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.

[floatingoncloud]

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 <3

[ExamSolutions_Maths]

Great to have a quick response. Thank you.

[ExamSolutions_Maths]

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

[mahrinahmed]

9 years later, and still relevant. THANK YOU.

[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

[christinewen]

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.

[YoussefMedhatAboutaleb]

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

[Anudari]

love you and your channel !!!! <3 saved me sooo much time and hassle

[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!!

[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.

[mahrinahmed]

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.

[ihaashyhd]

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

[lucky_em]

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

[joelbeeby]

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

[Michael-cqqk]

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

[ExamSolutions_Maths]

M2. Check out your exam board on my site.

[raindrop]

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

[josheeee]

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

[ryanathwal]

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