Physics - Mechanics: Applications of Newton's Second Law (3 of 20) incline with 2 blocks

[@viuku0]

Thanks for all your hard work you’re still helping people 10 years later 🙏

[@MichelvanBiezen]

Thank you. We are glad that our videos are still helping students around the world. 🙂

[@thehrchannel8983]

Please don't ever delete your videos! I will need them to review for my exam! :)

[@sohailkhan-eu6uo]

i wanna like but u got 69 likes

[@karalgh643]

lol how did ur exam go

[@kylefink3929]

@@karalgh643 he probably got his bachelors already

[@frankdimeglio8216]

In carefully considering what is THE EARTH/ground, what is THE SUN, AND the fact that the stars AND PLANETS are POINTS in the night sky, we know that E=MC2 is CLEARLY and necessarily proven to be F=ma IN BALANCE; as ELECTROMAGNETISM/energy is gravity !!! (Gravity is ELECTROMAGNETISM/energy.) Consider what is the speed of light (c) ON BALANCE. Great. Now, very importantly, outer "space" involves full inertia; AND it is fully invisible AND black. Think. BALANCE and completeness go hand in hand !!! E=MC2 is CLEARLY AND necessarily proven to be F=ma ON BALANCE, as ELECTROMAGNETISM/energy is gravity !!! Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, as ELECTROMAGNETISM/energy is gravity; as E=MC2 is CLEARLY and necessarily F=ma IN BALANCE. By Frank DiMeglio

[@mooney003]

@@frankdimeglio8216 Hell y e a h. That's cool. Bad ass

[@3zoabdullah333]

your a savior to people like me who sleep during the university classes

[@MichelvanBiezen]

( As a student I used to fall asleep in many of my classes as well in college) 🙂

[@MichelvanBiezen]

LOL!

[@donnyltd6476]

His explanations are so intuitive, ahm almost scared to believe it could be that simple

[@TheHotOnes0]

You're the man!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! The only reason why I'm succeeding in Missouri S&T is because of you!

[@javirousgrant5020]

Great Job! You break it down so much easier than my physics teacher.

[@lwenjie16]

This channel is the reason why i pass my mechanics

[@rudrashiva]

Well some people are using KZbin the right way and some only to review stuffs, what is going on their life etc... This is what KZbin supposed to do, education is the strongest weapon, who knows there will be next great scientist in the making, who will solve the world greatest mystery and that is "time"

[@techguy9560]

You have set another ember for me in physics. My teacher destroyed it since the start of the semester. THANK YOU!

[@issahereuk3833]

everything I see videos like this, old and fun, I ask myself, isn't this like time traveling to the past? I mean it is insane how we get to see this 8 years after it is made when back then I didn't even know what physics was, today looking at this and questioning what this guy looks like today and what happened, why did he stop posting updates it is just insane how technology evolved thanks to all the ones who've seeked knowledge or is seeking it and thanks to the ones teaching and spreading it, just a wholesome world we live in you just gotta look around

[@MichelvanBiezen]

Thank you for your comment. We are actually still posting a video every day and have done so now for almost 10 years.

[@issahereuk3833]

@@MichelvanBiezen oh appreciated sir actually I'm new here, been watching a few videos but looking at the way you teach and explain giving all the possible shortcuts you are to remain undefeated with the high quality education you are offering, I wish you the best of life obviously one of the best teachers I've got taught by, hopefully I'll be here everyday throughout the journey

[@MichelvanBiezen]

Thank you and welcome to the channel!

[@kevintruong9699]

Thank you so much, you save my life. My professor is really sucked in explaining things

[@atirahwafa522]

i like the way you explain because it is more simple compared to what i have learnt in my university

[@jordanyong4814]

I really appricate the videos about mechanics you upload. They are first class, it gets into my head straight away! I wish you was my maths teacher... I was wondering if you could do a couple tutorials on how to do Moments (without vectors) such as calculating the magnitude of the turning effect if a force applied to a rigid body; problems about bodies in equilibrium and solving problems about non-uniform bodies, statics of a particle - such as a force acting on a particle on an inclined plane at and angle (not parallel to the plane which you have uploaded already). I hope you consider doing some more A-level maths tutorials. Many thanks.

[@MichelvanBiezen]

Toby, If you could give me an example or two of what types of problems you are looking for it will give me a better idea of what you are looking for. I will be doing more physics videos in the future, so it will help figure out what material to cover. Thanks.

[@MichelvanBiezen]

Toby, Many of the types of problems you are requesting are already there. Look in the 2 physics - mechanics play lists. I will continue to add to the list, but that will take a little while.

[@MichelvanBiezen]

Udin, There are other videos in the same playlist that cover inclines with friction

[@Markism07]

+Michel van Biezen really you wish you would draw a force diagram because i dont understand half the stuff youre talking about when you dont include it. im trying to draw one right now and im having trouble. :/

[@frankdimeglio8216]

E=mc2 is F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity IS ELECTROMAGNETISM/energy, AS E=MC2 IS F=MA. The Earth (A PLANET) is a MIDDLE DISTANCE form that is in BALANCED relation to the Sun AND the speed of light (c), AS the stars AND PLANETS are POINTS in the night sky; AS E=MC2 IS F=MA; AS TIME DILATION ultimately proves ON BALANCE that ELECTROMAGNETISM/energy is gravity. Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=MA; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=mc2 is F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. MOREOVER, a given PLANET (INCLUDING WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS F=ma, E=mc2, AND what is perpetual motion; AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. By Frank DiMeglio

[@frankdimeglio8216]

@@Markism07 E=mc2 is F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity IS ELECTROMAGNETISM/energy, AS E=MC2 IS F=MA. The Earth (A PLANET) is a MIDDLE DISTANCE form that is in BALANCED relation to the Sun AND the speed of light (c), AS the stars AND PLANETS are POINTS in the night sky; AS E=MC2 IS F=MA; AS TIME DILATION ultimately proves ON BALANCE that ELECTROMAGNETISM/energy is gravity. Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=MA; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=mc2 is F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. MOREOVER, a given PLANET (INCLUDING WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS F=ma, E=mc2, AND what is perpetual motion; AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. By Frank DiMeglio

[@satwikchakraborty7062]

you can preferably take g as 10 as it will simplify all your calculations and simplify the whole solving analogy too.

[@MichelvanBiezen]

Yes, but most books do not simplify like that, they use 9.8.

[@christianbonifacio6751]

Thank you Sir michel for the lesson you taught. Because of it, I was able to make it. I passed the ECE licensure exam.you are doing a lot of help to others.hope you will continue what you are doing.once again thank you sir!

[@MichelvanBiezen]

Great news! Thank you for sharing

[@SnoopyxX24]

my physics teacher suuuucks. glad I found you videos! such a big help

[@Andrea-uo5tx]

Hello! I hope you get to see this but thank you! Just got a full ride and your videos have been saving me! :)

[@MichelvanBiezen]

Awesome!

[@ataberdi2064]

Keep on doing, great meticulous explanations, i appreciate effort and your passion on teaching.❤

[@MichelvanBiezen]

Thank you! 😃

[@zaimnaqvi8893]

I wish i had you for my physics teacher!

[@ATMChilledMusic]

this guy is a god. my physics teacher makes this problem 10x more cancer and complicated using two separate systems instead of what this guy did, which was one.

[@giljorge7479]

Thanks!! Realy helped me understant these kinds of problems

[@MichelvanBiezen]

Glad it helped!

[@Garcia.Emilio]

I gotta Say i love your video. Very helpful for my physics course. hopefully i would apply it in my engineering career.

[@yashsharma4756]

I actually have one doubt , won't T1= minus mgsintheta + ma since it is in the opposite direction to the motion . Otherwise this was a great video , thanks so much

[@MichelvanBiezen]

No, that is not how you should look at it. Think of it in terms of what T does. It holds m1 against gravity AND T also accelerates it. Both terms are positive. You can also draw a free body diagram and use F = ma to get the same correct result.

[@violettamoclisa1846]

You are the best teacher!

[@check-inmedia3891]

Appreciate this, I've understood these problems easily

[@MichelvanBiezen]

You are welcome. 🙂

[@jailenerivera7649]

These videos are life savers!

[@loveyou75050]

At 10:00 Why are you adding the two forces? m1*g*sin theta is in the opposite direction of the motion. Thanks!

[@MichelvanBiezen]

The tension in the string is always equal to the force required to hold the body against the force of gravity PLUS the force required to accelerate it against gravity. (Only when the object accelerates downward (like with m2) will you subtract that term).

[@swagger2y]

You're such an awesome teacher

[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

[@lukehunter2362]

Don't you guys wish their was more time in the day to learn this subject more deeply? Its extremely intriguing when he teaches it

[@b0rdn-da-k1ng84]

sir you are an angel i cant thank you enough

[@bogdantureac4662]

You are an amazing teacher, beats my college professor for which I pay money

[@MichelvanBiezen]

Thank you. Glad you found our videos.

[@Alvarez0994]

shouldn't the parallel component @2:08 be m_1g/sin(30) ?isnt that component technically the hypotenuse of the triangle?

[@MichelvanBiezen]

Alex, If you slide the parallel component down so that it connects the tips of the vector mg and the vector mg cos(theta) then you have a triangle where mg is the hypotenuse and mg sin(theta) is the opposite side to the angle theta, which is the same angle as the angle of the inclined plane.

[@riyavarghese1292]

Hello sir, I was wondering what would happen if you cut the 10kg block?

[@talhaardc2687]

I can't understand when you are solving for T1. Isn't T1 and m1gsinQ oppusite direction,so why isnt T1=m1gsinQ-m1a?

[@MichelvanBiezen]

The best way to think about it is to look and see what T1 is doing. If nothing was moving or accelerating, (static situation) then T1 would just be keeping m1 from sliding and T1 would be equal to mg sin(theta). But T1 is also accelerating m1 up the incline and Newton's second law tells us that F = ma thus T1 must be increased by m1a. Thus T1 = mg sin(theta) + m1a

[@uwemkingsley8809]

Why would you put friction as zero? Does it mean there’s no friction in the system?

[@MichelvanBiezen]

In real life of course there is always friction. But we start problems like this in a more simple manner, by leaving off the friction forces (and yes there is no friction anywhere in the system). Later we add the additional complication of friction.

[@AndreyaChiwaya-ib4uj]

This is more helpful thank you..requesting for.more modified examples

[@MichelvanBiezen]

We have dozens of examples like that on Newton's second law applications.

[@vikramsrikishan5181]

Which video has this situation with friction?

[@PollyHarrison]

Hi! Am I right in thinking that you can essentially ignore the force NORMAL to the structure, m1*g*cos(30), ONLY if the coefficient of friction is nil when looking for F-net? Whereas if I had some friction, I'd need to include m1*g*cos(30)*friction in the relevant direction in my F-net calculation?

[@MichelvanBiezen]

The component of the weight perpendicular to the slanted surface is cancelled out by the normal force pushing back with an equal amount of force in the opposite direction (Newton's third law). However that said, it is used to calculate the friction force if there is a coefficient of friction between the 2 surfaces.

[@ronerickson8083]

Newton's Second Law of Motion needs a slight correction since they did not know about squaring a number in Newton's era we can assume that this may have been overlooked. As such (F=ma^2) is a natural correction that can be easily proofed through calculation, formulation, and experimentation. This correction is 33% more accurate if you do not accept it proof it yourself.

[@Love-vx4ld]

This is amazing sir, keep it up!

[@MichelvanBiezen]

Glad you found our videos. 🙂

[@zuhairalsaffar7001]

I like your way of teaching with my respect and thanks

[@lmarcel]

what happens if the coefficient of friction isn't 0 but let's say like, 0.15 or something like that

[@MichelvanBiezen]

We have examples in this playlist that show that with and without friction.

[@determinedgenerationtv5291]

sir that way you were using to find acceleration of the system is not so effective because its differing from when you find the acceleration by resolving the forces separately

[@MichelvanBiezen]

We have examples of both methods. I find this method to be the most effective and easy to learn for the students. .

[@udinanuar7202]

How to solve the overall questions if coefficient of friction is not equal to zero but equal to 0.2?

[@sukhmanisuri5272]

What would you do if you were solving for acceleration but the coefficient of friction was unknown?

[@MichelvanBiezen]

That would depend on what else is known / given.

[@billarwas8631]

Thank you so much for this video - incredibly helpful and clear

[@jesussaquin6266]

Guys quick question will I learn this in statics or mechanics and lab???

[@MichelvanBiezen]

You will learn this in a physics mechanics course.

[@jesussaquin6266]

@@MichelvanBiezen mechanics and lab touches on this then statics expands on it? There are two courses at my school one is mechanics the other is statics

[@MichelvanBiezen]

That sounds like two mechanical engineering classes. (typically statics comes first and then dynamics).

[@nimshichiwila7852]

I don't know how can I put it but its nice thing and we thanking you because its when helping revising .

[@swatiparida153]

Thank you so much for such explanations ......u made it really easy for me man.....

[@hayasakakazuha4144]

Sir, I hope you can help me with my problem. How can I find the slope angle if the only given in the problem are the weight of the two boxes?

[@MichelvanBiezen]

They should have also told you what the boxes are doing (moving, staying still, accelerating) and if there is friction or not.

[@hayasakakazuha4144]

@@MichelvanBiezen the box on an inclined plane is being pulled at a constant speed up a frictionless inclined plane by the box hanging on the chord, sir.

[@hayasakakazuha4144]

@@MichelvanBiezen A 30N block is pulled at a constant speed up a frictionless inclined plane by a weight of 10N hanging from a chord attached to a block and passing over a frictionless pulley at the top of the plane. Find a) the slope angle of the plane, b) the tension in the chord, c) the normal force on the block by the plane.

[@MichelvanBiezen]

If they are moving at a constant speed, there is no acceleration and therefore there is no net force. The weight of the hanging block (M2g) is equal to the component of the weight of the block on the incline along the incline. M1g sin(theta) M2g = M1g sin(theta) sin(theta) = M2/M1 and theta = sin^-1 (M2/M1)

[@MichelvanBiezen]

theta = sin^-1 (10/30)

[@carlo5665]

Sir one doubt how can there be two diff tension in the same wire

[@MichelvanBiezen]

In the example in this video, the tension is the same throughout the wire. But if the pulley had mass or some resistance to its motion, then the tension on both side of the pulley would be different.

[@carlo5665]

@@MichelvanBiezen really appreciate that and i will looking forward for more of your videos

[@irwinisraeltomas7097]

what if they exchange their given mass? can i say m1g - m2g? because my 'a' positive is going to the left and my 'a' negative is going to the right?

[@MichelvanBiezen]

+irwin israel Tomas I am not sure what you mean by "what if they exchange their given mass?" If you put m2 on the incline and m1 is hanging from the rope, you have a very different problem. You would have to do more than just interchanging m1g and m2g. I always recommend that you don't make those types of "quick" guesses, but work out the problem as stated in a systematic fashion as shown in the example.

[@yao9346]

When trying to find the tension, why don’t we consider the total mass since the m in ma is m1+m2? Can I just use one single mass? I thought they accelerate as a system so we can’t separate them...

[@MichelvanBiezen]

This playlist will shed some more light on that: PHYSICS 4.8 FREE BODY DIAGRAMS

[@gurash2419]

hello respected sir its excellent lec. but can u make little more clear screen tq

[@thounaojamanilsinghluwang1464]

Sir can you plize add for finding individual acceleration for each masses

[@MichelvanBiezen]

We have lots of examples like that in the playlist. (See the home page).

[@thounaojamanilsinghluwang1464]

Michel van Biezen thank you very much sir for your time these videos are very helpful to clear basic concepts thank you very much sir

[@bollyfan1330]

Interesting question would be if the wedge was massless and sitting on a frictionless flat surface. The 10 kg mass touched the vertical face of wedge. There was a coefficient of friction of mu for each of the masses with the wedge. How would system move? Compute accelerations.

[@bollyfan1330]

m1 a_rope sin(theta) = (T - mu N1) sin(theta) - m1 g + N1 cos(theta) ----------- FBD m1 in vertical direction, a_rope = acceleration in direction of rope pull (up and right) m1 (a_rope cos(theta) + a_wedge) = (T - mu N1) cos(theta) - N1 sin(theta) ----------- FBD m1 in horizontal direction, a_wedge = acceleration of wedge horizontally (right) N1 sin(theta) + mu N1 cos(theta) = N2 + T cos(theta) ----------- FBD massless wedge in horizontal direction, "T cos(theta)" term comes from pulley N2 = m2 a_wedge ----------- FBD m2 in horizontal direction m2 g - T - mu N2 = m2 a_rope ----------- FBD m2 in vertical direction, a_rope = acceleration in direction pulling rope (downward) There are 5 variables (T, N1, N2, a_rope, a_wedge) and 5 equations, so you can then solve for a_rope and a_wedge to know how system will move. If you want to compute the effective weight of the system on the flat surface, which equals Nw (normal reaction of wedge on flat surface): Nw = N1 cos(theta) - mu N1 sin(theta) + mu N2 + T sin(theta) + T ----------- FBD massless wedge in vertical direction, "T sin(theta) + T" term comes from pulley

[@thailandfutsal5508]

Why we can assembly it in one equation cause force that act on the system is not the same axis

[@MichelvanBiezen]

Because we are only looking for the net force acting on the syatem along the line of motion of the system

[@thailandfutsal5508]

But force is in the different axis

[@MichelvanBiezen]

The pulley changes the direction of the string. So we consider the forces ALONG THE DIRECTION OF MOTION FOR EACH PART OF THE SYSTEM

[@ihnnojeffersonsurigao469]

How do you know if it's a sinθ or cosθ?

[@MichelvanBiezen]

By using the definitions: sin = opposite side / hypotenuse cos = adjacent side / hypotenuse

[@catherineb.merchant4632]

Hello Michael. Looking at sin 30 degree which is located in quadrant three, I thought we should have had negative 0.5 instead of positive 0.5 which will make the acceleration 7.62 instead of 3.27. PLEASE PROVIDE CLARITY. Thanks..

[@MichelvanBiezen]

The signs of the forces are referenced to the assumed direction of the acceleration. + for the forces aiding the acceleration and - for the forces opposing the acceleration.

[@catherineb.merchant4632]

THANKS!

[@ww8720]

what about if there is a spring at where m2 is?

[@MichelvanBiezen]

It depends on the question. It can be a simple harmonic motion problem or a static problem, again depending on how it is worded.

[@katyushah.beilschmidt6705]

why is the sin and cos there ( i know trig but I just don't understand-isn't the velocity going down the slope the adjacent?)

[@engineeringfamily6187]

we don't get u so please explain what u want to be told clearly

[@ravibharti8751]

Sir,why you not included T1 and T2 as the part of the net force.though you have taken m1g sine theta and m2g why not T1 and T2

[@MichelvanBiezen]

Those are internal forces to the system and do NOT contribute any force to the acceleration of the SYSTEM

[@ravibharti8751]

lot of thanks for replying.Sir,but T1 and T2 holding the block together if it would have not been it just fall with acceleration m1g sine theta and m2g.

[@MichelvanBiezen]

Correct, the strings holds the block together making into a single system such that the blocks move at the same speed and have the same acceleration. That is why we can solve it as a single system.

[@nguonsokunthea4600]

Awesome!!! I really enjoyed your lecture.

[@bhaktichokshi]

What should I do if there is friction?

[@MichelvanBiezen]

There are lots of examples in the playlists where there is friction as well.

[@michaellim4125]

if there's a moment of intertia(bc of the pulley) do i have to add the mass of the into the total mass? c

[@Hiep156]

Hey, what happens if both of these objects have the same mass? Does it mean that the acceleration is zero? And does it mean that the speed is also zero?

[@MichelvanBiezen]

No, because one is on the slope and one is hanging. If they had the same mass the block on the slope would still be accelerated upward unless the friction is too large.

[@Hiep156]

@@MichelvanBiezen Let's assume there is no inclined plane just a normal table with horizontal direction, then will the acceleration be zero (both objects with the same mass)? Also, if we take another example where the object on the table in horizontal direction has bigger mass than the one which is hanging on the right side of the system, then which direction will the whole system go, to the right or to the left? Just consider the pulley to be massless.

[@multiakc9232]

thankyou very much for associatig us...........may god live u long ameen

[@DubKicker101]

This man is a genius.

[@9031olivia]

My professor asked a similar problem, but after solving for tension and acceleration he wants us to find the position of block one when it stops sliding up the ramp after block two hits the ground. I'm looking through the 5 kinematic equations, but I'm getting stuck on all of them. :(

[@MichelvanBiezen]

+Olivia Wentworth You can solve that in two ways: 1) use conservation of energy 2) use equations of kinematics. After the hanging block hits the ground it no longer pulls on the string. Then find the net force on the block on the incline, use that to find the acceleration and then use the equation: V^2 = Vo^2 + 2*a*x

[@rehanaaktar8145]

Love your tactics sir

[@kayzzchilu8686]

Thanks Sir am really understanding the concept here especially on the part of tension... thanks

[@thepunisher1951]

Very nice video! Exam helper!

[@user-if9od1zs4b]

Sir,May I ask some question?If the question state μs=0.50.That means,there is friction force exist,right?

[@MichelvanBiezen]

Yes, and the "s" means static friction

[@user-if9od1zs4b]

@@MichelvanBiezen thank you,sir.

[@nemesis6636]

Man, I wanna ask one thing, are these university level questions in America??

[@MichelvanBiezen]

Both high school and university. The difference is that the problems are more difficult at the university level.

[@nemesis6636]

Ohh. Sir actually, I am preparing for JEE advanced. Could you suggest me on how can I approach difficult questions as I have tried to solve them, I have succeed 4 out of every 10 questions. And those questions are just too out of the box. For reference, you can google 2020 iit jee advance paper.

[@MichelvanBiezen]

Did you find our JEE advanced test question videos? You can find them from the home page.

[@nemesis6636]

@@MichelvanBiezen Thanks a lot sir. Will help me a lot for my exam.

[@sadiatahir3825]

Why you are not considering tension?

[@MichelvanBiezen]

Tension is a force INTERNAL to the system and thus does not add to the acceleration. Only EXTERNAL forces acting on the WHOLE system affect the acceleration.

[@sadiatahir3825]

Okay Sir Thank you. You are The best.

[@j0mezzy]

Since we know T1=T2 Can we solve T1 and T2 then set them equal to each other to find acceleration ? I tried it but I got an extremely large Acceleration, doesn't seem right...

[@MichelvanBiezen]

+Jomel Sagsagat No, you can't solve it that way.

[@j0mezzy]

Why is that

[@laetitia7402]

Thank you so much for your videos.🙏🏽

[@albertgan2378]

What if you're asked for the m2 given m1 kinetic friction and a theta

[@MichelvanBiezen]

+Tatsuya Shiba I am not sure what you are asking. Are you asking what would happen if you switch m1 and m2?

[@albertgan2378]

+Michel van Biezen thanks i found the answer , i watched your other videos :)

[@doodelay]

Why is it Fnet/8kg + 10kg and not Fnet/(8kg(sin30) + 10 kg)? It seems that the mass of 8kg would be reduced by the angle of incline.

[@carultch]

The mass is not reduced by the incline. Only the effect of the weight is reduced due to the incline. The weight still acts 8 kg mass in its full amount, but part of the weight is supported by the normal force from the incline. The inertia never ceased to exist, and the whole 8 kg mass still has to be accelerated.

[@meryemnuriyeva2591]

if there is friction how would change tension force?

[@MichelvanBiezen]

+Meryem Nuriyeva There are many examples in the videos with and without friction.

[@meryemnuriyeva2591]

ok i will watch. thank you)))

[@anastasiaclaude8256]

Is T1 always equal to T2?

[@MichelvanBiezen]

Yes, IF the pulley has no mass and no friction. (see the moment of inertial examples where the pulley has mass and therefore the tension is not the same on both sides of the pulley)

[@anastasiaclaude8256]

@@MichelvanBiezen I understand.Thank you so much

[@thailandfutsal5508]

Why we don’t need to plus it likes vector

[@MichelvanBiezen]

There are different approaches. In one approach you can draw a free body diagram around each component and determine the vectors, and then add them (solve them simultaneously), or you can take the whole system together as shown in the video.

[@irwinisraeltomas7097]

thanks but i understand it now because of your other vid

[@arush366]

Which grade physics is this in the US?

[@MichelvanBiezen]

Either 12th grade in high school or first year in college.

[@mosesnamupala6392]

This video helped me a lot. Thank you.

[@gopikrishnabachina9232]

just use the formula T= [m1m2/(m1+m2)]*(1+sin30)*g

[@thailandfutsal5508]

Why do we don t think tension in rope

[@MichelvanBiezen]

The tension in the rope is an internal force to the system and does not affect the acceleration of the system.

[@thailandfutsal5508]

Hello ,thank you

[@ibrahimsahak1696]

Thank you sir , helpful for my math exam

[@MichelvanBiezen]

Happy to help

[@mojgandehghani6969]

Good job! Thank you Michel!

[@gerardleotaud9909]

why did m1cos30 cancel out?

[@MichelvanBiezen]

Gerard, Because it is directed perpendicular to the incline and is thus cancelled out by the normal force (equal in magnitude and opposite in direction).

[@zeyad544]

Because it equals the normal force .. which make no acceleration.

[@MopTopRock]

What if mass in both blocks are same?

[@MichelvanBiezen]

The acceleration would be smaller and the tension in the strings would be less. Try to work out the problem with the two masses being equal and see what you get.

[@sinethembamashiloane9908]

Yo thanks you I really needed this

[@MichelvanBiezen]

Glad it helped.

[@nouraa4874]

YOU ARE THE BEST !! THANKS A LOT ()"

[@amandavermette7183]

How/where did he get 9.8m/s squared?

[@MichelvanBiezen]

That is g = 9.8 m/sec^2 which is the acceleration due to gravity.

[@오서혁-b3t]

i just wondered that u wrote a=Fnet/Mt. my question is why did u use mg - mg theita. actually, Fnet is mgcos theita... I'm just confused...

[@MichelvanBiezen]

There is Fnet in the vertical direction: N = m2g cos theta and then there is Fnet in the direction of motion parallel to the incline Fnet = m1g - m2g sin theta

[@ihavetwoos]

Timer, danko is thank you 🙏🏿

[@MichelvanBiezen]

You are welcome.

[@graceb2404]

Forever my hero

[@11emergencymerger68]

This lesson name is Pesawat kerja..

[@roshanmujeeb]

This question was easy. Do tougher ones

[@surendrakverma555]

Very good 🙏🙏😊😊

[@MichelvanBiezen]

Thanks a lot 😊

[@TheArtofEngineering]

I truly love you!!!!

[@MichelvanBiezen]

Hi, I looked at your channel and it looks really interesting. You have a lot to offer. Do keep it up. I remember when our most watched video hit 100 views, and we celebrated. Keep it going.

[@TheArtofEngineering]

@@MichelvanBiezen 🙏 yes I’m presently in my first year of Engineering (full time) will graduate (if I make it) at retirement age!😊). I taught Engineer studies at high school for the last 2 yrs and your videos were so helpful when I had to work out truss analysis. Now the videos are assisting with Physical Modelling 1! You are a great lecturer. Thanks for the encouragement.... the plan is to create (when time permits) STEM resources for middle and high school. Cheers from Oz!