Physics - Mechanics: Applications of Newton's Second Law (17 of 20) video has error, will be remade

[@MichelvanBiezen]

I appreciate the comment. Glad you like the videos.

[@richardwcy1995718]

Prof Michel, you have no idea how much you've helped me through physics! Best tutorials for physics! very clear understanding!

[@MichelvanBiezen]

Richard, Thanks for letting me know. Good luck in your studies.

[@richardwcy1995718]

I am reviewing all the physics videos right now! this is helping so much for the final in a few days

[@frankdimeglio8216]

Michel van Biezen is a superior and an honest instructor. Very highly recommended. Bold and confident. Super hard worker. Understands top down thinking in physics.

[@MichelvanBiezen]

Wow, thank you!

[@joeyborja423]

Oh this is I would call a sad ending in the series. Looking forward to the remake. Thanks! More power and stay safe!

[@NM-zb6pd]

Acceleration relations: a1=a3+ay2(vertical) & a3=ax2(horizontal) for m2 to not move down ay2 = 0 so a1=a3 Force equations: 1. F-R=m3a3 2. R=m2ax2=m2a3 3. T=m1a1 4. m2g-T=m2ay2=0 since ay2=0 so T=m2g 5.m1a1=m2g from 3 & 4 so a1=m2g/m1 6. F=m2a3+m3a3=(m2+m3)a3 so a3=F/(m2+m3) 7. F/(m2+m3) = m2g/m1 from 5 & 6 & since a1=a3 =>F=m2(m2+m3)g/m1=5(5+20)9.8/5=25*9.8=245N

[@physicshacks6349]

T= m1a T= m2g Dividing = a=( m2/m1)g Fnet= total mass × acceleration = ( m1 + m2+ m3) × ( m2 g/m1) = (20+5+5)× ( 9.8) = 30×9.8 Newtons = 294 newtons

[@inishasubedi7470]

Professor Michael I passed physics exam through your videos, Your videos are so helpful, I was searching for the corrected viedeo no 17 of newtons application

[@MichelvanBiezen]

Glad we were able to help. I haven't had the time to reformulate this video. Thanks for the reminder.

[@garethm3171]

Looking at the Physics - Mechanics playlist videos 18,19 & 20 are missing.

[@stevekhan4063]

. So couldnt you perhaps find acceleration by T=m1a T=m2g T=T m1a=m2g m1 and m2 same mass cancel out therefore a=g 9.8 insteas of 4.9 ?

[@angadkhurana1866]

Sir why can't you solve it this way? When m1 and m2 are not moving with respect to the pulley then m2g=T(tension in the pulley) And on m1 since T is acting towards the right some force must act to the left of m1 counteracting it so a force equal to T should act on the left of m1which is equal to m2g= 5g. Now we know that force/mass=acceleration so the acceleration of m1 should be 5g/5 which is g=(9.8) and as the acceleration of m1 =m2=m3 the net force should be(m1+m2+M3)*a = 294 N Can you please tell where am I going wrong

[@shlokdave6360]

You are getting your reference frames mixed up. M1 and M2 cant be at rest wrt pulley in an inertial frame because of gravity. The entire system is being accelerated. One way to approach this with newtons second law is to find the acceleration which can negate the one of M1 under normal circumstances in the rest frame. For that first calculate the natural acceleration of a wrt a rest frame and just give the entire system of M1+M2+M3 the same acceleration. That will make the effective acceleration of M2 and M1 wrt M3 to be zero in the rest frame. If you want to consider the pulleys frame of reference, please remember that it has to be under acceleration as a whole for M1 and M2 to be stationary wrt the pulley. Either approach leads to the same answer.

[@you2tooyou2too]

You also have to accelerate M2 at the same rate horizontally as M1 & M3, but only after M3 moves to make contact with M2 ! Which would beg the Cf(M2,M3), except the slide of M2 is also 0), so M2 must eventually also be accelerated horizontally. He got there, but w/o mentioning the required delay step in the force.

[@NAxxen18]

That was very helpful , I'm a jee aspirants

[@MichelvanBiezen]

You will find these type of trickly problems on the JEE tests. 🙂

[@Rohitkumar-yo8js]

sir.please correct me if I am wrong,here for calculating accleration you took m1 and m2 as a system ,but we cannot take that as a system because accleration of m1 is horizontal direction an m2 is in vertical direction..the accleration will be m2*g/m1.

[@MichelvanBiezen]

you can call them one system together if you only calculate the magnitude of the acceleration.

[@lacceybird]

Hi I think the acceleration of the system is 9.8 m/s^2 because we know F=ma and this must balance the tension pulling on m1 so that m2 is not accelerating down and if m2 is not accelerating downward we know that tension must equal m2g so F=m1a can be set equal to T=m2g m2g/m1 = a and the masses cancel out so you get 9.8m/s^2

[@MichelvanBiezen]

The answer in the video is correct.

[@Steven-ff4wl]

@@MichelvanBiezen I'm sorry, it is not. 9.8 m/s^2 is the correct answer.

[@charlesbromberick4247]

I think for this problem to make sense, mass m2 MUST rest without friction on the side of m3 such that the tension in the string is the pure weight of 49, and sideways acceleration force is transmitted through the mass interface and NOT the string; and then my answer would be 30 x 9.8 = 294. (If it doesn´t rest on the cart, the tension becomes the vector sum of 5a and 5g or 5sqrt(aa + gg), and equating this to 5a implies g=0 and the problem becomes trivial.)

[@MichelvanBiezen]

I re-evaluated this problem and I found that it cannot be solved as is shown. We will make a new one with slightly different masses to make it work.

[@charlesbromberick4247]

@@MichelvanBiezen Great - I´ll look forward to the modified version. (I´m betting your idea will work out beautifully, with the top mass greater than hanging mass and suffient space so it can hang at an angle.)

[@charlesbromberick4247]

@@MichelvanBiezen Just for the fun of it, I made two small modifications to your problem: I made the hanging mass 3, and I made the rope over the pulley short so the mass wouldn´t hit the cart. It worked outreal well, and I got a force of 205.8 newtons

[@MichelvanBiezen]

Yes, that is the way the problem can be solved. We plan on shooting the video this weekend.

[@charlesbromberick4247]

@@MichelvanBiezen Fantastic - I think it will be a really neat problem - thanks

[@шавкатнаджметдинов]

Sir you are really helpfull i love your videos!

[@sparsh684]

Why did we include the mass of m1 ? Wouldn't there be double the force ( both the tension and the applied force) acting on it causing double the acceleration in mass m1. Sir, could you elaborate what you mean by "If Vy = 0 then the Fnet accelerates all three masses". I have searched the entire net and haven't found anything.

[@MichelvanBiezen]

If the two smaller masses do not accelerate relative to the third large mass on wheels, then all 3 masses will act as a single system. Note that there are a number of examples of these types of problems, starting with simpler ones that will help you understand this concept.

[@sparsh684]

What would be the free body diagram of Mass M1? In the x-axis it already has the force of tension accelerating it at 4.9m/s2. Since it is already ACCELERATING wouldn't we need to accelerate only mass M2 and M3 so that all have the same acceleration of 4.9 m/s2. Where should I look to understand this concept much clearly ? I have basically watched most of your videos and this is the only one I am having problem with. Any help is appreciated.

[@ahsanali9931]

@@MichelvanBiezen please sir explain why m2 is non inertial reference frame?

[@charlesbromberick4247]

Very nice analysis of a very intresting problem. Tks

[@MichelvanBiezen]

Glad you liked it! Still need to redo this problem. I suspect there is something wrong, but I haven't had the time to analyze it.

[@charlesbromberick4247]

@@MichelvanBiezen I think the push force is maybe only acting on two of the masses, the one on top being stationary with respect to outside world and the other two accelrating to the right.. I too am uneasy, but still love the problem.

[@charlesbromberick4247]

I´ll give it more thought as I try to fall asleep tonight - but I´m thinking it might be an impossible situation.

[@MichelvanBiezen]

From a conceptual perspective, if the system can be pushed hard enough so that the acceleration of the car equals the acceleration of the two blocks on the pulley, the hanging block will not go down. But I'll have to give is more thought.

[@charlesbromberick4247]

@@MichelvanBiezen jajaja - who says physics isn´t fun?!

[@maxmax0]

If the whole system has the acceleration pointing to the right, then, if looking at the m2 alone, there must be a force acting on m2 horizontally pointing to the right. Where does such a force come from? The only way to make sense of it seemly is to assume the pulley is small enough so that m2 is hanging vertically without a gap to the m3, namely m2 and m3 touch with each other.

[@MichelvanBiezen]

I have to rework this problem. Had not had the time yet.

[@maxmax0]

@@MichelvanBiezen Thank you Sir. It's a great video and problem. Looking forward to the reworking.

[@vaibhavbangia]

SIr, why are we adding the mass of m2 when calculating the force required? Why are we not adding the mass m3 and m1 only?

[@MichelvanBiezen]

Because the net force required to accelerate m1 and m2 can only be provided by the weight of m2.

[@Jallah-Tafari]

@@MichelvanBiezen Sir, your videos are awesome! Nonetheless, I also wonder why the inertia that must be overcome isn't solely in m1 and m3, because m2 isn't sitting on top of m3. Maybe I'm overlooking something.

[@Jallah-Tafari]

Here's another point of view: If m1 were isolated on a frictionless surface, if that surface began to move, friction is the only force that would cause the m1 to move; however, without friction, wouldn't the surface beneath slide without resistance? If that's the case, then wouldn't the force needed to keep the surface underneath m1 " up to speed " be ( m3 )( a )? It's been a longggg time since I've worked a problem like this, so I could very well be overlooking something.

[@nuavecmoi]

Any videos on gears?

[@MichelvanBiezen]

This is the only one related to gears: kzbin.info/www/bejne/r5KrdqGqgc-goc0

[@antoniussupriyantaanton5847]

Nice .whit you...is very good...ia understand...fisika

[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

[@sandeepkamboj413]

it's confusing ..there is no use of FBD. suppose M1 is kept on M3 and there is no string and friction between blocks is zero ..when we apply force on M3 only M3 will move with the acceleration F/M3..same thing applies here acceleration must be F/M3.if I am wrong please enlighten me.

[@MichelvanBiezen]

We need to redo this problem.

[@brandonschmidt8321]

If the acceleration is 4.9, then the net force on m1 would be 24.5 N. This would mean that the tension is 24.5 N. If the tension is 24.5 N on m2, then m2 would experience a net force downward and therefore a downward acceleration. No?

[@omaramanullah6197]

Brandon Schmidt that's exactly what I'm thinking.

[@sanskartiwari2996]

that was hell awesome! sir do you teach anywhere else or do you just lecture online

[@MichelvanBiezen]

Yes, I teach at a local college and a local university.

[@wasafitvonline9248]

sorry sir... I'm advanced science student and you contribute almost 90% on my physics subject.... but I can see here in Mechanics you didn't discussed about two things 1.coefficient of restitution 2. inelastic collision 3. rocket in space if you already discussed about them... please locate me sir

[@MichelvanBiezen]

We do now. All these topics are now included in the physics playlist.

[@procrastmh]

Sir, if there is no friction between m1 and m3, wouldn't m2 just be freefalling? considering m1 and m2 have the same mass

[@procrastmh]

which leads me to thinkinh that the acceleration is 9.8 and not 4.9, but one of your videos with this case (where the object seems to be freefalling) also has 4.9 as the answer

[@procrastmh]

I get it now. there is no net force acting on m1

[@procrastmh]

even if m1 has mass there is still no friction between m1 and m3

[@MichelvanBiezen]

That is correct.

[@Sdsisdsi]

Respected SIr If there is no friction between M1 and M3 then force F applied on M3 cannot have any affect on M1. Force of friction is a must to cause the motion of a body stacked on top of another. So i believe that this solution is not correct

[@dhruvg550]

And if m2 was sticking to to the trolley as well, would there be any changes? Assuming there is no friction.

[@MichelvanBiezen]

If there is no friction between m2 and the car, then there would be no difference.

[@VibhaMasti]

Why do we consider Fnet acting on all three masses when it is accelerating only the 20kg mass?

[@MichelvanBiezen]

If Vy = 0 then the Fnet accelerates all three masses.

[@Avatar7249]

plz sir , why the solution of this problem in reference of Serway is different ?? I'm so confused .. I wish you reply . Thank you , prof

[@MichelvanBiezen]

Is the problem EXACTLY the same? Which part of the solution does not agree?

[@Avatar7249]

Michel van Biezen only numbers are different ... but there's no acceleration for m2 as it doesn't go down .. I appreciate your effort, sir .

[@diamondglitter205]

if we apply acceleration bigger than 4.9 would m1 move to the left and lift m2

[@MichelvanBiezen]

diamond glitter Interesting question. It seems counter intuitive, but yes.

[@mdsrmunim4633]

Sorry sir you are wrong. In first portion while determining acceleration of two 5 kg block, you have used newton's law in a non inertial reference frame..... This problem can be solved by the concept of pseudoforce and correct answer is F=294 N. thanks

[@MichelvanBiezen]

You are correct. I made a wrong assumption. The force is indeed 294 N. Thanks for your comment. We will reshoot the video.

[@mdsrmunim4633]

@@MichelvanBiezen thank you for replying sir... I have learnt a lot from you.... 😊

[@ahsanali9931]

Please tell me why m2 is non inertial reference frame?

[@ahsanali9931]

@@mdsrmunim4633 please bro help me why m2is non inertial reference?

[@mdsrmunim4633]

@@ahsanali9931 for acceleration

[@KaptainTech]

Excellent work!

[@rexrivera8821]

sir, why is that the acceleration of M1 and M2 will be the same to the acceleration of the whole system? i thought they're different?

[@MichelvanBiezen]

Only the magnitudes are the same. (they do travel in different directions). But the magnitudes have to be the same since they are connected by the same string.

[@rexrivera8821]

+Michel van Biezen ah! so, it means, same magnitude same acceleration sir?

[@MichelvanBiezen]

Only if we are just calculating the magnitude only. If you want to express it as a vector quantity, you'll have to add the direction as well.

[@rexrivera8821]

+Michel van Biezen okay sir. thanks alot :)

[@pensivist]

Was this exercise ever remade?

[@MichelvanBiezen]

Not yet. Thanks for the reminder.

[@erc333]

I don't completely see why you can ignore the tension on m1...

[@ClumpypooCP]

tension in m1 is equal and opposite to the tension in m2. Therefore, they cancel each other out and you can ignore it.

[@stevekhan4063]

Im curious of what would happen if M2 is at an angle ?

[@nicholaslau3194]

steve khan Derive perpendicular forces of M2 (m2g sin(angle)) this would be the force required to put M1 and M2 at rest. The force will be less than 147N

[@stevekhan4063]

no its actually more i just did it a few days ago..

[@MrStoryteller_me]

18, 19, 20 are missing :(

[@MichelvanBiezen]

+Shubham Chauhan Oh yes. I only did 17 and never went back to renumber them.......

[@MrStoryteller_me]

Sir then will you upload others also.. Please

[@vishnuthiru4535]

kzbin.info/www/bejne/oqKkhYiHZ7lkoNE

[@ibrahimnazm]

ده‌ست خۆش

[@mostakimrabbi538]

sir 18, 19, 20 ?

[@thailandfutsal5508]

Why f=m1+m2+m3 *a

[@nickliii1i1i1]

this one is too hard to imagine in reality.

[@physicshacks6349]

This is a concept of pseudo force