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

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In this video I will show you how to calculate the acceleration and tensions of 2 objects around a pulley on a wedge (One hanging off a pulley and the other sliding on the wedge.)

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@@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

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[@MichelvanBiezen]

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

[@MichelvanBiezen]

LOL!

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

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

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

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

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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.

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This channel is the reason why i pass my mechanics

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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

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Good job! Thank you Michel!

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I like your way of teaching with my respect and thanks

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This video helped me a lot. Thank you.

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

[@laetitia7402]

Thank you so much for your videos.🙏🏽

[@kayzzchilu8686]

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

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I wish i had you for my physics teacher!

[@kevintruong9699]

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

[@rehanaaktar8145]

Love your tactics sir

[@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

[@ataberdi2064]

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

[@MichelvanBiezen]

Thank you! 😃

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You are the best teacher!

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

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Awesome!

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Very nice video! Exam helper!

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Thanks!! Realy helped me understant these kinds of problems

[@MichelvanBiezen]

Glad it helped!

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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"

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[@MichelvanBiezen]

Glad you found our videos. 🙂

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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!

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

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[@MichelvanBiezen]

You are welcome. 🙂

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[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

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This man is a genius.

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[@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.

[@subhojitchatterjee1102]

very very helpful

[@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.

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Thank you sir , helpful for my math exam

[@MichelvanBiezen]

Happy to help

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life saver...

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Yo thanks you I really needed this

[@MichelvanBiezen]

Glad it helped.

[@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

[@Mathias6191]

THANK YOU

[@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.

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I truly love you!!!!

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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!

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U r the best

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Excellent. Thanks.

[@MichelvanBiezen]

Glad you liked it!

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love the videos. love the bow tie

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Love you for this

[@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).

[@riyavarghese1292]

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

[@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.

[@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.

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Forever my hero

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This lesson name is Pesawat kerja..

[@bidhatatripathi9956]

thank you so much!

[@MichelvanBiezen]

Glad it helped!

[@vikramsrikishan5181]

Which video has this situation with friction?

[@udinanuar7202]

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

[@physics67]

Thank you so much

[@MichelvanBiezen]

You're most welcome 🙂

[@liyuenmarajh]

Perfect thank you so much

[@MichelvanBiezen]

You're welcome 😊 Glad you found it helpful.

[@ibraahimyuusuf7005]

thank u

[@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

[@hedayetcareerinstitute2694]

Thanks sir

[@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.

[@ranajerjawi7690]

Ily thank u

[@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.

[@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

[@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

[@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.

[@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

[@mossesgregorie828]

you do well

[@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.

[@irwinisraeltomas7097]

thanks but i understand it now because of your other vid

[@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!

[@hasan14598]

Lovely❣️

[@MichelvanBiezen]

Thank you

[@bogdantureac4662]

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

[@MichelvanBiezen]

Thank you. Glad you found our videos.

[@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.

[@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

[@surendrakverma555]

Very good 🙏🙏😊😊

[@MichelvanBiezen]

Thanks a lot 😊

[@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.

[@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.

[@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. .

[@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.

[@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.

[@jabirmohammed2598]

Think you

[@MichelvanBiezen]

You are welcome.

[@physicsloverjoinmychannel2937]

But What About Is There1, Friction 2. The Third Ground MasS How We Calculete Shuch Like Question At Last But Not Least Thanks

[@MichelvanBiezen]

There are dozens of examples on this channel with and without friction in all kinds of different situations. They can all be found from the home page.

[@physicsloverjoinmychannel2937]

@@MichelvanBiezen ok can u put the link

[@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

[@roshanmujeeb]

This question was easy. Do tougher ones

[@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

[@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)

[@bhaktichokshi]

What should I do if there is friction?

[@MichelvanBiezen]

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