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

[@wywardwy]

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]

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

[@furfss]

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]

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

[@tombskater3000]

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

[@danishashar4861]

nah only dumb people struggle at physics

[@tombskater3000]

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]

Stefan Bekker you are damn right man

[@ryancassarini4557]

"my mind is not on this" same

[@thy7917]

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]

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

[@thy7917]

That is my opinion

[@danishashar4861]

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

[@thy7917]

+Danish Ashar wrong about?

[@thy7917]

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

[@NadeawJaiman]

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

[@kendradurussel8217]

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]

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

[@coltennabers634]

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.

[@PolyPixel]

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 KZbin! 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]

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]

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!

[@MichelvanBiezen]

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]

What happened on 7:00?

[@MichelvanBiezen]

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]

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]

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

[@Peter_1986]

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.

[@MichelvanBiezen]

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.

[@delon.thompson]

He wanted to start again respect for a perfectionist

[@cassac8049]

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.

[@valeriaoliver2593]

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.

[@MichelvanBiezen]

Thank you and we are glad we could help.

[@hiimsofiyah4744]

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

[@xrisku]

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

[@MichelvanBiezen]

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"

[@quamos99]

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

[@alphaspartan]

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!

[@giorgi1129]

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]

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

[@victoriah.4930]

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

[@jacobvandijk6525]

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

[@MichelvanBiezen]

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]

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

[@sudiXP]

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]

Thank you Mahammud, we really appreciate your comment.

[@MrDgketchum]

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

[@mccombsculpture]

I'm glad that's in there. Real people.

[@phynos8936]

Agreed.

[@rajumiah1000]

You are much better than my professor

[@jhankusona]

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

[@bhekithembadlamini7251]

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

[@chillywilly3397]

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

[@morrisombiro]

You make physics a lot easier ! THANKS SO MUCH !

[@robelfitiwi3344]

+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]

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

[@robelfitiwi3344]

+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]

+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]

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

[@sonicfastman123_2]

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

[@mankiddyman]

hahahahahahah

[@thedagger994]

hahahhahahahaha

[@tyn8058]

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!

[@kathygonzalez9706]

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! 👏👏👏👍👍👍

[@LifenLaugh381]

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

[@Star-si9uc]

Thank you for explaining it so simple.

[@chocolate1149]

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?

[@petermokola7186]

A block rests on a plane that is inclined at angle of 30degrees to the horizontal. Coefficient of kinetic friction between the block and the plane is 0.2 . if the block had a mass of 10kg and a man wishes to push it up the plane : (a) what is the direction of the frictional force? (b)what force parallel to the incline is necessary to keep it sliding up the plane ? (c) what horizontal force would be required to push it up the plane ?

[@appleukg]

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).

[@MrDoYouWannaBeOnTop]

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

[@Harrynorty]

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

[@MichelvanBiezen]

Glad to help

[@G8T.99]

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.

[@marcrogue5268]

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]

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).

[@desena1991]

He became so relatable at 6:55

[@bryanestradachiang7176]

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]

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

[@bhaktichokshi]

+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]

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]

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?

[@floriannecaparanga1498]

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

[@THUNGUNS]

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

[@MichelvanBiezen]

+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

[@funnycomedy9151]

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]

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

[@mahlonkalunga4415]

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

[@MichelvanBiezen]

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).

[@JamesonWhitton]

What direction is friction if the box on the incline has a greater mass than the hanging object?

[@MichelvanBiezen]

That depends on the direction of motion, or what the direction of the motion would be if there was no friction. The direction of friction typically is in the opposite direction to the motion if there was no friction.

[@JamesonWhitton]

Thank you!

[@dutsywhitaker455]

How would you go about doing the same problem but there is an angled force applied directly to m1 going up the ramp?

[@MichelvanBiezen]

We have similar examples like that in the playlists.

[@dutsywhitaker455]

@@MichelvanBiezen Thank you for the quick response. The closest video I can find isn’t a pulley system kzbin.info/www/bejne/fpvdZJhnrtqgo7M

[@MichelvanBiezen]

We probably should make some videos where we combine forces pushing at an angle while the system is connected over a pulley. We have them as separate examples now.

[@wrathandlaughter]

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]

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

[@MichelvanBiezen]

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.

[@kirbyrules99]

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

[@relaxingmomentswithnoxy4785]

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]

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

[@rollmosses1351]

This is comparatively simple. Mechanics is not that hard. Just have to be careful. Thank you sir, and please keep it up.

[@MichelvanBiezen]

That is a good way of putting it. "It is not that hard, but you just have to be careful". Great comment.

[@potatojam6519]

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

[@feriacientifica6139]

Very clear ! Congratulations and many thanks from Santiago, Chile

[@joelharris6158]

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

[@hesamnooraee7156]

your explainations are so clear sir . i became more powerful on solving problems . thank you so much sir

[@MichelvanBiezen]

Good for you. Keep up the hard work.

[@Gustavo-gy3dq]

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

[@MichelvanBiezen]

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)

[@II2023II]

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]

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]

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

[@ElectroWiz118]

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

[@MichelvanBiezen]

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]

@@MichelvanBiezen okay thank you that really helps

[@AlizKilla1]

Great video, easy to understand

[@el33tkrew]

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

[@MichelvanBiezen]

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.

[@franzguillermo2125]

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

[@MichelvanBiezen]

Great! Glad you found our videos! 🙂

[@lujainaakz1630]

+Michel van Biezen what if we have mk or ms and what m do you mean in the question?

[@MichelvanBiezen]

As a variable, m means "mass". As a unit m means "meter".

[@USMCRabbit]

if your mass on the ramp is larger than the hanging mass and this makes the mass on the ramp slides down the ramp instead of up the ramp, do you still do the acceleration the same? I have a problem that is this way and i keep getting a negative number for the acceleration. Would that be caused by the fact that instead of going up the ramp it is going down the ramp?

[@MichelvanBiezen]

I would call the direction of the acceleration positive down the incline and set Fnet = forces aiding the acceleration - forces opposing the acceleration.

[@USMCRabbit]

so if m1 is the mass on the incline and m2 is the hanging mass it should look like: (m1gsin-m1cosf-m2g)/(total mass)? (f=coefficient of friction)

[@MichelvanBiezen]

Yes but don't forget the g in the second term.

[@USMCRabbit]

thank you. I will try that out.

[@quamos99]

nice question

[@glecielopez8588]

I am confused of placing sine and cosine.

[@MichelvanBiezen]

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

[@yashpawarelitepawar]

+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

[@adegboyegasamuel342]

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]

We have several playlists with examples like that. See this video: kzbin.info/www/bejne/p6aXZ6uCabCDrLc in the playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS

[@diamondawg17]

Love to see professionals like yourself get frustrated at physics... thanks for leaving this in the video

[@MichelvanBiezen]

HaHa, yes, sometime the brain just locks up and the obvious is not seen.

[@herzikkimolog]

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]

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

[@themagesticunicorn7284]

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

[@MichelvanBiezen]

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]

​@@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

[@WonderingNobody]

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

[@sakshamarora1594]

Thanks a lot for the video!! Could you please explain, why is T=m2g-m2a and not T=-m2g-m2a?

[@MichelvanBiezen]

Note that we are calculating the magnitude of the tension (not the direction)

[@kritikasharma5436]

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]

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

[@jadidablomacmod3204]

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

[@MichelvanBiezen]

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.

[@umarani7468]

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

[@fightforfitness2256]

It's great. I have no other words to say !

[@stevekim5233]

When you solved for the acceleration, how come you didn't use -9.8 and you used 9.8? Do we not use the negative gravitational force when we do these problems? For example mg cos theta, mg sin theta, or w = mg and etc. thank you

[@MichelvanBiezen]

When solving these types of problems you take the magnitude of g. Note that the acceleration of the system (in the assumed direction) = (sum of the forces aiding the acceleration - sum of the forces opposing the acceleration ) / total mass.

[@Jaded_o]

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

[@aliciawilliams3762]

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

[@nicolledransfeldt6703]

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]

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.

[@jorgearocha6766]

hi Mr. Michael . congratz for your very helpfully chanel. i really like your work. im and arquitecture student. i would like to know how to calculate the mass "ma" without acceleration. thanks

[@bruceshimukayi3424]

This is an excellent explanation

[@MichelvanBiezen]

Thank you. Glad you liked it. 🙂

[@ibrahimabulwifa4296]

clear , easy . thank you for your explanation .

[@bhaktichokshi]

Won't the friction affect the tension of the inclined object?

[@MichelvanBiezen]

The concept of the tension is only applied to the strings (not the object). And yes, the friction force will affect the tension.

[@kelvinyuen3146]

hi, why isnt tension included in the equation?

[@MichelvanBiezen]

Tension is internal to the system. The system only accelerates due to a net force acting on the system.

[@BryannahNicole]

This video helped me so much! Thank you

[@peg1122]

Why dont masses cancel out in numerator and denominator

[@MichelvanBiezen]

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

[@melbudomo]

Now, i know angels do walk on earth. Thank you sir for guiding us, this is a tremendous help for us struggling students.

[@LilJollyJoker]

Really amazing jobs explaining it!

[@MichelvanBiezen]

Glad you think so!

[@Bayouassassin]

I understand the whole concept of finding the acceleration, but instead of finding acceleration how would you be able to find mass of m2 when acceleration is given instead.

[@MichelvanBiezen]

Peter, You would do the problem exactly the same way. Start with F = ma, then plug in all the forces that are aiding the acceleration (positive) and all the forces opposing the acceleration (negative). Plug in the acceleration and all the masses you know. Then solve for the one unknown mass.

[@josebertramtelmosojr.3147]

Im a fan professor, great teacher

[@MichelvanBiezen]

Thank you

[@MohammedWaseemAkram09]

sir u doing a fab work... u got struck fo a while... doesn't matters at all! i love the way u spreading knowledge!!! keep moving ahead... love u sir

[@bridgetngona1980]

l now understand better. THANK YOU Sir

[@hillaryjacobs8627]

I love this man.

[@passionatephysics3676]

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

[@crazythings1753]

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

[@MichelvanBiezen]

You are most very very very very bery welcome!

[@dineshdins2718]

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

[@MichelvanBiezen]

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).

[@khalidhuseyn7086]

something wrong with the audio? my left speaker is the only one working, great video regardless.

[@MichelvanBiezen]

Yes, our older videos were recorded in mono sound before we figured out what we were doing.

[@khalidhuseyn7086]

@@MichelvanBiezen Props to replying 7 years later haha

[@asuka-ryo]

I got a question where I need to find seperate acceleration for both masses 😭

[@MichelvanBiezen]

The magnitude of the acceleration is the same for both masses. Only the direction is different. For the block on the incline you have to separate the magnitude into x and y components.

[@asuka-ryo]

@@MichelvanBiezen Thank you so much, sir! I'll keep that in mind 👌

[@josetecson149]

is the coefficient of friction constant at all times in your figure even if the angle is increased?

[@MichelvanBiezen]

+Jose Tecson The coefficient of friction between two surfaces depend on the structure of the surfaces and not on the angle