[@phil5807]

❤

[@salma-amlas]

What if I prefer to resolve distances instead of forces? How would I do that?

[@bennybenson4642]

No idea bro, but I thought I should reply anyway

[@randhawaheed2720]

@@bennybenson4642 lmaoo

[@AlEx-ro7kd]

is my trigonometry so bad that i can't get how u get the components? shouldn't the 20gcos70 be 20g/cos70?

[@ExamSolutions_Maths]

Sorry no.

[@kiinaahplays9229]

Al Ex cos = opposite/hypotenuse so in this case, opposite is the component so the component would be cos x hypotenuse which is 20g x cos70

[@amattullahshajida8113]

Thankyou sir my prayer with you🙏

[@bennybenson4642]

you're a fucking God, thank you

[@zipporahhodder5088]

why isn't there a vertical component of the reaction at the top of the ladder? maybe its a stupid question sorry

[@mihalytorley5480]

because there is no friction between the ladder and the wall

[@randhawaheed2720]

I'm very late to this but he has shown the vertical component of reaction see 06:11

[@ssajid40]

is this M1?

[@ExamSolutions_Maths]

M2. Check out your exam board on my site.

[@lincc7777]

Still hard to me 😭

[@julistick]

from your calculation the final answer should be 3.63N instead of 35.7N

[@cooltmv]

Make sure to multiply 40 by 9.8 on the numerator as it's 40g (g=9.8)

[@julistick]

sorry mistake from me......

[@SabrinaXe]

Wait this wasn't physics? I don't even take mechanics but take Statistics. Not fair, mechanics is so much easier!!!