Physics PDF Worksheets: www.video-tutor.net/physics-basic-introduction.html Full-Length Videos & Exams: www.patreon.com/MathScienceTutor/collections Full 1 Hour 28 Minute Video: www.patreon.com/MathScienceTutor Direct Link to The Full Video: bit.ly/3QdBavj
I subscribed to you because I'm in organic chemistry and I found your videos extremely helpful, but I'm also in physics and just stumbled across these videos and I just need you to know you are literally the reason that I am passing any of my classes right now :") your videos are so much more helpful than my professors will ever be
@@steve4718 physics doesn’t change depending on where you’re at in school lmfao still gotta learn the basics in college (think abt the amt of people who haven’t had high school physics in years or at all)
@@yikes3442 yeah here in Germany they make us study these things and statistically 70% of the class has taken physics already in school meanwhile i come from a different country where the last time we had physics was in 10th grade during the lockdowns
this helped, my physics teacher doesn't do a lot of teaching (basically none at all) so having alternative explanations this simple and easily accessible is huge.
For those that want some more practice on pulley problems, I've uploaded a video where I solve such problems step-by-step! It might help some of you out there :) ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-y7g9gHchOIk.html
Hi, there's something I'm confused about. When considering the rotational inertia of the pulley and doing a problem with it, it becomes apparent that each side of the rope has its own unique tension that is not equal to each other. In this case, doesn't the method to solve problems without the consideration of pulleys violate this principle? What assumptions are made in order to solve it like this? If it's the existence of the pulley itself that causes the uneven tension, what exactly is the pulley doing to the string to essentially split apart its effects?
Oh my fucking god YOU ARE GOD These r da first few problems/lesson in My fuckin physics class. And SOMEHOW you PERFECTLY timed your past 2 uploads to when i LITERALLY fuckin needed them You are an underrated educational yt channel and you DESERVE all our respect.
Great video, although, my main problem in understanding was when solving for tension. When we made the tension equation, why couldn't we plug in the value at that moment itself?
when he was solving for the acceleration he used T= to make two equations to set the whole solving part up. at that point we did not know the tension, which is what he solved for later
Thank you so much for what you do! God bless you! And remember: Jesus is coming on a day we least expect. So plss use the time you have now to get right with Him and call upon Him, He loves us all. He will never turn a repentant heart away. Luke 15:7- "I tell you that in the same way there will be more rejoicing in heaven over one sinner who repents than over ninety-nine righteous persons who do not need to repent." May God bless everyone who reads this💜💜
so then a = (m2g - T)/m2 and a = (T - Mkm1g)/m1 also work if you have tension. That's pretty cool. But because we haven't found T, we're substituting the T instead.
I'm a bit confused. In the 1st example isn't the weight of the 5 kg box supposed to equal to the tension experienced by the string (T=m2g)? The box itself is always attached to the string so the forces must always be constant right? Physics is not my strongest subject 😢😂
No because they go in different directions T goes up and m2g goes down for exemple let's say you took 3 steps to the right and then 2 steps to the left the final spot from the start is 1 steps to the right because 3-2=1 so it should be T-m2g. Dont know if I explained it well hope you understood
Organic tutor help me out this question.... An elastic string is stretched between two fixed points which are 1 meter apart. An object weighing 3.7N is suspended at the middle of the stretched elastic causing the midpoint to drop by 0.11 meters. Calculate the: (a) angle of each part of the string to the vertical (b) Tension in each part of the string.
Do you have another example where your calculations are based off time, distance and positioning. I was given the task to do research about this kind of situation but with sensors, but I just can't find any.
i have the same question with different numbers and i applied the formula u recommended at 9:05 to solve for acceleration. However, the answer is not correct.
Why label ay as negative? If you label Tx as negative, given that the force in x direction, Tx is therefore negative, wouldn't that give you the right net acceleration, without having to artificially add a negative sign to the ay?
It’s based on a coordinate plane. To the right on the x-axis is positive, to the left on the x-axis is negative, upwards is positive on the y-axis, and downwards is negative on the y-axis. Since block m2 moves downwards, it is negative. Hope this helps!
@@fridamoreno9689 Hey, just realized that he labels ay as negative because in his coordinate plane, downward=ay and upward=-ay. Therefore, he says ax=ay since m1 follows a path to the right and then downward. But if upward=ay and downward=-ay, one would have to make the substitution ax=-ay, since m1 follows path to the right and then downward.
My physics is a little more complicated than your videos, because my prof is a little wild. However these videos have helped me pass like 4 classes so far 😂
