Thank you so much for helping me build a foundational understanding of physics with your detailed explanations. You take the time where other teachers don't to explain elementary things -- it is really making the difference for me.
For the last problem, I was wondering why you kept it at 30 degrees instead of plugging in 330 because the angle was under the horizontal? Thank you for the help!
Thank you for the video! For the last example, why is the sum of forces in the y-direction equal to 0? Since the force pulls 30 degrees below the x-axis, if the force keeps pulling, doesn't it go further and further in the y-direction? So how can we assume that the sum of forces in the y-direction is 0? Thanks in advance for your help!
You're thinking of it more as a torque (turning force) whereas this would be for example pulling with a rope from an angle of 30 degrees that stays constant. Unless stated that 30 degrees stays the same
Hi Chad! For the question at 31:00 it said she's accelerating 2m/s2 downward isn't it meant to be -2 in the calculations? If not please why ,thanks for your efforts ❤
first chem, and now physics. thank you
Very welcome.
Thank you so much for helping me build a foundational understanding of physics with your detailed explanations. You take the time where other teachers don't to explain elementary things -- it is really making the difference for me.
You're welcome and Thank You.
Thank you so much, Chad!!!! After spending 3 hours in a lecture, I did not understand this concept and you saved the day in one hour.
You're welcome - Glad the channel helped you!
For the last problem, I was wondering why you kept it at 30 degrees instead of plugging in 330 because the angle was under the horizontal? Thank you for the help!
how do we know that 50N is part of the y component? Is it because the question states that it is "below the horizontal"?
Yes, essentially, it points diagonally downwards, which means of the 50N part of that is in the y direction
Thank you for the video! For the last example, why is the sum of forces in the y-direction equal to 0? Since the force pulls 30 degrees below the x-axis, if the force keeps pulling, doesn't it go further and further in the y-direction? So how can we assume that the sum of forces in the y-direction is 0? Thanks in advance for your help!
You're thinking of it more as a torque (turning force) whereas this would be for example pulling with a rope from an angle of 30 degrees that stays constant. Unless stated that 30 degrees stays the same
24:49 So if normal force is 490 N and positive as you annotated, does the weight now become negative?
thank you so much Mr.
You're welcome
Hi Chad! For the question at 31:00 it said she's accelerating 2m/s2 downward isn't it meant to be -2 in the calculations? If not please why ,thanks for your efforts ❤
I talk about this from 30:15 onwards :)
@@ChadsPrep thank you , totally missed it 😂.
It happens to the best of us...
21:23 did we just treat a woman as an object, then? love your videos btw!