Even though my solution was't right but I realized a very important thing after watching this solution and that is physics is not pure mathematics,I need to think way deeper while doing physics than while doing algebra, the intuition(which didn't arrive to me while doing the problem) that Frictional force between the right set of two masses will become maximum first literally blew mind(thank you for blowing it) . I was just substituting values for things like F_f=N(mu) without thinking much about it, but stressed on the algebra, which was easy. I've learnt a very useful lesson today and from now on I'll be more conscious of the real situation that the problem is presenting to me, while doing physics. Thank you very much.
Brilliant solution! Absolutely loved the explanation and reasoning behind every single step taken to arrive at the solution. This took me more than 2 decades down the memory lane when I used to avidly study Physics by Resnick and Halliday ,in a small town in India, and solve some of the apparently difficult problems in mechanics from "I E Irodov's problems in general physics"
As it is usual in you, Sir: simple, easy, straight to the point, and much more in depth than I would ever could imagine. I do appreciate the time and effort that you display in showing us in such a simple way, something that could be very complicated. Long You Live!!!
I did it like this.. If no slipping a = F/(2M + 2m). And then f1 - T = ma And T-f2= ma Thus f1>f2 so f1 reaches the value umg first. Corresponding f2 value from these equations is umg - 2ma, which is the only driving force for M mass in the left. So I get a = (umg - 2ma)/ M and simplifying a = umg /(M + 2m). Putting this in my very first equation I get F = 2(M + m)umg/(M+2m).
Possibly , I did it this way where yu don't hv to consider the second frictional force, just work with the fbd of the first block system only and apply the slipping condition for that bcoz look closely, if the mass m in the front end starts slipping then automatically the mass m in the rear end will also start slipping as the string connecting them shud always remain taut throughout the motion, hope this can help
Dr. Lewin, your lectures are always a delight. I hope the young people get as much from them as I do. The problems are challenging and inspiring, and your explanations of the solutions are clear and understandable. Too, the "ah-ha" moment for me is like stepping out into the warm sunshine on a fresh, spring day. ;-) Always your friend, Richard :D
Thanks 4 the vid :-) I would never have thought that I could understand four simultaneous equations and combined into one equation explaining the diagram so perfectly. Physics is mentally satisfying.
Months ago I had to study for a Physics exam at my university. Even though I was amazed and I enjoyed the lectures, I struggled with it. I passed the exam, but I think that if I had found your lectures early I would have done it better :) Greetings from Italy
This was a good blood bath. I demand from myself that I understand systems like this by just thinking, and therefore chose not to draw free body diagrams. I am a mechanical engineer by training, but having worked on IT the past 24 years has rusted my intuition. Therefore this took a lot of thinking and finding out that my first answer was wrong, before I got there. But I did eventually think it correctly. This practice was very useful. It's good to notice how difficult it can be to understand even a relatively simple system like this and how silly mistakes you make before you get all interactions straight in your head.
esa062, I did the same thing. I assumed wrongly that the 2 frictional forces and the string tension were all equal, and quickly arrived at a result where M fell completely out of the picture. Made no sense to me, so I went back and did it more carefully, getting the right result. Very useful practice, indeed. It's been 48 years since I graduated with a minor in physics, and boy, have I gotten rusty. I thank Dr. Lewin for these challenges.
Got the equations... Assumed friction the same and committed the blunder Thank you sir! Learnt a lot through this question! I have an exam in a few days and this was very helpful!! 😊 Waiting for next question
I still actually don't get it. You see, at 7:40, it would make more sense that the tensions are different, and not frictions, so as to provide desirable acceleration. Friction is constant*mass*g. Since all 3 are the same for both masses, friction just cannot be different, instead tension should be different. So friction is the same, and tension for mass 2 is larger than tension for mass 1.
Sir when u have posted it , I have not revised the Newton laws completely , and I said as soon as I completely revise that , I will do it .. today I saw the ques. After some mistakes I have solved it , I came here to see the last screenshot of final ans. And it's correct ... But now I will see ur entire soln. for more analysis .. :-) .thanks.a lot sir ..
Yaay! I got the correct answer. Sir, I will be sitting for JEE Advanced this week. What would have been the case if there were two coeff. of friction, mu1 and mu2 respectively? Do we need to take case and solve?
Yayy! I was right. I got the right answer at the first try itself. No other answer ever passed my mind n I'm happy abt that😊. However, I am sure that each one of us enjoyed solving n that's the real spirit. Thank you Sir.
If μ=infinite is like the small masses are glued on the large masses, so the force to start moving the system on the bottom frictionless surface is: F=2(m+M)α, where α is any arbitrary acceleration, but not infinite (in the sense of too large).
This is the kind of problem that would be on an olympiad, which is really helpful because i am preparing for one and this is showing the methodology gor solving stuff like this
I found it very interesting that if M = 0, then F = μmg, and as M → ∞, F → 2μmg. In other words F is always between μmg and 2μmg, no matter how big M is.
i thought about this problem for a while and i was thinking that both the frictions are one and the same! that's the real tricky part. thanks for the solution professor.
it's NOT tricky. Free-body diagram for the small mass on the right: Fr(right) - T = 0, small mass on the left T - Fr(left) = ma add up the 2 eqs and assume that F(r)=F(l), then ma=0.
This problem can be solved much more simply, without simultaneous equations. You do this by selecting your F=MA sets differently. First, M on the left only. Then M&m on the left along with m on the right, then all the masses. (There is no reason to "cut" the string. Masses m are guaranteed to act together.) Friction max at either place is umg. Amax left (due to friction on left) is umg/M. Amax right (due to friction on right) is umg/(M+2m). It is obvious friction on right is the limiting (lesser acceleration allowed) one, so we multiply the Amax right by all the mass present. This requires no simultaneous equations! By treating the problem as a "chain", we draw our diagrams around the serial sets of links in sequence, making the "cuts" at the friction surfaces and the main tension. We don't even need to worry about T! The crux move is getting A, and that was gotten in 2 single (unknown) equations, and we discarded the (obviously) larger one.
Sir, great explanation. Thanks a lot. Here is my solution: If we consider all the four masses and string as a system then, F = 2(m+M)a ......(1) If we consider the two upper masses m and the left mass M as as system then we see that the acceleration of this system is 'a' towards right. The only external force that accounts for this acceleration is the friction between the right lower and upper masses. Thus f = (2m+M)a ......(2) And taking the left lower mass M as a system, it accelerates with 'a' units and the force that accounts for this acceleration is the friction between the left lower and upper masses. Thus f' = Ma ......(3) Clearly f > f' ( from (2) and (3)) From (1) we get that as F increases, 'a' also increases. From (2) and (3) we get that as 'a' increases, f and f' also increase. However, since f > f', f reaches the limiting value of static friction earlier than f'. Thus slipping occurs first between the right blocks as F increases. Thus the instant when slipping occurs, f = umg ......(4) From, (1), (2), and (4), we get umg = (2m+M)F_max/2(M+m) Thus F_max = 2umg(M+m)/(2m+M)
There is absolutely no need to assert that frictions are different, in fact the answers explicitly says it. Problem very simple: F = 2(m+M)a "motion of whole system due to F"; T=(m+m)a "left side masses moving due to tension"; and T=m(mug-a) "motion of mass m due to tension and friction". The simple algebra to solve the system of linear equations.
Standard physics question for jee in india But your explanation is very nice to solve it without much equation and calculations Love from India Your MIT lectures are amazing
i only used 3 equations since i concluded early on that the friction on the left will be smaller than on the right and later adding the condition that the friction on the right has to reach the max value Frmax= umg however during calculation i only used 2 equations because i combined some equations further eq1: whole system; F=2(M+m)a eq2: top right; Fr-T=ma eq3: left; T=(M+m)a eq 2 and 3: Fr=(M+2m)a=umg => a=(umg)/(M+2m) put that into eq1 and i get F= (2(M+m)umg)/(M+2m)
There's an eigth equation. Consider the system m2 & m1, you get f=2ma=fr1-fr2. Of course, however, as you said, only four are simultaneously necessary. The key is to be able to formulate each individual mass using the formulae, hence why four are needed, since there are four distinct masses.
You need four equations to be solved as a system, because there are four unknowns (T, T2, F, and a), which stand for the overall tension that pulls the assembly, the tension in the string, the friction force that isn't at its maximum possible value, and the acceleration of the assembly respectively.
I found the answer by setting 3 "f = ma" equations noticing as first statement that the acceleration must be the same for all the masses One is for the system as a whole finding the F we needed F=2(m+M)a One for the 3 masses "pulled" by the first M knowing that the first frictional force is the only force acting on that 3 mass body F1 = (2m+M)a
From an engineer's perspective: Wagon #1 has mass M Wagon #2 has mass 2m Wagon #3 has mass M The break point of the wagon couplings is mgμ. What is the max acceleration that the locomotive may attempt to impart to the train? 1) Clearly, the threatened coupling is the first one. 3) It will break if we attempt to accelerate Wagons #2+#3 with more than { mgμ/(2m+M) } 4) The tractive force of the loco required to generate this acceleration for the whole lot is obviously F=( M + 2m + M) times the above acceleration. A little more ambitious..and snap!
Before solving I thought it was an easy problem which just needed application simple basic concepts(which it needed) but was trapped in the trap and thought Frictional force on 1= Frictional force on 2. Btw thanks to u and Josh for presenting this problem.
I got this correct answer after a wrong turn, but, it might help others to see how I solved it. I envisioned the forces of friction, Fr1 and Fr2, that Walter talks about, as if they were additional ropes with known breaking points. The Fr2 rope only pulls a single large mass, M, but the Fr1 rope pulls two small masses, m, and one large mass, M. So, it became apparent that Fr1 will 'break' before Fr2 because it is pulling a larger load. With this framework in mind the problem became a great deal simpler ... for me at least.
What makes the frictions of the two identical masses m's connected with string T, and on top of identical surfaces, mu's , and to o identical M's DIFFERENT? !! (assuming the string is rigid of course)
With Equation (2), why are we eliminating F and considering only T since mass m + M for Object 1 is also accelerating towards force F and will have same Force as object 2, no?
Fortunately, I was able to do this, since when we were taught friction for the first time, our teacher STRESSED multiple times on the fact that friction can have any value between 0 and its maximum value.
@@lecturesbywalterlewin.they9259 sir you are very good teacher because you showing the practical of everything in physics but in india they do not contain practical of all things
7:40 I don't get how the frictions must be different. Both frictions are constant*weight, and since the constant and mass is same for both masses, friction is the same. It therefore makes more sense that the tension must be different, and not friction.
As i understand that Fr1 and Fr2 seems equals at first glance,but Fr2 adjusted itself lower than Fr1 so Fr1 gets maximum itself at first.considering Fr2 also as (mu*m*g)[which is not unreasonable to think in first look] will be a big blunder..did i get that right?i I also like to add something to context if we look to the two small blocks in the top with mass little 'm',as we know frictional force = (co-efficient of friction)*(normal reaction),then friction will be same on both which is not,in this case that formula doesn't hold,does that mean that formula for frictional force[[i.e. frictional force = (co-efficient of friction)*(normal reaction)]] will not always work??
If T is already larger than Fr2 on the small mass (m) on the left would then it not be moving across the surface of large mass M on the left and therefore not moving the large mass under it???
Sir, how do we exactly define electric potential at a point? It was OK to me till I read abt how conductor kept in changing magnetic field itself behaves like a battery and potential difference calculated between two pts comes different when different paths are taken. I think it is confusing to refer it as potential in case of non conservative forces.
What, if the force of friction is also greater than the tension on the second m and force(tension) is basically applied to the whole system of (m+M) which would give us T=(m+M)a then we get the same answer but in a more reasonable sense
are you talking about ignited mind ? ya he is good but his way of teaching is very different from prof. Walter lewin,prof. Lewin explain concept with real experiments where he explain with the help of story telling and non working models,he is also good.
The reason that Fr1 > Fr2 is because when the Force is in right direction, the big mass 1 will slide creating the friction force on small mass 1 that is in direction of F force of course, but the tension force of that small mass 1 is low compared with Fr1 because that small mass is attracting the small mass 2 by the tension force of this small mass 2 from the spring. In Imaginative way, the tension force of the small mass 2 is playing the crucial role on that mass because is affecting from the small mass 1...which this concludes that the friction force 2 or Fr2 is created only by the small mass 1 which is attracting the small mass 2 by the tension force of this mass from the spring, onlyy based on the F direction...I only talked by the situation in a Imaginative way and I hope that this is understandable because The Professor treated a problem that disturb our imagination for better and not taking it intuitively without advancing it in the critical part which is the most needed...So I say it again that we students are the luckiest ones to have him in this world because only he turns on the light inside every student's perspective for understanding the dynamics of life...As we see, the best subject is taught by the best Professor and this means that situations in this reality system can be imagined very good, indeed but in a understandable way.
For a small block m, if the tension is > or < the friction then the forces are unbalanced so the small block will move relative to the surface, block M, therefore slipping? So you want the tension on the small blocks to equal the friction on the small blocks? Why not?
I regret that you have not discussed what happens when M=m; M=0 and m>0; M>0 and m=0. All with mu=1.0, or perhaps 0.5. That is like baking a nice cake and then not eating it :). Otherwise very nice problem and very clean solution, thank you.
you forgot M= infinity and m = infinity to substitute extremes in an eq is ONLY meaningful if the result is then obvious. I did that. If the result is not at all obvious it's meaningless to put extremes in an eq. But it takes experience (which you do not have yet) to see which substitutions are meaningless. you learn nothing more than my soln if mu=1 and if mu=0.5. Nothing special is happening that could be predicted. mu=1 then F=2mg*(m+M)/(2m+M). What have you learned now? mu=0.5 then F=mg*(m+M)/(2m+M) what have you learned now? choose m=0 then F=0 what have you learned now? chose M=0 that gives you a result that is not at all obvious. F is then mu*mg what have you learned now . . . . NOTHING!
Ja ik ben diverse malen in Groningen geweest in de jaren (1936-1966) dat ik in Nederland woonde. Ik was daar vorige maand nog met m'n vrouw. We waren in Nederland op vakantie. Het Groninger Kunst museum is heel fantastisch.
Can we do this problem in a slightly different way by assuming frictional forces to be equal and instead of that adding up psuedo forces appropriately(Taking the lower right block to be the frame of reference)
*It's NOT a limiting case.* The friction on the right mass m MUST be larger than on the left mass m that's a requirement for the system as a whole to move with a given acceleration. PERIOD!
+Lectures by Walter Lewin. They will make you ♥ Physics. Yes!, btw i was saying u(myu)mg is the limiting case for fric. on the smaller block and i did the mistake by taking u(myu)mg as fric.force for both the blocks but now i got it it would be different. processor ur questions in a way clears the hidden concepts.thanks and keep posting.....😀
If we draw the free body diagram of M on left side, the only force on it is (mu)mg ( max friction force ) . So it's acceleration will be umg/M . Where lies the mistake?
sir I understood that the fr1 will be larger than fr2, but I still can't understand why fr1 will first reach its max. value,but not that both fr1 and fr2 will reach their max. value simultaneously, can you please help?🙂
Firstly " this is called strip learning" and secondly--- Is there really any such string which has negligible mass and is very strong? I would like to use that for my running "Brain":) well I'm just Joking!!!
Why didn't you consider the pulling force on the first mass m. And the frictional force usually opposes the motion. I think the T and frictional force should be sum for the force opposing the motion of mass m because of the applied force F
Sir how should i contact you ? I had an alternate solution to this problem and also a doubt regarding the force on two long straight current carrying conductors (Ampere). I want to know why they attract each other when the current is in the same direction and repel when they're in the opposite direction.
Professor shouldn't the friction be always opposite to the motion? Why is the the friction on the little box 1 the same side of the moviment? Is is because it's opposite to the traction?
Hey Dr. Lewin I got the answer correct my username was wet swole if you check the other video but you still posted my answer. I left my answer as ((Mumg)/(M+2m))+ umg And though it seems different, after I set the denominator of umg then the equation is the same as yours. Do you count unsimplified answers as incorrect? Thanks
I do not understand how the friction forces can be different. The coefficients and normal forces are the same, why shouldn't the friction forces be the same? Thank you for the explanation in advance.
For static friction specifically, it is a constraint force that is not necessarily equal to mu*N. The mu*N is a maximum value of static friction forces, rather than a guarantee that static friction forces will equal that value. Static friction can be zero, as it is for an object sitting on a surface with no force attempting to slide it.
If a mass stands still on an incline (angle alpha), then the frictional force along the incline is mg*sin(alpha). If a mass is sliding down that incline then the frictional force is mu*mg*cos(alpha). mu is the kinetic friction coeff. *PLEASE watch my 8.01 lecture in which I cover this in GREAT detail* *+ demos.*
Yes sir that's the concept which I was asking for the mass m on left on which the tension t becomes larger than friction even before reaching the limiting value of friction umg where u is coefficient of friction. So please clarify my doubt. please reply.
pseudo forces are in rotating frames of reference. They can be very confusing. They are NOT needed as I always work in the Lab reference frame. In the lab there is a centripetal force in the rotating frame there is a centrifugal force (pseudo force) which is equal in magnitude but in opposite direction.
Professor, are fr1 and fr2 the instantaneous frictional forces? Because, the maximum value of both, fr1 and fr2 are the same -- mu times the normal reaction....
>>>>the maximum value of both, fr1 and fr2 are the same>>>> yes but that is NOT relevant. What counts is that fr1 MUST always be larger than fr2. Thus fr1 is the first that reaches a maximum
I want to thank you for doing all of this work. I would have thought that you would have given up teaching for your love of Art History, and I really cannot see how you keep up with all of the answers coming in from around the World. I have a difference of opinion in your explanation, and I hate to make a foolish statement to the entire World and I would never want to hurt your feelings. I am very hesitant in sending it to you because I respect your status and position. I would never want to offend you. I did get the correct answer though, and I thought it was easy. And it makes no difference in the answer at all. Here we go (taking very deep breath) Dear Professor Lewin, I have achieved the correct answer, but I must take issue on your explanation of the answer. I have no degree in Physics; I have worked with it all of my adult life in designing and building of equipment used in lifting, pulling and the like. You stated in the video, at 5 minutes and 23 seconds, that the tension on the string is the same, and I agree that it is exactly the same on the line connecting the 2 smaller blocks, but also on the line that is applying the force to the system as long as the force is smaller than the force of friction on the blocks. But then at 8 minutes and 30 seconds you say that the frictional forces on the 2 smaller blocks are different. Remember that we are speaking of a hypothetical problem here, and not one of the real world where things are never perfect, but in the realm of this problem, we can work with perfection. The mass of the 2 smaller blocks is identical in this problem, and the force of gravity is exerting the same amount of acceleration on both blocks. Therefore the normal force on exerted on both of the larger blocks is identical. The frictional force on both of the smaller blocks on the larger blocks is identical. In the Real World, I will agree with you that one of the 2 smaller blocks has less friction than the other due to tolerances in measurement, but again this is not the real world. The reference to the system for observing the 2 (m+M) blocks as they float on the frictionless surface causes the confusion, I am sure, with your students. I picture this as though we are on a lake, and one boat is towing another. As force is applied to the original system they will both accelerate equally along the frictionless surface with nothing but the string connecting the 2 smaller blocks together. Much like a couple of boats being pulled slowly through a calm lake, one pulling the other. But if you change your reference from being on the frictionless surface (the lake) and stand atop the large block that is being towed, you eliminate some of your unknowns in the system. Now we have no movement of the system while the applied forge is less than the frictional force of one of the small blocks, and if you were to add force gauges to each of the 2 lines connecting the system, you would find exactly the same reading on both gauges. As the applied force approaches the force of friction of one of the small blocks (they are both equal) the entire system is static, and there is no movement. But, as the applied force reaches the force of friction, one of the 2 small blocks will move, most likely the first one because of momentum; however, it could be the rear small block that moves, and it really does not matter. In conclusion, since this is a hypothetical problem, then both of the frictional forces are exactly the same, but in reality and exposed to experimentation, you are exactly correct that there are differences in the friction between the blocks. Sincerely, Jeremy Pierce Amarillo, TX
Professor, couldn't you explain why I am wrong to say that if the friction force 2 is smaller than friction force 1 then it reaches its maximum first? Thank you.
Prof.walter i solved it but i thought that the object (small m) will be affected by force equal (ma) because when we pull the big object M: the small will be affected by the same accelration in opposite direction And thank you
Lectures by Walter Lewin. They will make you ♥ Physics. in first thank you to reply me but what i mean is when you ride a car and car start accelerate you will get back by the same acceleration why does it happen in this case ? I listen it in your lecture 8.01
when the car accelerates you don't get taken back by the same acceleration. it's the inertia of rest that's doing the job. your body tries to resist motion i.e. change of state.
What do you think about following problem? The uniform rod of lenght L and mass M lies on the plane. It can move on the plane without friction. In the direction perpendicular to the rod a small block of mass m is moving with the velocity v. At which distance x from the midle (of the rod) the rod should be hitted by the block in order to rotational and translational energies of the rod after the collision are equal? Let us assume that collision is elastic.
I have got another one, which I discussed on a forum with some physicists and did not get a clear answer. The problem seems to be very simple and goes as follows. A block of mass M can move on a horizontal plane without friction. A sphere of mass m is moving with the velocity v (in a vertical plane) and hits a horizontal surface of the block at the incident angle alpha. What is the angle of reflection of the sphere and what is the speed of the block after collision? Collision is elastic. We have here two equations: conservation of kinetic energy and conservation of momentum in a horizontal direction. Momentum in vertical direction is not conserved because of interaction with the horizontal surface on which the block is moving. So we have two equations and three unknowns (vertical and horizontal velocities of the sphere after collision and the velocity of the block). My conclusion is that the answer depends on the details of interaction between the sphere and the block, but maybe I am wrong.
The equation F=mu*N is not set up to work as a proper vector equation, whether it be a cross product or dot product. It is a scalar equation, with an implied understanding that the direction of F is not the same as the direction of N. F is in the direction to oppose existing sliding motion (kinetic friction) or in the direction to oppose attempted sliding motion from an initially stationary pair of contacting objects (static friction).
bloodbath is apt, and I nearly exsanguinated! If you had not posted a day early, you may have had one less subscriber! I never challenged that the two frictional forces were different (why didn't I listen when you demonstrated the same logic on weight change and tension in lecture 7????)Still, once again ,an important step forward for me.A great task from your contributor but a wonderful logic lesson from you.Thank you
Everyone (and there were many) who assumed that the friction was the same on both sides VIOLATED F=ma. As a result they ended up with a garbage solution that made no sense.
Sir, I have a question for you. I follow your explanation very well, but there is one question that keeps on bugging me. Where do you get the expression that Ff1 =mu*m*g? If it is from the normal force acting on the small mass on the right, you can get the same expression for the Ff2 = mu*m*g from the left small body. Now, I have a contradiction.
And I can see that from your equations. But if I just apply the same logic that I think you applied for the Ff1 in Ff2, I get a different solution. I am just trying to understand as to why.
you make a *GIGANTIC mistake* It seems that you have forgotten all about the physics of friction. I suggest you watch my 8.01 lectures in which I cover "friction". It will open your eyes and remove your confusion.
I used only 2 equations: Take both the masses on the left plus top mass on the right. Net force on this system is f = mu*m*g = (M + 2*m)*a. This implies a = mu*m*g/(M + 2*m).................. This is the maximum acceleration. Now take all 4 masses as one system. Net force on this system is F = 2*(M + m)*a. Plug in the max acceleration here. That gives F = 2*mu*m*g*(M+m)/(M + 2*m)
Even though my solution was't right but I realized a very important thing after watching this solution and that is physics is not pure mathematics,I need to think way deeper while doing physics than while doing algebra, the intuition(which didn't arrive to me while doing the problem) that Frictional force between the right set of two masses will become maximum first literally blew mind(thank you for blowing it) . I was just substituting values for things like F_f=N(mu) without thinking much about it, but stressed on the algebra, which was easy.
I've learnt a very useful lesson today and from now on I'll be more conscious of the real situation that the problem is presenting to me, while doing physics.
Thank you very much.
Brilliant solution! Absolutely loved the explanation and reasoning behind every single step taken to arrive at the solution. This took me more than 2 decades down the memory lane when I used to avidly study Physics by Resnick and Halliday ,in a small town in India, and solve some of the apparently difficult problems in mechanics from "I E Irodov's problems in general physics"
As it is usual in you, Sir: simple, easy, straight to the point, and much more in depth than I would ever could imagine. I do appreciate the time and effort that you display in showing us in such a simple way, something that could be very complicated. Long You Live!!!
I did it like this..
If no slipping a = F/(2M + 2m).
And then f1 - T = ma
And T-f2= ma
Thus f1>f2 so f1 reaches the value umg first. Corresponding f2 value from these equations is umg - 2ma, which is the only driving force for M mass in the left. So I get a = (umg - 2ma)/ M and simplifying a = umg /(M + 2m). Putting this in my very first equation I get F = 2(M + m)umg/(M+2m).
Today i did that in same way
Thank you sir. Actually I didn't realize that the two frictional forces are not same. Thanks for such a great explanation.☺
:)
Possibly , I did it this way where yu don't hv to consider the second frictional force, just work with the fbd of the first block system only and apply the slipping condition for that bcoz look closely, if the mass m in the front end starts slipping then automatically the mass m in the rear end will also start slipping as the string connecting them shud always remain taut throughout the motion, hope this can help
@@rajdeeplahiri8842 yes u r right brooo
I also do the same thing by assuming whole system and omited the f2 😂😂😂😂
Simple method 😂😂😂✌️
They are THE SAME, he made a bad mistake!
Thank you for the problem Professor! It was really fun and challenging to work on it during the week!
Dr. Lewin, your lectures are always a delight. I hope the young people get as much from them as I do. The problems are challenging and inspiring, and your explanations of the solutions are clear and understandable. Too, the "ah-ha" moment for me is like stepping out into the warm sunshine on a fresh, spring day. ;-)
Always your friend,
Richard :D
:)
Are you the one who did the really cool trick with the chalk where you draw a line but it makes dots?
ua-cam.com/video/raurl4s0pjU/v-deo.html
Had been struggling with this one for 4 days. Thank you so much for the solution sir! Only one of my friends could solve this correctly.
:)
Thanks 4 the vid :-) I would never have thought that I could understand four simultaneous equations and combined into one equation explaining the diagram so perfectly. Physics is mentally satisfying.
:)
Months ago I had to study for a Physics exam at my university. Even though I was amazed and I enjoyed the lectures, I struggled with it. I passed the exam, but I think that if I had found your lectures early I would have done it better :)
Greetings from Italy
:)
I too did not see that the 2 frictions were different. More learnings. thankyou
A very neat problem indeed, professor. Would you be inclined to think of a challenging problem related to electromagnetism?
Now I get it , nice question sir , really enjoyed it :)
Sir thanks for your valuable time and efforts and making the concept easier to us
:)
This was a good blood bath. I demand from myself that I understand systems like this by just thinking, and therefore chose not to draw free body diagrams. I am a mechanical engineer by training, but having worked on IT the past 24 years has rusted my intuition. Therefore this took a lot of thinking and finding out that my first answer was wrong, before I got there. But I did eventually think it correctly. This practice was very useful. It's good to notice how difficult it can be to understand even a relatively simple system like this and how silly mistakes you make before you get all interactions straight in your head.
you will get a 2nd chance next week.
esa062, I did the same thing. I assumed wrongly that the 2 frictional forces and the string tension were all equal, and quickly arrived at a result where M fell completely out of the picture. Made no sense to me, so I went back and did it more carefully, getting the right result. Very useful practice, indeed. It's been 48 years since I graduated with a minor in physics, and boy, have I gotten rusty. I thank Dr. Lewin for these challenges.
Yay that's great!! Could you tell us the topic for the next problem please?
Newtonian mechanics
Thank you
I really liked this problem :) Thanks for posting it!
:)
I wish we where taught something in high school . I find the ways of looking at things very interesting.
Got the equations... Assumed friction the same and committed the blunder
Thank you sir! Learnt a lot through this question! I have an exam in a few days and this was very helpful!! 😊
Waiting for next question
:)
I still actually don't get it. You see, at 7:40, it would make more sense that the tensions are different, and not frictions, so as to provide desirable acceleration. Friction is constant*mass*g. Since all 3 are the same for both masses, friction just cannot be different, instead tension should be different. So friction is the same, and tension for mass 2 is larger than tension for mass 1.
Sir when u have posted it , I have not revised the Newton laws completely , and I said as soon as I completely revise that , I will do it .. today I saw the ques. After some mistakes I have solved it , I came here to see the last screenshot of final ans. And it's correct ... But now I will see ur entire soln. for more analysis .. :-) .thanks.a lot sir ..
Yaay! I got the correct answer. Sir, I will be sitting for JEE Advanced this week. What would have been the case if there were two coeff. of friction, mu1 and mu2 respectively? Do we need to take case and solve?
living legend...:-) :-) :-) love you sir..
Yayy! I was right. I got the right answer at the first try itself. No other answer ever passed my mind n I'm happy abt that😊. However, I am sure that each one of us enjoyed solving n that's the real spirit. Thank you Sir.
Thank you for putting me out of my frustration :)
:)
If μ=infinite is like the small masses are glued on the large masses, so the force to start moving the system on the bottom frictionless surface is: F=2(m+M)α, where α is any arbitrary acceleration, but not infinite (in the sense of too large).
We wouldn't call it friction at that point, but yes, you are correct. We'd call it adhesion.
This is the kind of problem that would be on an olympiad, which is really helpful because i am preparing for one and this is showing the methodology gor solving stuff like this
I found it very interesting that if M = 0, then F = μmg, and as M → ∞, F → 2μmg. In other words F is always between μmg and 2μmg, no matter how big M is.
i thought about this problem for a while and i was thinking that both the frictions are one and the same! that's the real tricky part. thanks for the solution professor.
it's NOT tricky. Free-body diagram for the small mass on the right:
Fr(right) - T = 0, small mass on the left T - Fr(left) = ma
add up the 2 eqs and assume that F(r)=F(l), then ma=0.
F1 is greater because it pulls 3 blocks where as F2 pulls 2 blocks at same acceleration !
Thanks a lot for making me have a more clear concept
:)
Thanks sir for putting me out of frustration and happy Diwali 2 u you sir
This problem can be solved much more simply, without simultaneous equations. You do this by selecting your F=MA sets differently. First, M on the left only. Then M&m on the left along with m on the right, then all the masses. (There is no reason to "cut" the string. Masses m are guaranteed to act together.) Friction max at either place is umg. Amax left (due to friction on left) is umg/M. Amax right (due to friction on right) is umg/(M+2m). It is obvious friction on right is the limiting (lesser acceleration allowed) one, so we multiply the Amax right by all the mass present. This requires no simultaneous equations!
By treating the problem as a "chain", we draw our diagrams around the serial sets of links in sequence, making the "cuts" at the friction surfaces and the main tension. We don't even need to worry about T! The crux move is getting A, and that was gotten in 2 single (unknown) equations, and we discarded the (obviously) larger one.
As a Chinese student,I've watched many of your vedioes,wish that could meet you someday.(I am sorry for my grammar problem). -Professor Lewin
Here's how to say that with correct grammar:
"As a Chinese student, I've watched many of your videos. I wish that I could meet you someday."
Sir, great explanation. Thanks a lot. Here is my solution:
If we consider all the four masses and string as a system then,
F = 2(m+M)a ......(1)
If we consider the two upper masses m and the left mass M as as system then we see that the acceleration of this system is 'a' towards right. The only external force that accounts for this acceleration is the friction between the right lower and upper masses.
Thus f = (2m+M)a ......(2)
And taking the left lower mass M as a system, it accelerates with 'a' units and the force that accounts for this acceleration is the friction between the left lower and upper masses.
Thus f' = Ma ......(3)
Clearly f > f' ( from (2) and (3))
From (1) we get that as F increases, 'a' also increases.
From (2) and (3) we get that as 'a' increases, f and f' also increase. However, since f > f', f reaches the limiting value of static friction earlier than f'. Thus slipping occurs first between the right blocks as F increases.
Thus the instant when slipping occurs,
f = umg ......(4)
From, (1), (2), and (4), we get
umg = (2m+M)F_max/2(M+m)
Thus F_max = 2umg(M+m)/(2m+M)
There is absolutely no need to assert that frictions are different, in fact the answers explicitly says it. Problem very simple: F = 2(m+M)a "motion of whole system due to F"; T=(m+m)a "left side masses moving due to tension"; and T=m(mug-a) "motion of mass m due to tension and friction". The simple algebra to solve the system of linear equations.
Standard physics question for jee in india
But your explanation is very nice to solve it without much equation and calculations
Love from India
Your MIT lectures are amazing
i got F=2 mu*(M+m) .
now I understand where I was missing.
thank you
You amaze us always
sir happy diwali from india and also happy halloween ,👻👻👻
heppee diwali 2u2
I luv it ! This’s amazing for me thank u 🧑🚀😊
You’re welcome 😊
i only used 3 equations
since i concluded early on that the friction on the left will be smaller than on the right
and later adding the condition that the friction on the right has to reach the max value
Frmax= umg
however during calculation i only used 2 equations because i combined some equations further
eq1: whole system; F=2(M+m)a
eq2: top right; Fr-T=ma
eq3: left; T=(M+m)a
eq 2 and 3: Fr=(M+2m)a=umg => a=(umg)/(M+2m)
put that into eq1 and i get F= (2(M+m)umg)/(M+2m)
There's an eigth equation. Consider the system m2 & m1, you get f=2ma=fr1-fr2. Of course, however, as you said, only four are simultaneously necessary. The key is to be able to formulate each individual mass using the formulae, hence why four are needed, since there are four distinct masses.
ask google or Quora
You need four equations to be solved as a system, because there are four unknowns (T, T2, F, and a), which stand for the overall tension that pulls the assembly, the tension in the string, the friction force that isn't at its maximum possible value, and the acceleration of the assembly respectively.
I found the answer by setting 3 "f = ma" equations noticing as first statement that the acceleration must be the same for all the masses
One is for the system as a whole finding the F we needed
F=2(m+M)a
One for the 3 masses "pulled" by the first M knowing that the first frictional force is the only force acting on that 3 mass body
F1 = (2m+M)a
From an engineer's perspective:
Wagon #1 has mass M
Wagon #2 has mass 2m
Wagon #3 has mass M
The break point of the wagon couplings is mgμ.
What is the max acceleration that the locomotive may attempt to impart to the train?
1) Clearly, the threatened coupling is the first one.
3) It will break if we attempt to accelerate Wagons #2+#3 with more than { mgμ/(2m+M) }
4) The tractive force of the loco required to generate this acceleration for the whole lot is obviously F=( M + 2m + M) times the above acceleration.
A little more ambitious..and snap!
didn`t go without the glasses hein lol ... Thanx Sir ^^
I recorded this 6 weeks ago!
Before solving I thought it was an easy problem which just needed application simple basic concepts(which it needed) but was trapped in the trap and thought Frictional force on 1= Frictional force on 2. Btw thanks to u and Josh for presenting this problem.
:)
you are really great
I got this correct answer after a wrong turn, but, it might help others to see how I solved it. I envisioned the forces of friction, Fr1 and Fr2, that Walter talks about, as if they were additional ropes with known breaking points. The Fr2 rope only pulls a single large mass, M, but the Fr1 rope pulls two small masses, m, and one large mass, M. So, it became apparent that Fr1 will 'break' before Fr2 because it is pulling a larger load. With this framework in mind the problem became a great deal simpler ... for me at least.
What makes the frictions of the two identical masses m's connected with string T, and on top of identical surfaces, mu's , and to o identical M's DIFFERENT? !! (assuming the string is rigid of course)
With Equation (2), why are we eliminating F and considering only T since mass m + M for Object 1 is also accelerating towards force F and will have same Force as object 2, no?
Fortunately, I was able to do this, since when we were taught friction for the first time, our teacher STRESSED multiple times on the fact that friction can have any value between 0 and its maximum value.
:)
I love you, Professor
Thanks sir for this informative video... Happy diwali.
heppee diwali 2u2
thankyou sir :)
@@lecturesbywalterlewin.they9259 sir you are very good teacher because you showing the practical of everything in physics but in india they do not contain practical of all things
7:40 I don't get how the frictions must be different. Both frictions are constant*weight, and since the constant and mass is same for both masses, friction is the same. It therefore makes more sense that the tension must be different, and not friction.
Static friction is not necessarily equal to mu*N. It could be as high as this value, but it could also be as low as zero.
As i understand that Fr1 and Fr2 seems equals at first glance,but Fr2 adjusted itself lower than Fr1 so Fr1 gets maximum itself at first.considering Fr2 also as (mu*m*g)[which is not unreasonable to think in first look] will be a big blunder..did i get that right?i
I also like to add something to context if we look to the two small blocks in the top with mass little 'm',as we know frictional force = (co-efficient of friction)*(normal reaction),then friction will be same on both which is not,in this case that formula doesn't hold,does that mean that formula for frictional force[[i.e. frictional force = (co-efficient of friction)*(normal reaction)]] will not always work??
>>>frictional force = (co-efficient of friction)*(normal reaction)>>>
WRONG
MAXIMUM friction = (co-efficient of friction)*(normal reaction)
CORRECT
Lectures by Walter Lewin. They will make you ♥ Physics. Thanks professor for clearing my doubt...I am grateful to you for many reasons..
If T is already larger than Fr2 on the small mass (m) on the left would then it not be moving across the surface of large mass M on the left and therefore not moving the large mass under it???
Well , such questions are very usual for those who solved irodov , I have solved for jee adv.
excellent application of FBD
Sir, how do we exactly define electric potential at a point? It was OK to me till I read abt how conductor kept in changing magnetic field itself behaves like a battery and potential difference calculated between two pts comes different when different paths are taken. I think it is confusing to refer it as potential in case of non conservative forces.
watch my 8.02 lectures - or ask googel
Lectures by Walter Lewin. They will make you ♥ Physics.
Ok Sir. Thank you for replying.
What, if the force of friction is also greater than the tension on the second m and force(tension) is basically applied to the whole system of (m+M) which would give us T=(m+M)a then we get the same answer but in a more reasonable sense
I've seen an Indian Walter lewin, a young boy... on UA-cam channel IGNITED INDIA..
:)
are you talking about ignited mind ? ya he is good but his way of teaching is very different from prof. Walter lewin,prof. Lewin explain concept with real experiments where he explain with the help of story telling and non working models,he is also good.
Dominus Providebit Indian, not Pakistani. Youth in Pakistan is persuaded by their priests to go into terrorism.
The reason that Fr1 > Fr2 is because when the Force is in right direction, the big mass 1 will slide creating the friction force on small mass 1 that is in direction of F force of course, but the tension force of that small mass 1 is low compared with Fr1 because that small mass is attracting the small mass 2 by the tension force of this small mass 2 from the spring. In Imaginative way, the tension force of the small mass 2 is playing the crucial role on that mass because is affecting from the small mass 1...which this concludes that the friction force 2 or Fr2 is created only by the small mass 1 which is attracting the small mass 2 by the tension force of this mass from the spring, onlyy based on the F direction...I only talked by the situation in a Imaginative way and I hope that this is understandable because The Professor treated a problem that disturb our imagination for better and not taking it intuitively without advancing it in the critical part which is the most needed...So I say it again that we students are the luckiest ones to have him in this world because only he turns on the light inside every student's perspective for understanding the dynamics of life...As we see, the best subject is taught by the best Professor and this means that situations in this reality system can be imagined very good, indeed but in a understandable way.
For a small block m, if the tension is > or < the friction then the forces are unbalanced so the small block will move relative to the surface, block M, therefore slipping? So you want the tension on the small blocks to equal the friction on the small blocks? Why not?
I was asking myself exactly the same questions. IF T Is larger Than Ftr block should move.
I regret that you have not discussed what happens when M=m; M=0 and m>0; M>0 and m=0. All with mu=1.0, or perhaps 0.5. That is like baking a nice cake and then not eating it :). Otherwise very nice problem and very clean solution, thank you.
you forgot M= infinity and m = infinity
to substitute extremes in an eq is ONLY meaningful if the result is then obvious. I did that. If the result is not at all obvious it's meaningless to put extremes in an eq. But it takes experience (which you do not have yet) to see which substitutions are meaningless. you learn nothing more than my soln if mu=1 and if mu=0.5. Nothing special is happening that could be predicted. mu=1 then F=2mg*(m+M)/(2m+M). What have you learned now? mu=0.5 then F=mg*(m+M)/(2m+M) what have you learned now? choose m=0 then F=0 what have you learned now? chose M=0 that gives you a result that is not at all obvious. F is then mu*mg what have you learned now . . . . NOTHING!
Thank you Prof.
Iam just asking about equation 2 , how is it possible to accelerate both masses at the same time?
I answered your question in my solution. Watch it again.
Free-body diagrams!
Dear Walter, have you ever been to Groningen? i have an almost certain memory that i've seen you, when i was a young boy.
Ja ik ben diverse malen in Groningen geweest in de jaren (1936-1966) dat ik in Nederland woonde. Ik was daar vorige maand nog met m'n vrouw. We waren in Nederland op vakantie. Het Groninger Kunst museum is heel fantastisch.
Can we do this problem in a slightly different way by assuming frictional forces to be equal and instead of that adding up psuedo forces appropriately(Taking the lower right block to be the frame of reference)
>>> assuming frictional forces to be equal>>>
They are NOT equal! That's KEY here!
Lectures by Walter Lewin. They will make you ♥ Physics. Yeah i realized that when i Drew the FBD just now .THANKS!
i also did the same considering both fric. forces equal....but now i understood it is the limiting case.
*It's NOT a limiting case.* The friction on the right mass m MUST be larger than on the left mass m that's a requirement for the system as a whole to move with a given acceleration. PERIOD!
+Lectures by Walter Lewin. They will make you ♥ Physics. Yes!, btw i was saying u(myu)mg is the limiting case for fric. on the smaller block and i did the mistake by taking u(myu)mg as fric.force for both the blocks but now i got it it would be different. processor ur questions in a way clears the hidden concepts.thanks and keep posting.....😀
If we draw the free body diagram of M on left side, the only force on it is (mu)mg ( max friction force ) . So it's acceleration will be umg/M . Where lies the mistake?
sir I understood that the fr1 will be larger than fr2, but I still can't understand why fr1 will first reach its max. value,but not that both fr1 and fr2 will reach their max. value simultaneously, can you please help?🙂
Could you do problems with multiple pulleys and multiple masses hung and being accelerated.
do Problem #31. That has all the relevant Physics.
Firstly " this is called strip learning"
and secondly--- Is there really any such string which has negligible mass and is very strong?
I would like to use that for my running "Brain":) well I'm just Joking!!!
Why didn't you consider the pulling force on the first mass m. And the frictional force usually opposes the motion. I think the T and frictional force should be sum for the force opposing the motion of mass m because of the applied force F
You are my physics god
:)
I got the right answer by different approach but fr1 and fr2 I got that now
Sir how should i contact you ? I had an alternate solution to this problem and also a doubt regarding the force on two long straight current carrying conductors (Ampere). I want to know why they attract each other when the current is in the same direction and repel when they're in the opposite direction.
use google
Right now i'm on the physics lesson and we have this problem
Professor shouldn't the friction be always opposite to the motion? Why is the the friction on the little box 1 the same side of the moviment? Is is because it's opposite to the traction?
F=ma
Hey Dr. Lewin I got the answer correct my username was wet swole if you check the other video but you still posted my answer. I left my answer as ((Mumg)/(M+2m))+ umg
And though it seems different, after I set the denominator of umg then the equation is the same as yours. Do you count unsimplified answers as incorrect?
Thanks
:)
I will take that as a yes haha!
hurray I got it right!
"you may DROWN in the zoo of F=ma." Not only is the professor an exalted excellent educator but a great wordsmith also.
:)
Hi! I do not understand why if Fr2 is smaller than T, the small mass does not slip over the larger mass. thanks
watch my solution
I do not understand how the friction forces can be different. The coefficients and normal forces are the same, why shouldn't the friction forces be the same? Thank you for the explanation in advance.
For static friction specifically, it is a constraint force that is not necessarily equal to mu*N. The mu*N is a maximum value of static friction forces, rather than a guarantee that static friction forces will equal that value. Static friction can be zero, as it is for an object sitting on a surface with no force attempting to slide it.
Sir what will be the answer when coefficient of friction(u) is not same and between left blocks it is less than (M/(2m+M)) times between right.
ask google
Sir, will the smaller mass on the left experience always the friction smaller than applied force howsoever small the applied force may be?
If a mass stands still on an incline (angle alpha), then the frictional force along the incline is mg*sin(alpha). If a mass is sliding down that incline then the frictional force is mu*mg*cos(alpha). mu is the kinetic friction coeff. *PLEASE watch my 8.01 lecture in which I cover this in GREAT detail* *+ demos.*
Yes sir that's the concept which I was asking for the mass m on left on which the tension t becomes larger than friction even before reaching the limiting value of friction umg where u is coefficient of friction. So please clarify my doubt. please reply.
please please i beg you please reply.
Nice solution sir
:)
that was great........man I'm rusty
sir why aren't we considering the pseudo force which is acting on small mass 'm' due to the applied force F.
pseudo forces are in rotating frames of reference. They can be very confusing. They are NOT needed as I always work in the Lab reference frame. In the lab there is a centripetal force in the rotating frame there is a centrifugal force (pseudo force) which is equal in magnitude but in opposite direction.
Professor, are fr1 and fr2 the instantaneous frictional forces? Because, the maximum value of both, fr1 and fr2 are the same -- mu times the normal reaction....
>>>>the maximum value of both, fr1 and fr2 are the same>>>>
yes but that is NOT relevant. What counts is that fr1 MUST always be larger than fr2.
Thus fr1 is the first that reaches a maximum
Thank you professor...
Happy halloween BTW...
😊
I want to thank you for doing all of this work. I would have thought that you would have given up teaching for your love of Art History, and I really cannot see how you keep up with all of the answers coming in from around the World.
I have a difference of opinion in your explanation, and I hate to make a foolish statement to the entire World and I would never want to hurt your feelings. I am very hesitant in sending it to you because I respect your status and position. I would never want to offend you. I did get the correct answer though, and I thought it was easy. And it makes no difference in the answer at all.
Here we go (taking very deep breath)
Dear Professor Lewin,
I have achieved the correct answer, but I must take issue on your explanation of the answer. I have no degree in Physics; I have worked with it all of my adult life in designing and building of equipment used in lifting, pulling and the like. You stated in the video, at 5 minutes and 23 seconds, that the tension on the string is the same, and I agree that it is exactly the same on the line connecting the 2 smaller blocks, but also on the line that is applying the force to the system as long as the force is smaller than the force of friction on the blocks. But then at 8 minutes and 30 seconds you say that the frictional forces on the 2 smaller blocks are different. Remember that we are speaking of a hypothetical problem here, and not one of the real world where things are never perfect, but in the realm of this problem, we can work with perfection. The mass of the 2 smaller blocks is identical in this problem, and the force of gravity is exerting the same amount of acceleration on both blocks. Therefore the normal force on exerted on both of the larger blocks is identical. The frictional force on both of the smaller blocks on the larger blocks is identical. In the Real World, I will agree with you that one of the 2 smaller blocks has less friction than the other due to tolerances in measurement, but again this is not the real world.
The reference to the system for observing the 2 (m+M) blocks as they float on the frictionless surface causes the confusion, I am sure, with your students. I picture this as though we are on a lake, and one boat is towing another. As force is applied to the original system they will both accelerate equally along the frictionless surface with nothing but the string connecting the 2 smaller blocks together. Much like a couple of boats being pulled slowly through a calm lake, one pulling the other. But if you change your reference from being on the frictionless surface (the lake) and stand atop the large block that is being towed, you eliminate some of your unknowns in the system. Now we have no movement of the system while the applied forge is less than the frictional force of one of the small blocks, and if you were to add force gauges to each of the 2 lines connecting the system, you would find exactly the same reading on both gauges. As the applied force approaches the force of friction of one of the small blocks (they are both equal) the entire system is static, and there is no movement. But, as the applied force reaches the force of friction, one of the 2 small blocks will move, most likely the first one because of momentum; however, it could be the rear small block that moves, and it really does not matter.
In conclusion, since this is a hypothetical problem, then both of the frictional forces are exactly the same, but in reality and exposed to experimentation, you are exactly correct that there are differences in the friction between the blocks.
Sincerely,
Jeremy Pierce
Amarillo, TX
I respect your opinion. However, your solution is wrong and mine is correct.
I hope we are still friends.
happy diwali sir.......from india
heppee diwali 2u2
omg! you replied me , today is my sleepless night........ big fan of you......
sir can indian join mit for research work or ph.d....?
Professor, couldn't you explain why I am wrong to say that if the friction force 2 is smaller than friction force 1 then it reaches its maximum first? Thank you.
I cannot add to the clarity of my solution. It's up to you to figure out what you did wrong - it's not up to me.
Prof.walter i solved it but i thought that the object (small m) will be affected by force equal (ma) because when we pull the big object M: the small will be affected by the same accelration in opposite direction
And thank you
>>>when we pull the big object M: the small will be affected by the same accelration in opposite direction >>>
NO, it's in the SAME direction
Lectures by Walter Lewin. They will make you ♥ Physics. in first thank you to reply me but what i mean is when you ride a car and car start accelerate you will get back by the same acceleration why does it happen in this case ? I listen it in your lecture 8.01
when the car accelerates you don't get taken back by the same acceleration. it's the inertia of rest that's doing the job. your body tries to resist motion i.e. change of state.
What do you think about following problem? The uniform rod of lenght L and mass M lies on the plane. It can move on the plane without friction. In the direction perpendicular to the rod a small block of mass m is moving with the velocity v. At which distance x from the midle (of the rod) the rod should be hitted by the block in order to rotational and translational energies of the rod after the collision are equal? Let us assume that collision is elastic.
use google
I have got another one, which I discussed on a forum with some physicists and did not get a clear answer. The problem seems to be very simple and goes as follows. A block of mass M can move on a horizontal plane without friction. A sphere of mass m is moving with the velocity v (in a vertical plane) and hits a horizontal surface of the block at the incident angle alpha. What is the angle of reflection of the sphere and what is the speed of the block after collision? Collision is elastic. We have here two equations: conservation of kinetic energy and conservation of momentum in a horizontal direction. Momentum in vertical direction is not conserved because of interaction with the horizontal surface on which the block is moving. So we have two equations and three unknowns (vertical and horizontal velocities of the sphere after collision and the velocity of the block). My conclusion is that the answer depends on the details of interaction between the sphere and the block, but maybe I am wrong.
ask Quora!
If friction is a vector = mu×mg then what causes the direction of the Ff not to be parallel to mg. Sorry for asking all these questions professor?
watch my 8.01 lectures on friction or use google
The equation F=mu*N is not set up to work as a proper vector equation, whether it be a cross product or dot product. It is a scalar equation, with an implied understanding that the direction of F is not the same as the direction of N. F is in the direction to oppose existing sliding motion (kinetic friction) or in the direction to oppose attempted sliding motion from an initially stationary pair of contacting objects (static friction).
I got tension is equal to exactly 1/2 of Fmax. Is that correct?
Yes. It has to be, because it equals the net mass on half the mass of the assembly.
bloodbath is apt, and I nearly exsanguinated! If you had not posted a day early, you may have had one less subscriber! I never challenged that the two frictional forces were different (why didn't I listen when you demonstrated the same logic on weight change and tension in lecture 7????)Still, once again ,an important step forward for me.A great task from your contributor but a wonderful logic lesson from you.Thank you
Everyone (and there were many) who assumed that the friction was the same on both sides VIOLATED F=ma. As a result they ended up with a garbage solution that made no sense.
Lectures by Walter Lewin. They will make you ♥ Physics. Humbled, but wiser.
Sir, I have a question for you. I follow your explanation very well, but there is one question that keeps on bugging me. Where do you get the expression that Ff1 =mu*m*g? If it is from the normal force acting on the small mass on the right, you can get the same expression for the Ff2 = mu*m*g from the left small body. Now, I have a contradiction.
I have no time to add to the clarity of my solution. Ask some friends to help you. This is high school physics. good luck
thank you, Dr. Lewin. If anyone out there could reply to my question, it will be greatly appreciated.
ask one of your friends. This is a high school problem. I cannot add to the clarity of my solution. >
NO !! it's smaller than that.
And I can see that from your equations. But if I just apply the same logic that I think you applied for the Ff1 in Ff2, I get a different solution. I am just trying to understand as to why.
you make a *GIGANTIC mistake* It seems that you have forgotten all about the physics of friction. I suggest you watch my 8.01 lectures in which I cover "friction". It will open your eyes and remove your confusion.
When will you give the next problem?
Sat Oct 28
I used only 2 equations:
Take both the masses on the left plus top mass on the right.
Net force on this system is f = mu*m*g = (M + 2*m)*a.
This implies a = mu*m*g/(M + 2*m).................. This is the maximum acceleration.
Now take all 4 masses as one system.
Net force on this system is F = 2*(M + m)*a.
Plug in the max acceleration here.
That gives F = 2*mu*m*g*(M+m)/(M + 2*m)
you used 3
Watch my video near the end.
I showed you that you can do it with 2 eqs + #5 which is 3 eqs.
what if we add a string between the two large masses as well :D
I see your glasses😂😂. Thank you for your solution
I taped this solution 6 weeks ago LONG before my cataract surgery
Lectures by Walter Lewin. They will make you ♥ Physics. Respect!