Sir but why mgsin(theta) was considered as positive when it was in the negative x direction (left)? Can it be fs-mgsin(theta)? Will ans be different in this case?
Does this body satisfy rotational equilibrium? Because friction is away from CG.. and it will cause a torque with respect to an axis passing through CG of the Blok.. so linear resultant force is zero.. but rotation about CG is not zero
please believe that I am serious about needing a legit real-world answer to this question, as I tried to follow the video and answer the question myself, but understand I had a party to celebrate my 67% final grade in Alg II since I had somehow passed this gut-wrenching class and wouldn't be forced to take and fail heinously any advanced math courses. I was lost before this gentleman finished his opening sentence -it was English but not my English. So it's with the greatest respect for everyone who speaks THAT^, that I ask for the answer to this: if Ed needs to live in a box with dog Sam & 2 cats Tim and Tom, and he wants to live under an overpass, what the greatest angle his box wouldn't slide down without being secured and with the not insubstantial weight of four occupants plus belongings? is the answer the same as in the video - 11°? I would think 11° is more of a gentle rise and closer to lying down than standing up which is how I decided the angle of the incline couldn't be over 44° but would be safest (i.e. the box will be stable, resting on the incline and normal movements/shifting of occupants won't cause box to slide down into near constant traffic.) at a max angle of 33°: that's just a smidge (= .75°) smaller than the angle that is the midpoint between half of 90° and half of half of 90° or 33.75=45+22.5÷2 . That's how my un-mathy brain decided that 33° was furthest from lying flat without starting to need to stand, since you'd slide anyway. THAT makes sense to me but is it a good answer to the question posed? I'll appreciate all relevant replies which aren't meant to make me feel stupid (my last IQ was 157, and I know more now than I did) and that leave out all unnecessary comments regarding the question. thanks so very much to whomever finds this more than 4 years after the vid originally posted! 🙇🏼♀️.
The reason you guys don't get 100m views plus is your ignorance of not explaining well towards the end. We need a real UA-camr who will solve things with proper explanation.
Explained it the same way as my professor, except you actually make it understandable! Thank you!
Glad it helped!
😄
Your work goes a long way, we are forever grateful
Americans are very pragmatic in mathematics and physics. your explanations are clear and good pedagoqique.
Explained it better than my professor, huge thanks!
Thank you man for the great explanation of every step needed. You saved me so much time and stress with this video and im gratefull and i salute you.
thank you so much... its 3am and my shit is due by 5
you’re a real one for this😭😭
thank you so much for posting this it really helped.
this video came in clutch. firm handshakes 🤝🤝
You got this
Thank you for explaining so well.
Got that aaaa-haaaa moment
So what can I do if I want the object to move for max. acceleration with ange X
thank you for this si, very clear explanation :)
Every time I put tan(0.2) in the calculator I get .003. How did you get 11.3?
Use arctan (which is the same as tan^-1). Don't use tan.
@@randomcarrot7970 Thank you so much!!
@@domenicocomita2950 no problem!
@@randomcarrot7970 You just saved me hours worth of headache. Thank you
How about if you already have your weight and coefficient, and are just trying to find the angle?
That’s what the videos covers. The angle is atan(mu)
@@PhysicsNinja Great thank you!! :)
Thank you for the great explanation!
Thank u so much great explanation👍
Sir but why mgsin(theta) was considered as positive when it was in the negative x direction (left)? Can it be fs-mgsin(theta)? Will ans be different in this case?
no, if you just add fs from his equation or mgsin(theta) from yours then you find that both parts are equal to one another
What if you used a pully system with slightly less weight pulling with you. At the FS point goes to pully with an = or > block helping pull.
sir how about we are given a task to solve for the unknown mass given only the maximum angle of inclination? any idea sir?
Does this body satisfy rotational equilibrium? Because friction is away from CG.. and it will cause a torque with respect to an axis passing through CG of the Blok.. so linear resultant force is zero.. but rotation about CG is not zero
Thank you
Hi, if I have two friction forces which is the static and kinetic what should I use?
For this problem use static because we are trying to find out the max angle the instant before it starts moving
why is wy cos instead of sin? usually i see x axis = cos and y axis = sin
thx u helped me a lot :)
Thank u for ur explanation
Super helpful!!
What kind of Julio Profe are you? Idk, but thanks for helping me in my Physics homework!
please believe that I am serious about needing a legit real-world answer to this question, as I tried to follow the video and answer the question myself, but understand I had a party to celebrate my 67% final grade in Alg II since I had somehow passed this gut-wrenching class and wouldn't be forced to take and fail heinously any advanced math courses. I was lost before this gentleman finished his opening sentence -it was English but not my English. So it's with the greatest respect for everyone who speaks THAT^, that I ask for the answer to this: if Ed needs to live in a box with dog Sam & 2 cats Tim and Tom, and he wants to live under an overpass, what the greatest angle his box wouldn't slide down without being secured and with the not insubstantial weight of four occupants plus belongings? is the answer the same as in the video - 11°? I would think 11° is more of a gentle rise and closer to lying down than standing up which is how I decided the angle of the incline couldn't be over 44° but would be safest (i.e. the box will be stable, resting on the incline and normal movements/shifting of occupants won't cause box to slide down into near constant traffic.) at a max angle of 33°: that's just a smidge (= .75°) smaller than the angle that is the midpoint between half of 90° and half of half of 90° or 33.75=45+22.5÷2 . That's how my un-mathy brain decided that 33° was furthest from lying flat without starting to need to stand, since you'd slide anyway. THAT makes sense to me but is it a good answer to the question posed? I'll appreciate all relevant replies which aren't meant to make me feel stupid (my last IQ was 157, and I know more now than I did) and that leave out all unnecessary comments regarding the question. thanks so very much to whomever finds this more than 4 years after the vid originally posted! 🙇🏼♀️.
Thanks!!!
The reason you guys don't get 100m views plus is your ignorance of not explaining well towards the end. We need a real UA-camr who will solve things with proper explanation.