Physics - Pully on an Incline (2 of 2) With Friction

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  • Опубликовано: 2 окт 2024
  • Visit ilectureonline.com for more math and science lectures!
    This lecture will cover Newton's Second Law: F = ma.
    Problem Text:
    A 5kg mass is placed on an incline with friction making an angle of 30 degrees with the horizontal. A rope is attached and positioned over a pulley at the top of the incline. A 6kg mass is suspended from the free end. What is the acceleration of the system. Assume the coefficient of friction is 0.2.

Комментарии • 364

  • @furfss
    @furfss 9 лет назад +170

    I appreciate that they left in the part where he gets frustrated with his calculations not adding up. It's encouraging to know that even the best can struggle with Physics.

    • @tombskater3000
      @tombskater3000 8 лет назад +4

      +Stefan Bekker I did cringe quite hard tho. Physics cringe compilation may 2016(ap test times of course)???

    • @tombskater3000
      @tombskater3000 8 лет назад

      ha gay Oh, so you've never cringed after making a mistake?

    • @danishashar4861
      @danishashar4861 8 лет назад +3

      nah only dumb people struggle at physics

    • @tombskater3000
      @tombskater3000 8 лет назад +1

      AliGay Nice guys... I don't get too heated at a fellow student(or even teacher) for making a mistake or there, especially if for that one mistake there's a million times they've solved,plugged, memorized, derived and experimented flawlessly.
      Also, why are you watching a simple tutoring video about algebraic physics if you weren't 'struggling' a bit? I mean cm'on. This is easy stuff(This paragraph is satire boys. Think Samuel Clemens on this one.).

    • @mohammedawadaljafary2730
      @mohammedawadaljafary2730 7 лет назад +1

      Stefan Bekker you are damn right man

  • @ryancassarini4557
    @ryancassarini4557 8 лет назад +256

    "my mind is not on this"
    same

  • @wywardwy
    @wywardwy 7 лет назад +179

    I love that you didn't edit the frustration out. Shows that it's normal to be drained by a problem and you just gotta keep plugging away at it.

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад +46

      Yes, I too get confused at times, until I can get my brain unstuck.

  • @thy7917
    @thy7917 8 лет назад +175

    wow sir, your explaination is simply the best of all, even better than khan academy. your pace of teaching isn't too fast nor too slow and the way you explain makes everything so much understandable in the name of physic. Please continue to upload videos. Thank you so much!.

    • @danishashar4861
      @danishashar4861 8 лет назад

      +Thy Nguyen Nah fam. Khan academy is a new level. You can't compare

    • @thy7917
      @thy7917 8 лет назад +4

      That is my opinion

    • @danishashar4861
      @danishashar4861 8 лет назад +1

      +Thy Nguyen well your opinion is wrong. don't give wrong opinions

    • @thy7917
      @thy7917 8 лет назад +1

      +Danish Ashar wrong about?

    • @thy7917
      @thy7917 8 лет назад +1

      +Danish Ashar kid, if you dnt like what i say you can fk off :)

  • @PolyPixel
    @PolyPixel 7 лет назад +18

    Professor van Biezen, your videos flow so effortlessly, at such a smooth and steady pace, and I have been watching one after the other, so it was not until the moment of frustration we see in this video that I finally wake up to all the work that you and your helpers must put into making one great video after the other. There are so many videos, and they are so good!! Yours is the best science/math channel on RUclips! I want to thank you for taking the time (and these must have taken you SO much time) that you have put into sharing your excellent teaching style and knowledge with us all, and the effort that you have made to do so. I don't know why you do this other than to help those of us who are struggling to learn. Thank you so much!!! Kudos to you, Sir!

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад +10

      PolyPixel,Thank you for this very heartwarming comment. Yes, between my wife and I we have invested 7500 hours of our time into this venture and it has grown far beyond our expectation. Every one of us has been given a skill and we believe that a big part of our life and existence is to use that skill to give back to society in response to the blessing of life that we have received. It is great to know that these videos have helped many thousands of students around the world.

    • @PolyPixel
      @PolyPixel 7 лет назад +2

      G-d bless you and your wife and family. I owe you a debt of gratitude. All the best for continued success in your teaching endeavors and everything!

  • @xrisku
    @xrisku 6 лет назад +43

    at 7 minutes, that is me with every physics problem. :P

    • @MichelvanBiezen
      @MichelvanBiezen  6 лет назад +17

      That was before we learned how to edit videos. Now we'd just edit it out. But it is kind of funny. XD - Editing "Crew"

  • @MichelvanBiezen
    @MichelvanBiezen  10 лет назад +10

    If the pulley has mass you must solve the problem using the moment of inertia of the pulley. Instead of F - m x a you must use Torque = moment of inertia x angular acceleration. Look at the videos on moment of inertia. There are some examples where the pulley has mass.

    • @Peter_1986
      @Peter_1986 10 лет назад

      What happened on 7:00?

    • @MichelvanBiezen
      @MichelvanBiezen  10 лет назад +1

      Laurelindo
      It looks like the "editor" didn't take out that part of the video. (oops) This is one of the early videos that we made. Notice how they didn't all go that smooth.

    • @trevdc2434
      @trevdc2434 10 лет назад +2

      Laurelindo LOL he forgot to multiply the numerator by gravity. He didn't mess up on the board, he just plugged it in to his calculator wrong.

    • @davidalearmonth
      @davidalearmonth 8 лет назад

      Quick editing tip. Even in RUclips, you can fairly easily trim out the bit in the middle where you stumbled on the calculator. Thanks for the videos!

    • @Peter_1986
      @Peter_1986 8 лет назад +4

      David Learmonth
      I actually kinda like bloopers like that, I am not bothered by them at all and gladly watch the events unfold! =D
      They also show that even professors can make mistakes.

  • @hiimsofiyah4744
    @hiimsofiyah4744 7 лет назад +20

    6:53 my reaction everytime i press my calculator.
    nevertheless, the best account for teaching physics so far!

  • @jacobvandijk6525
    @jacobvandijk6525 10 лет назад +96

    I love this moment of despair! We are all human, aren't we? ;-)

    • @MichelvanBiezen
      @MichelvanBiezen  10 лет назад +33

      Sjaak,
      I wouldn't elevate it to the level of "dispair", but you can see that I have my moments that my brain gets stuck and I can't see where to go next. That makes understanding it afterwards so much better. Yes, we are all human.......

    • @lujainaakz1630
      @lujainaakz1630 6 лет назад

      :) I agree with you Dr, we must keep going and trying until we success

  • @victoriah.4930
    @victoriah.4930 6 лет назад +2

    I like how you explain this. You're going at the perfect pace and it's really helpful. Thank you!

  • @rajumiah1000
    @rajumiah1000 9 лет назад +11

    You are much better than my professor

  • @LifenLaugh381
    @LifenLaugh381 6 лет назад +2

    By far the best video on physics that expalins everyhting......m ready for this part of the test (TEEHEE)

  • @chillywilly3397
    @chillywilly3397 4 года назад

    good job sticking with the problem. I saw it...I'm glad you took a second to see it for yourself.

  • @tyn8058
    @tyn8058 6 лет назад

    just wanted to say thank you for explanations. You have helped me understand physics better than any other book or teacher. Keep up the great work!

  • @sudiXP
    @sudiXP 6 лет назад +1

    Dear professor Van Biezen, thank you and your, beautiful supporting, wife. I wish I could express my gratitude, for helping hundreds of thousands if not millions of students like myself, even more. Your efforts in making these videos and the encouragement your wife gave you is a symbol of your good intentions. I wish you both nothing but happiness and success.

    • @MichelvanBiezen
      @MichelvanBiezen  6 лет назад

      Thank you Mahammud, we really appreciate your comment.

  • @Star-si9uc
    @Star-si9uc 8 лет назад +7

    Thank you for explaining it so simple.

  • @delonthompson2132
    @delonthompson2132 5 лет назад +18

    He wanted to start again respect for a perfectionist

  • @chocolate1149
    @chocolate1149 7 лет назад +1

    Mass m = 90.8 kg sits on an inclined plane that makes an angle α = 25.3o with the horizontal. A massless string is tied to m, passes over a frictionless, massless pulley, and is tied to mass M = 56.8 kg hanging on the other end. Assume:
    - the coefficient of kinetic friction between m and the surface of the plane is μs = 0.269.
    - the string is parallel to the surface of the incline.
    If m is initially moving up the incline find the acceleration (magnitude in m/s2 and direction) of the system. (To give direction: give the answer as positive if the acceleration of m is up the incline, and negative if the acceleration is down the incline.)
    it gives 0.31 but it says its wrong, do you know why?

  • @NadeawJaiman
    @NadeawJaiman 8 лет назад +35

    well at 7:00 I learnt not only Newton's Law but also "Perfection is Imperfection"
    the dopest Professor! Thanks ü

  • @G8T.99
    @G8T.99 7 лет назад +2

    Enlightening lecture!
    However I would like to recommend using the value of 'g' as 10 m/s^2 just for easier calculations and explanation.
    Thanks.

  • @kirbyrules99
    @kirbyrules99 8 лет назад +1

    Really useful lessons, please keep on going sir you are aiding the education of a lot of students!

  • @funnycomedy9151
    @funnycomedy9151 4 года назад +1

    Thanks very much sir.you have help me so much for am having exams of the same things.... watching from the university of Zambia

    • @MichelvanBiezen
      @MichelvanBiezen  4 года назад

      Wow, it is great to welcome you to the channel!

  • @joelharris6158
    @joelharris6158 10 лет назад

    Love you videos! Just came across them with a physics exam tomorrow... Big help!

  • @kendradurussel8217
    @kendradurussel8217 2 года назад +14

    I just wanted to say thank you for making these videos. You may have single-handedly saved my Physics grade. You make it so much easier to understand than my professor and the way you break it down and explain it makes sense to me. You've earned a follow from me for sure.

    • @MichelvanBiezen
      @MichelvanBiezen  2 года назад +2

      Thank you. We are glad you found our videos. :)

  • @bruceshimukayi3424
    @bruceshimukayi3424 Год назад +1

    This is an excellent explanation

  • @crazythings1753
    @crazythings1753 3 года назад +2

    it is very very very very very very very clear thanks very much

    • @MichelvanBiezen
      @MichelvanBiezen  3 года назад +1

      You are most very very very very bery welcome!

  • @AlizKilla1
    @AlizKilla1 4 года назад +2

    Great video, easy to understand

  • @WonderingNobody
    @WonderingNobody 7 лет назад +1

    You make everything very easy..Thank you so much.

  • @Jaded_o
    @Jaded_o 6 лет назад +1

    Thank you very much ,
    You make everything easy,
    you are the best.

  • @adegboyegasamuel342
    @adegboyegasamuel342 3 года назад +1

    You are too great sir I love the fact that u portrait a good quality teaching sir
    and that moment of despair change my way of thinking it really great u didn't edit it sir 🥰😋
    A question on the tension
    I can't picture a situation that the pulley that will have a mass and how it will affect the tension on the system

    • @MichelvanBiezen
      @MichelvanBiezen  3 года назад +1

      We have several playlists with examples like that. See this video: ruclips.net/video/qu46xL4KPvU/видео.html in the playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS

  • @hillaryjacobs8627
    @hillaryjacobs8627 5 лет назад +2

    I love this man.

  • @BryannahNicole
    @BryannahNicole 6 лет назад +1

    This video helped me so much! Thank you

  • @ibrahimabulwifa4296
    @ibrahimabulwifa4296 5 лет назад +1

    clear , easy . thank you for your explanation .

  • @franzguillermo2125
    @franzguillermo2125 Год назад +1

    Thank you very much for this wonderful video! help me a lot on figuring out this type of problem..

  • @umarani7468
    @umarani7468 7 лет назад +1

    sir it is very clear for me now I can solve any pully problem thank u sir

  • @jhankusona
    @jhankusona 7 лет назад +7

    Thank you Sir.It helped me a lottt.You made the difficult things very easy to understand.Am grateful to you.

  • @quamos99
    @quamos99 8 лет назад +9

    i love the way you explain. really my mechanics became a lot easier.

  • @passionatephysics3676
    @passionatephysics3676 7 лет назад +1

    Thanks u so much sir.... God always with u... Stay blessed.

  • @adrenvargiz9704
    @adrenvargiz9704 3 года назад

    Really great explanation to help with reviewing!

  • @MrDgketchum
    @MrDgketchum 10 лет назад +11

    forgot to edit out the short period of frustration at 7:02

  • @coltennabers634
    @coltennabers634 9 лет назад +6

    You absolutely blow my professor out of the water. You are so much more clear with all the steps you take and it makes understanding the topics much easier. Thanks so much.

  • @YonasMoges-qp7uc
    @YonasMoges-qp7uc Год назад +1

    I really appreciate you thank you

  • @robelfitiwi3344
    @robelfitiwi3344 8 лет назад +8

    +Michel van Biezen hey but there is a slight confusion. you gave a coefficient of friction right? but you need to make it clear whether it is a static or kinetic. for the question we need a static coefficient.

    • @MichelvanBiezen
      @MichelvanBiezen  8 лет назад +1

      +Robel Fitiwi Since the objects are in motion, this is kinetic coefficient of friction.

    • @robelfitiwi3344
      @robelfitiwi3344 8 лет назад +1

      +Michel van Biezen , okay if you take it as "kinetic " , try to use the static friction if it makes the object move . Unfortunately, it won't make it move .

    • @robelfitiwi3344
      @robelfitiwi3344 8 лет назад +1

      +Michel van Biezen which makes using kinetic friction zero percent necessary.
      I mean you literally can't get acceleration because it won't move .

    • @wowfmomf6126
      @wowfmomf6126 6 лет назад

      Well just give it a little push with your hand and the system will move, and the rest is just as calculated.

  • @floriannecaparanga1498
    @floriannecaparanga1498 3 года назад +3

    "My mind is not on this.." Me: Same energy. BUt thank you so much for this videoooo

  • @glecielopez8588
    @glecielopez8588 9 лет назад +4

    I am confused of placing sine and cosine.

    • @MichelvanBiezen
      @MichelvanBiezen  9 лет назад +1

      Glecie Lopez
      There are playlists on the channel that can help you understand:
      Take a look at:
      TRIGONOMETRY BASICS
      TRIGONOMETRY - THE RIGHT TRIANGLE

    • @yashpawarelitepawar
      @yashpawarelitepawar 9 лет назад

      +Glecie Lopez consider a vector, find the angle (theta) with respect to the x axis, then take resultant of the vector, the one with angle theta is costheta and the (90-theta) one is sintheta

  • @potatojam6519
    @potatojam6519 7 лет назад +14

    I seriously think all he needs is a real calculator...

  • @Jallah-Tafari
    @Jallah-Tafari Год назад +1

    Good sir, I owe you!!!

  • @sonicfastman123_2
    @sonicfastman123_2 6 лет назад +7

    great video, my left ear is very thankful for your help

  • @vaiterius
    @vaiterius 2 года назад +1

    Thank you sir!

  • @victorpenya6132
    @victorpenya6132 3 года назад +1

    Beautiful man

  • @MichelvanBiezen
    @MichelvanBiezen  10 лет назад +3

    Murathan,
    g = 9.8 m/sec^2
    g is the acceleration due to gravity, experienced by all objects on the surface of the earth.
    Weight = mass x g
    or w = mg
    weight is a force and therefore, mg is the force pulling an object with mass m to the Earth.

  • @appleukg
    @appleukg 4 года назад +1

    how do we judge the direction of motion before starting the problem? We get a different equation if the motion is assumed to be in the opposite direction - i.e. if the mass on the incline is going down. in that case a = (m1+m2 sin theta - m1cos theta * mu) * g / (m1 + m2).

  • @Miasmayhem
    @Miasmayhem 7 месяцев назад +1

    You are great! Thank you

  • @svarsvarzai2869
    @svarsvarzai2869 6 лет назад +1

    MvB thanks

  • @hutingting
    @hutingting Год назад +1

    here because of physical 100 quest 3 ...relearning the pully system

  • @masonthompsen248
    @masonthompsen248 5 лет назад +1

    What about when the pulley itself has mass?

    • @MichelvanBiezen
      @MichelvanBiezen  5 лет назад +1

      Those examples are covered in the playlists involving moment of inertia.

  • @alphaspartan
    @alphaspartan 5 лет назад +3

    First of all, thank you for the great explanation. Second of all, the bit where you have a calculator issue was hilarious and I applaud you for leaving it in. Bravo good sir. I learned something today!

  • @jonp.6131
    @jonp.6131 4 года назад +3

    It's okay, Michael, we have all lost our cool when doing this

  • @nachomama55
    @nachomama55 7 лет назад

    Thanks again.

  • @kathygonzalez9706
    @kathygonzalez9706 7 лет назад +3

    you sir, are awesome! I almost cried watching your videos, I really appreciate people like you who explains things so simple and such organized manner! This is what makes learning fun! kuddos to professors like you! 👏👏👏👍👍👍

  • @abundios8211
    @abundios8211 10 лет назад +3

    why couldnt i find your videos sooner....

  • @II2023II
    @II2023II 2 года назад +1

    To know the direction of acceleration, i need to calculate mg for hanging mass then calculate the mg sin theta for another mass, the biggest weight will be the direction of acceleration?
    Is that true, Could you please explain it to me?

    • @MichelvanBiezen
      @MichelvanBiezen  2 года назад +1

      Yes, you need to compare the weight of the hanging mass to the mg sin(theta) of the mass on the slope. Off course you also need to take into account the friction force. Or you can just assume a direction, work out the problem and then see if you get a realistic answer.

    • @wrathandlaughter
      @wrathandlaughter 2 года назад

      @@MichelvanBiezen still replying after 9 years!? you're such a G.O.A.T. my guy!

  • @MrDoYouWannaBeOnTop
    @MrDoYouWannaBeOnTop 9 лет назад +3

    what is we were asked to find the tension of the string?

  • @vanessamcgrew4486
    @vanessamcgrew4486 7 лет назад +2

    i got 2.291 m/s^2 for acceleration :(

  • @morrisombiro
    @morrisombiro 8 лет назад +3

    You make physics a lot easier ! THANKS SO MUCH !

  • @cassac8049
    @cassac8049 3 года назад +2

    His videos are so amazing, and it is very refreshing to know the reasons behind equations! My physics teacher barely explains anything, and jumps all over the place without telling us why.

  • @kritikasharma5436
    @kritikasharma5436 5 лет назад

    this video is amazing.Can i ask why did you subtract force of friction although it is acting in the same direction as that of acceleration? Thanks!

    • @MichelvanBiezen
      @MichelvanBiezen  5 лет назад

      If you look at the drawing, you will see that the direction of the friction forces is opposite to the direction of the acceleration.

  • @cassylacute
    @cassylacute 7 лет назад +3

    no matter how i do it on my calculator I'm getting 9.6 what am i doing wrong

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад

      For what result? a of T?

    • @cassylacute
      @cassylacute 7 лет назад

      It was for A bit I got it my calculator was on a different mode thank you. Do I need to find the t value?

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад

      Only if you are interested in finding the tension.

  • @THUNGUNS
    @THUNGUNS 8 лет назад +2

    Why is it umgcostheta, not umgsintheta? I thought the force of friction is in the x direction.

    • @MichelvanBiezen
      @MichelvanBiezen  8 лет назад +1

      +Thun Guns Pro
      The definition of the friction force is: F fr = N * mu where N is the normal force and mu is the coefficient of friction.
      The normal force on an inclined plane is equal to the perpendicular component of the weight of the object, the reactionary force to the component. Thus F fr = N * mu = mg cos (theta) * mu

  • @bryanestradachiang7176
    @bryanestradachiang7176 Год назад +1

    I am working in an experiment where I placed a car on an inclined, the car is attached to a string and the string passes through a pulley, the string is holding a bigger mass that is pulling the car to the highest point of the inclined. But we measured the tension of the string with a sensor as we held the car to avoid moving. Now I have to calculate the Tension applying Newton's First Law and compare the theoretical Tension with the measured Tension. Where I am struggling with it is with my diagram, I don't know if I have to include the force I am applying onto the car. Any tip would be appreciated.

    • @MichelvanBiezen
      @MichelvanBiezen  Год назад +1

      That seems like a dangerous experiment. You should be able to find the information you need from these videos.

  • @marcrogue5268
    @marcrogue5268 7 лет назад +2

    Hello professor, I just started taking dynamics and since we have to break up forces into their x and y components we also need to account for the x and y component of the Friction force, which never really happened with kinematics

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад

      On an inclined plane, the components of the forces are written in terms of the parallel and perpendicular components (relative to the surface of the plane).

  • @jadidablomacmod3204
    @jadidablomacmod3204 Год назад +1

    Why is friction force = m1gCOS theta and not m1gSIN theta? Is the friction force on the y axis? Why not on x axis?

    • @MichelvanBiezen
      @MichelvanBiezen  Год назад +1

      Neither. The friction force is parallel to the slanted surface. And by definition the friction force = normal force to the surface x coefficient of friction. We have videos that go into more detail on the friction force.

  • @matthewmoran943
    @matthewmoran943 8 лет назад

    How come in this video, you used both m1g(sin theta)-m1g(cos theta), but in the video before (without friction), you only used m1g (sin theta)?

    • @MichelvanBiezen
      @MichelvanBiezen  8 лет назад +1

      +Matthew Moran
      m1g(cos theta) * mu is the friction force opposing the acceleration of the system

    • @matthewmoran943
      @matthewmoran943 8 лет назад +1

      +Michel van Biezen oh ok. Thanks!

  • @Gustavo-gy3dq
    @Gustavo-gy3dq 7 месяцев назад +1

    Im confused wouldn't there be another force which is the Tension force, also would the tension force be equal to m2g?

    • @MichelvanBiezen
      @MichelvanBiezen  7 месяцев назад +1

      Tension is internal to the system and does not affect the acceleration of the system. (It is not an outside force acting on the system)

  • @LillianRig
    @LillianRig Год назад +1

    I know I need to be paying attention but all I can focus on is how your voice sounds like gru from despicable me, also thanks 4 the vid it was helpful

    • @MichelvanBiezen
      @MichelvanBiezen  Год назад +1

      You are not the first viewer who has mentioned that. 🙂

  • @giorgi1129
    @giorgi1129 Год назад +2

    I was able to perfectly understand both parts. These videos are excellent, and you are an excellent teacher. Please continue to enrich the science education of others the way you did here. Physics and Math and their related fields are often seen as too daunting, and it's heroes like you that make them accessible to those who would otherwise believe they don't have the brains to do them.

    • @MichelvanBiezen
      @MichelvanBiezen  Год назад +1

      Thank you! 😃 We are glad you find these videos helpful!

  • @penguinpenguinpenguin
    @penguinpenguinpenguin Год назад +2

    thanks i didnt study for my test

  • @saharsaf802
    @saharsaf802 5 лет назад

    AMAZING !

  • @ElectroWiz118
    @ElectroWiz118 9 месяцев назад +1

    What would the equation look like if the mass of m1 was bigger than m2

    • @MichelvanBiezen
      @MichelvanBiezen  9 месяцев назад +1

      It depends on how much bigger. If it is big enough to cause an acceleration in the other direction ,you would simply reverse the order Aiding force = m1gsin(theta) Opposing force m2g and the friction force.

    • @ElectroWiz118
      @ElectroWiz118 9 месяцев назад +1

      @@MichelvanBiezen okay thank you that really helps

  • @bhaktichokshi
    @bhaktichokshi 7 лет назад +1

    +Michel van Biezen; And what if the object was at rest and now moving. Ther will be static friction too, right? In that situation, what should I do?

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад +1

      If the object is at rest, we only use the static coefficient of friction. Once the object begins to slide, then the static coefficient of friction is no longer used and we only use the static coefficient of friction.

    • @zackcarman7845
      @zackcarman7845 7 лет назад

      Michel van Biezen Do you mean we don't use the static coefficient of friction and use the kinetic coefficient of friction once the object begins to slide?

  • @relaxingmomentswithnoxy4785
    @relaxingmomentswithnoxy4785 2 года назад +1

    2022 am still this is great thank you so, but if the camera will go close to the board so we can see clearly.

    • @MichelvanBiezen
      @MichelvanBiezen  2 года назад +1

      Yes, we made some improvements in recent years, moving the camera closer to the board, unlike this older video.

  • @el33tkrew
    @el33tkrew 2 года назад +1

    What if we find the acceleration, and the Frictional Force? How would we find that?

    • @MichelvanBiezen
      @MichelvanBiezen  2 года назад +2

      If you are given the acceleration and you want to find the frictional force, you would solve it in the EXACT SAME WAY. Set up the equation as shown and then you'll have the coefficient of friction as the unknown that you must solve for instead of the acceleration.

  • @bridgetngona1980
    @bridgetngona1980 5 лет назад +1

    l now understand better. THANK YOU Sir

  • @aliciawilliams3762
    @aliciawilliams3762 4 года назад +1

    yeah. i am feeling the same but i have much of the semester left. I have to do better.

  • @dianays8248
    @dianays8248 7 лет назад +1

    why are the forces subtracted?

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад +1

      The net force = (all the forces aiding the acceleration) - (all the forces opposing the acceleration)

    • @dianays8248
      @dianays8248 7 лет назад +1

      thank youu!

  • @dineshdins2718
    @dineshdins2718 6 лет назад +1

    why?are we not multiplying the total mass x9.81

    • @MichelvanBiezen
      @MichelvanBiezen  6 лет назад +1

      Always start from the basic equations. In this case we use Newton's second law (F = ma) which means that a = F/m which means we divide the net force applied by the total mass (not the weight).

  • @indoororchidsandtropicals358
    @indoororchidsandtropicals358 4 года назад +1

    I want to know when, as a biologist, I am ever going to use mechanics. Thanks.

    • @MichelvanBiezen
      @MichelvanBiezen  4 года назад +1

      Only people who go into medicine, (doctors, physical therapists, etc.), or people that go into designing artificial limbs, medical equipment ....

  • @LuisConradoAlcala
    @LuisConradoAlcala 7 лет назад +1

    Fucking awesome

  • @grapecraftfactions7675
    @grapecraftfactions7675 6 лет назад +1

    your answer is wrong, correct one is 3.89m/s2

    • @MichelvanBiezen
      @MichelvanBiezen  6 лет назад +1

      The answer in the video is correct. Thanks for checking.

  • @peg1122
    @peg1122 2 года назад +1

    Why dont masses cancel out in numerator and denominator

    • @MichelvanBiezen
      @MichelvanBiezen  2 года назад +2

      You cannot cancel per the algebraic rules. (M - m) / (M + m) cannot be reduced

  • @bhekithembadlamini7251
    @bhekithembadlamini7251 7 лет назад +1

    Thanks Michel, your way of explaining has improved my understanding.

  • @wrathandlaughter
    @wrathandlaughter 2 года назад +1

    also i had a question, in vectors Ax = ACos(theta) and Ay = ASin(theta), why is x and y switched with Sin and Cos now?

    • @wrathandlaughter
      @wrathandlaughter 2 года назад +1

      i found the answer myself, it's an upside down triangle to match side with side, ha!

    • @MichelvanBiezen
      @MichelvanBiezen  2 года назад +1

      These are not Ax and Ay vectors. The vector components of the weight (mg) are parallel and perpendicular to the incline. Look at an example when the angle is 80 degrees.

  • @nicolledransfeldt6703
    @nicolledransfeldt6703 8 лет назад +1

    Awesome, thank you for the help! And I read every comment and nobody commented on the event that occurred directly after the frustration..Was that a spider you killed???

    • @MichelvanBiezen
      @MichelvanBiezen  8 лет назад +3

      I think it was a little fly. Usually we edit this out, but this one slipped through the editing process. As you can see, I do get stuck sometimes as well. Understanding why we get stuck and how to get unstuck is a valuable tool for education.

  • @valeriaoliver2593
    @valeriaoliver2593 5 лет назад +1

    I would like to thank you because I was in a crying panick attack because I did not understand anything and your video was the light of my night. I wish you the best of blessings in your life and in your families life. Thank you.

  • @themagesticunicorn7284
    @themagesticunicorn7284 3 года назад +1

    Can you please explain why the acceleration is not negative, it's going in the downward direction!!

    • @MichelvanBiezen
      @MichelvanBiezen  3 года назад +1

      Since one block goes up the incline and the other block moves downward, it makes more sense to express the acceleration in terms of the magnitude and show the direction with the curved arrow.

    • @themagesticunicorn7284
      @themagesticunicorn7284 3 года назад +1

      ​@@MichelvanBiezen Alright, I understand! Thank you. Also, I just took my physics midterm and I got the highest grade and I really want to thank you because it's all possible due to your videos! Cheers and keep doing what you're doing :D

  • @charlesbromberick4247
    @charlesbromberick4247 Год назад +1

    I want to see incline plane pull at angle to the plane.

  • @mahlonkalunga4415
    @mahlonkalunga4415 6 лет назад +1

    Wow Michael!
    Well explained.
    But why did you multiply the costheta with the friction only and not the whole sum of the forces?

    • @MichelvanBiezen
      @MichelvanBiezen  6 лет назад +1

      In the numerator, we have 3 forces. One that aids the acceleration, the other 2 oppose the acceleration. One of those 2 is the friction force. The definition of the friction force is that it is equal to the normal force (mg cos(theta)) x coefficient of friction (u).

  • @herzikkimolog
    @herzikkimolog 7 лет назад +1

    In which grade you teach these in the US and how old the students usually are? I am making a comparison about this with another country's education.

    • @MichelvanBiezen
      @MichelvanBiezen  7 лет назад +1

      It depends on the individual and the school. This usually taught in high school for the first time.

  • @Harrynorty
    @Harrynorty Год назад +1

    Thanks bro. U make it so easy when u break it down to just each force opposing the other without adding extra stuff. bless

  • @helloworld4147
    @helloworld4147 8 лет назад +1

    Very helpful Sir. Thanks you very much.