8.01x - Lect 25 - Static Equilibrium, Stability, Rope Warker

[@tuinbolado1]

Professor Walter Lewin, I am a mechanical engineer from Brasil and I teach phisycs as a voluteer, i would like to thank you for all that you have done for education. You have no idea how your lectures are inspire me. You are really making difference in this crazy word. Thanks a lot for help me to help others.

[@lecturesbywalterlewin.they9259]

+Antonio Romero Thank you Antonio for your kind words. Comments like that are very rewarding. My 50 years of teaching in a somewhat eccentric way has paid off. About 5 million people watch my lectures yearly.

[@albertdowson5436]

@@lecturesbywalterlewin.they9259 Sir, thanks a lot

[@ramjb]

@@lecturesbywalterlewin.they9259 I'm showing this lectures to my girlfriend (who she always insisted she's maths and physics challenged and that she wouldn't get any of it no matter what because back at school she wouldn't understand anything), and she's getting everything. She gets lost in the way the equations are calculated, of course, but she grasps the fundamental ideas you're trying to transmit. It's hillarious (And really satisfying) to see her react to your lectures, she's always like "wow, I REALLY understood the reasoning behind that!". My answer is that there are no challenged students. Only bad teachers. And you truly are an excellent teacher. what's me, I'm a failed physics student myself. The university I attended to had truly horrible teachers (they just would walk in, utter the lesson like a prayer, and walk out with no regards to whether anyone was following or understanding them, no matter about making them entertaining), and as a result I ended up quitting it and moving on to something else (computer engineering). I really wish I had teachers like you back then. I'd have loved the experience instead of hating it, and today I'd have a degree on what's always been something I've loved.

[@kartikkumbhare6999]

00:00 static equilibrium 34:30 stability 43:37 rope walker

[@sef2273]

I’ll often hold a ladder for someone to stabilize it as they go up it. I’ve always done it not just by hand, but by standing on the bottom rung itself. Glad to see my method proven in this experiment .

[@yssacnton95]

I don't know how to express a gratitude, this gyro was causing a head ache to me! Bloom! This makes it lot easier to understand and have imagination

[@Dr10Jeeps]

I am a university psychology professor who loves physics. Dr. Lewin is simply amazing in his ability to convey complex matters in an entertaining and (relatively) simple manner. On a lighter note , I want some of the MIT chalk! I also find myself in lectures wanting to say "I have here...."

[@mortezakhoshbin]

I HAVE NO WORDS TO Thank you. im about to finish your lectures for second time they are so intersting i really enjoy.thaaanks

[@lecturesbywalterlewin.they9259]

:)

[@CaptainCalculus]

At the 48:00...wouldn't the angular momentum of the wheel also provide stability, in exactly the same way a bicycle is stable only when it is moving?

[@idealthinker101]

Yes. But, imagine if the rope and wheel are both very smooth (zero friction coefficient), and the wheel only slided on the rope, Then too the rope-walker must be stable because of lowered Centre-of-mass.

[@Percussionfirema]

thank you professor Walter Lewin, it's a awesome lecture. your demonstration always inspire me. Thank you

[@pascalkayosde8357]

Goof job. Highly motivating. Creative,pragamatic and creative.....,.learning made easy! Another Richard Feyman

[@TheJohncouch]

Hello again! I am trying to understand why at 6:15 for the cross product you have the product of the magnitudes along with the cosine and NOT sine. Is there a simple trick I am missing here? I would have went with sine from the cross product formula

[@TheJohncouch]

I should clarify that I understand you explain that the cross products refer to the lengths in the same moment you are calculating the cross products, I just don’t see how that is consistent with the cross product formula. This is where my confusion lies

[@bastian8001]

The sine of the angle between 0,5l and Mg is the same as the cosine of alpha. So instead of 0,5l•Mg•sin(beta) you can also write 0,5l•Mg•cos(alpha), which you do because you only want one angle in your equation.

[@ashitbarank]

Sir your mesmerising presence makes us love Physics. I envy the students of MIT

[@idealthinker101]

To be specific - Students of 8.01, 8.02 and 8.03 courses.

[@vishalfgm]

being a jee aspirant it is very useful for me

[@rainerwahnsinn3262]

At 6:42 he lost a g in the process?! EDIT: He later noticed it at 7:46.

[@michael73931]

Dude lol I was about to ask this question.

[@thienthanhtranoan6723]

21:30, teacher, can you tell me the reasons why T2/T1 depend on how many turns of the rope, i don’t see it in the equations.

[@lecturesbywalterlewin.they9259]

if theta_o is 5 times 360 degrees then the rope goes 5 times around the rod.

[@pantheraonca3635]

Professor Lewin, at 16:56, is it true that friction can never exceed the maximum friction? You say that there comes a time when friction is larger than maximum friction.

[@lecturesbywalterlewin.they9259]

>>>You say that there comes a time when friction is larger than maximum friction.>>> I could not have said that. How many minutes into the lecture did I say that? You must have taken this out of context.

[@lecturesbywalterlewin.they9259]

I say "If the maximum friction goes up" That's very different from what you wrote. The max fr is N*mu and if N goes up the max fr goes up.

[@pantheraonca3635]

Can you please clarify the situation at 16:56?

[@lecturesbywalterlewin.they9259]

I listened to it again. Every thing I said is correct. Watch it again in the context of equations I refer to on the black board. I cannot improve on it.

[@pantheraonca3635]

What do you mean by (at 17:01) "now there comes a time that this force becomes larger than the maximum friction and then the ladder will start to slide". What exactly is "this force"? This is what is confusing me.

[@ismailsevimli6424]

How can we equalize cot(a)=2Us, this equation holds just when we don't have another object going along the ladder, but there we have another object and how can we substitute 2Us to its place and reach to the conclusion that d

[@lecturesbywalterlewin.they9259]

I cannot add to the clarity of my lecture - if you prefer to do it differently, be my guest

[@Rudra71124]

A great teacher, envy the blessed students who get to learn from him 🙏🙂

[@varunpatel1012]

At 39:00 I flattened on demonstration..🤩🤩

[@JinyuGuo]

Dear professor Walter, at 25:18, you mentioned in page 361 of the textbook, there is the derivation of T2/T1. I have read this part in the book, and I can understand the most part of the derivation. But a key step \int_{T_1}^{T_2} \frac{dT}{T} confuses me, what is the meaning of this step? Can you explain it to me? In my perspective, dT is the increment tension of a small segment, and the whole increment tension should be T2 -T1 = \int_{\theta_1}^{\theta_2}{\mu_sTd\theta}, am I right? but I can not link my equation with \int_{T_1}^{T_2} \frac{dT}{T} given in book. Could you please help me? thank you a lot, I am looking forward to see you reply, Best regards.

[@JingXu-d4n]

This is not the right way to solve for an integral from a differential equation...T in the equation is not a constant, you should not leave it alone when doing an integral.

[@xXxBladeStormxXx]

At 4:21 why is there no friction at point P if there is a normal force? Wouldn't there be a friction force in the upward direction? Edit: someone already asked this question 4 years ago and you answered. Thanks!

[@sundaymanali5854]

6:27 what happened to g?

[@sundaymanali5854]

nevermind keep watching ppl. dont get confused like me lol

[@hubertpierre9507]

Hartelijk dank voor een prachtige uitleg 👍

[@lecturesbywalterlewin.they9259]

graag gedaan

[@aida8263]

Why is it missing the normal force acted upon the person on the ladder at minute 15:16?

[@amologusmogusmogumogu2535]

3:17 What if there was friction at p ?

[@carultch]

If there were friction (i.e. traction) at point P, it would help keep the ladder from sliding, so the critical angle where it starts to slide would be lower. It is a conservative assumption to neglect friction at point P. The ladder's sliding angle is also more sensitive to the traction at its footing, than the traction at the top of the ladder.

[@SwordSeraph]

What does it mean for a torque to have a direction? I understand its mathematical significance in figuring out a torque's sign in the torque balance, but what does it mean physically for the torque to go out of the board? Does the torque generate an angular acceleration in that direction?

[@lecturesbywalterlewin.they9259]

A torque is a cross product. tau=rXF, r and F are vectors. Thus the torque is perpendicular to both r and F. Watch my lectures! Thus by convention if you rotate a corkscrew clockwise the torque is a vector pointing in the direction in which the screw moves as it goes into the cork.

[@semseddincolak6217]

Sword Seraph

[@paulproofmath323]

This is the most important question ever. The Prof I guess couldn't understand it.

[@cocoa1996]

Which book are they using in this class? I want to see the derivation at ~25:00

[@lecturesbywalterlewin.they9259]

8.01 Physics Hans C. Ohanian 2nd edition W.W. Norton & Company ISBN 0-393-95748-9 8.02 Physics for Scientists & Engineers by Douglas C. Giancoli. Prentice Hall Third Edition ISBN 0-13-021517-18 8.03 Vibrations and Waves by Anthony French CRC Press ISBN 9780748744473 8.03 Electromagnetic Vibrations, Waves and Radiation by Bekefi and Barrett. The MIT Press ISBN 0-262-52047-8

[@cocoa1996]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you so much professor :)

[@BroadeningHorizonsos]

at 24:14 why we divided rope in little parts to calculate friction as friction is independent of area of contact?

[@lecturesbywalterlewin.they9259]

Please try to understand this from my lecture. I cannot be more clear.

[@carultch]

Because the friction is applied continuously across the rope-cylinder contact. Every millimeter of rope will have progressively greater tension, as you track it from hold-side to load-side. You ultimately divide the contact arc length into an infinite number of parts, which means you'll use integral calculus, to get the cumulative result. The equation you are deriving is called the Capstan equation. Capstan being a nautical term for the simple machine used for lifting the anchor, which involves a friction drum to hold it as the crew spins the capstan.

[@hodsgod]

Why don't the students ever ask questions, is this just for video? Surely some students don't understand everything first time?

[@lecturesbywalterlewin.they9259]

they meet with recitation instructors (classes of 25 students) twice a week and can then ask any question.

[@stabmaa]

Horiander

[@amologusmogusmogumogu2535]

PLEASE HELP ......At 1:58 , he drew two dotted lines but only marked one of them as b , why ? Why isn't the other one important? 4:04 , WHY is ther a normal force nq , WHY does it exist ?

[@carultch]

For the line he marked as b, he is assigning the variable b to the perpendicular distance between the two equal and opposite forces (not to be confused with equal and opposite forces in N's 3rd law) acting on this object. The other distance is not relevant, because it is parallel to both forces. You could in concept produce a radius vector between the two forces at the points they are applied, and take a cross product accordingly, and you will get the same result as if we only considered the distance perpendicular to the two forces.

[@carultch]

The normal force exists at BOTH points P and Q in that problem, because if it weren't present, the ladder would crush through the wall and floor. The normal force is the force that is necessary to keep one object from penetrating through another object. Normal means perpendicular in the term "normal force".

[@surendrakverma555]

Excellent informative lecture . Thanks and Regards 🙏🙏🙏

[@Originalimoc]

I was afraid he'd say "Now, I want to demonstrate that to you." at 19:05 XD

[@manassrivastava7448]

Sir in 6:32 you made a derivation of critical angle at which ladder can stand without standing. I do not understand why did you take l/2 in mg.cos a.l/2. Please explain me

[@lecturesbywalterlewin.they9259]

I can't explain it any better than I did in class. The torque is rXF and that requires simple trigonometry

[@neartosun]

Hello dear Lewin . How i can find the page 361? 25:16

[@lecturesbywalterlewin.they9259]

8.01 Physics Hans C. Ohanian 2nd edition W.W. Norton & Company ISBN 0-393-95748-9

[@yash29210]

I have a newer version of the book that you have but I couldn't find what you told at 25:18.........

[@lecturesbywalterlewin.they9259]

:)

[@yash29210]

What??

[@nevenjakopovic2934]

I'm sure it's something very obvious, but I just can't see the geometry at (A) 5:45 and (similar!) at time (B) 6:15. Torque (vectors): τ = r x F; so magnitude is τ = r * F * sin(angle r->F) I drew the situation and looked at the angles (going completely "by the book", i.e. evaluating angles from the direction vector towards the force vector, I used sin(180-α) = sin(α) for QC and sin(90-α)=cos(α) for QP). I got the same result, but with a lot more work. Prof. Lewin just takes the orthogonal projection of r and I can't see: why does it work? I've re-watched the lecture on vectors again but it did not help me here.

[@nevenjakopovic2934]

Thank you, professor, I'll take a closer look why that is true.👍

[@lecturesbywalterlewin.they9259]

google cross products.

[@nevenjakopovic2934]

In hindsight, this is what I've probably should have done first. However, this wasn't tremendously helpful either since the common treatment is to just show the usual basics. We can use Google, but should not stop thinking about a problem. :) Perhaps the most clear way to see it is true is to evaluate that AxB is the area of the parallelogram with some angle θ. If this angle is changed to 90 degrees, we get a rectangle, and then its area (which is the same as of the parallelogram) is given by a vector multiplied by the orthogonal projection of the other vector.

[@omozaas]

Neven Jakopovic .... torque simply means that we should multiply .....perpendicular distance from "point Q" to the "Line of action of force".....with the force Tau = RFsin(angle).... simply means that multiply​ F with the perpendicular distance starting from selected point to the line of force F..

[@surendarkumarg1988]

Why a tightrope walker spreads her hands while walking on rope ? How it helps to increase her stability ?

[@lecturesbywalterlewin.they9259]

increase of moment of inertia increases stability. Even better would be to hold a long stick in your hand way longer than your arms. Many tightrope walkers do that.

[@md65000]

I would have liked to see ladder example go one step further: If you put a bunch of weight on the bottom step of a ladder, then can a person climb past the center of mass to the top without the ladder falling?

[@rakshitsohlot7461]

Professor So where should I refer for the derivation

[@rakshitsohlot7461]

Professor why did not you take the force exerted by the man on the ladder to walk up the ladder in the problem on 10:24

[@lecturesbywalterlewin.they9259]

*watch past **10:24*

[@rakshitsohlot7461]

Lectures by Walter Lewin. They will make you ♥ Physics. Professor I did watch past 10:24 but didnt understand may you explain a little bit

[@lecturesbywalterlewin.they9259]

I cannot add to the clarity of this lecture. Watch it again!

[@MrKurier1]

Hello Professor Lewin Can you explain me in one or two sentences what is difference between maximum friction force and just friction force? I understand the equations but in this problem I dont very see that, mainly in 16:10 when you said that friction force and maximum friction force are higher... well if friction force is higher doesnt that mean that the ladder need to overcome more friction force so it is more stable .... ?

[@lecturesbywalterlewin.they9259]

If you place an object on an incline and you slowly increase the angle of the incline. The object will at first stay put. That means the frictional force balances the component of gravity along the slope. Increase the angle further and it still stays put. That means the frictional force balances the component of gravity along the slope. The frictional force has been growing and growing. There comes a time that this force is mu*N, then it can no longer get any larger. That is the maximum frictional force.

[@MrKurier1]

Thank you

[@intomd3730]

Hi professor. How can I see the derivation at minute 25:00?

[@lecturesbywalterlewin.they9259]

+manuel di nucci As I mention, it's in our book. Ohanian.

[@vishnugovind4634]

sir in 10.01 if angle is less the friction will be more than how can object fall (friction oposes motion)?

[@arjunvg4719]

24:33 sir where can I find the derivation of this equation ?

[@lecturesbywalterlewin.they9259]

it' in the book I used for the course. 8.01 Physics Hans C. Ohanian Physics Volume 1 2nd edition W.W. Norton & Company ISBN 0-393-95748-9

[@sharudeva]

What would have happened if the bicycle wheel (after giving a good spin with the motor) had been suspended onto the rope horizontally ( plane of wheel parallel to floor) ?

[@sharudeva]

Will it move like the mass in this pic ? tinyurl.com/z8d9ub2 // the mass represents the bicycle wheel.

[@lecturesbywalterlewin.they9259]

you should be able to answer this on your own.

[@sharudeva]

tinyurl.com/z8d9ub2 is this correct ?

[@lecturesbywalterlewin.they9259]

tell me in words. I prefer not to open websites

[@sharudeva]

Will it move like a conical pendulum, with the plane of the bicycle parallel to floor ?

[@lakshitkava]

Sir can i get those lecture notes(handouts) so it may help me to prove thise result

[@hushhushbee1647]

notes link is in description

[@prakharbhalla9461]

sir, if net force on body is zero but there is net torque not equal to zero then this net torque at 2:15 will be equal to (moment of inertia about center of mass *angular acceleration) which is same about any point we choose. is it true?

[@lecturesbywalterlewin.they9259]

net force zero. net torque not zero the object will start to rotate without translating.

[@prakharbhalla9461]

but what will be the angular acceleration. will it be (b*f)/moment of inertia about center of mass.

[@lecturesbywalterlewin.they9259]

if the object is not restrained, but completely free to move it will start to rotate about its center of mass. tau=I*alpha alpha =dw/dt, I is moment of inertia about the center of mass. watch my lectures.

[@prakharbhalla9461]

which lecture should i go through again sir as i have watched all your lectures and thanks for the reply.

[@chetankochar3056]

Professor, What are those rings you wear? I also saw them in an interview of you.

[@simpletiwari362]

Can anyone tell me the derivation of the relation between Ti and T2 in rod example of advantage of friction.

[@thelilbronco]

Hallo Professor, First thank you for putting these lectures online. These are really brillant. I just have a question. In lecture 8 showed that the fricton is indipendend of the mass and the surface area. But why does the frictional force increase with the number of roatiations around the pole? Is there another effect beside that you increase surface of frictional Touch? Because you use the coloumb friction coeifficent, i guess that there is the same problem as lecture 8. (Not like the tires) Thank you :)

[@dzonnigrf]

Hello, professor. Why isn't there a vertical component in the point P (the ladder example)?

[@lecturesbywalterlewin.they9259]

how many minutes into the lecture?

[@dzonnigrf]

4:30 Why is there no friction at point P? I don't see how the reaction force of the wall can only have horizontal component.

[@lecturesbywalterlewin.they9259]

I have assumed here that the wall is frictionless. Be my guest and add friction nd solve the problem again. It's not very difficult at all.

[@dzonnigrf]

Lectures by Walter Lewin. They will make you ♥ Physics. Thanks, professor.

[@papugamer7686]

Is there a mistake at 6:05 or are my eyes mistaking after watching so many of your lectures in a row. Because you said there its in the blackboard so its positive but you marked negative (-).

[@lecturesbywalterlewin.they9259]

you are mistaken

[@dr.sciencesc.d3088]

Hello Dr. Lewin, in the static equilibrium example with the person walking up the ladder, how come we do not consider the normal force from the contact between the person and the ladder or is this normal force combined with the normal force the ladder makes with the ground?

[@lecturesbywalterlewin.they9259]

+Dr. Science Sc.D If we want to know whether the ladder starts slipping we should only take into account forces that act on the ladder. That's the power of "free-body" diagrams.

[@dr.sciencesc.d3088]

Ah, thank you! The normal force acts on the person and not the ladder itself

[@prakharbhalla9461]

for a rectangular object in vertical plane if a small force is applied on top of it then a equal but opposite friction will act on the bottom surface. now to cancel the torque of these two forces the normal force from ground should shift parallel in the direction of force. Is this the way phenomenon occur or not?

[@lecturesbywalterlewin.they9259]

yes

[@prakharbhalla9461]

Thank you very much sir.

[@abdullahalsakka]

24:00 where can I find the derivation sir. Cause I don’t have the book 😪😪

[@lecturesbywalterlewin.they9259]

try to search the web

[@mfiorillo9191]

kzbin.info/www/bejne/fmG3aKBraJyMipI

[@tanmaychavan2583]

@@mfiorillo9191 Thanks :)

[@gauravdimri8009]

There is no silly question so please can you explain sir as why a body will be balanced if the center of mass lies below the point of suspension?

[@lecturesbywalterlewin.they9259]

question is not well defined. A pendulum at angle > zero degrees with mass m is not in balance.

[@ahmedhosny4755]

what does (e) refer to in the equa T1/T2=e^mu theta?

[@ahmedhosny4755]

؟؟

[@efeguleroglu]

euler's number (2.71828...)

[@praneethrajula1809]

In the Formula for balancing weight, should I take pi as 3.14 or angle in degree or radian? Please clear my doubt sir

[@carultch]

If pi is involved in an angle, it usually is a measurement in radians. It is common for formulas to only work in radians as the preferred angle unit. For instance the s=theta*r formula for arc length, and the Capstan equation for tension on both sides of a fixed drum.

[@commentor6439]

can't afford your book sir, where can I start for the tension 1 and 2 derivations

[@lecturesbywalterlewin.they9259]

try google

[@commentor6439]

Lectures by Walter Lewin. They will make you ♥ Physics. what is the device called so I can search?

[@vijayr1485]

I have found the Wikipedia page en.wikipedia.org/wiki/Capstan_equation, but the derivation is a little confusing. I'd like to hear your take on it Professor.

[@gprimeofx]

As a former sailor, this lecture gave me severe flashbacks 😄

[@johnbruhling8018]

Never cut toward yourself, always cut away!

[@johnbruhling8018]

(The irony does not escape me however, the lecture describes basically how it happened but still, safety first!)

[@johnbruhling8018]

The knife in your hand and in the pencil is just like the ladder. The 'digging in' is friction and the knife in hand experiences only that resistance. The angle that you create as you push the knife in is the sum of the vectors and changing the angle and force at which you push eventually overcame the maximum allowable (F of f > F of f max). (As always thank you for your lectures)

[@rakshitsohlot7461]

Hello professor which book did you refer in the video which edition

[@lecturesbywalterlewin.they9259]

8.01 Physics Hans C. Ohanian 2nd edition W.W. Norton & Company ISBN 0-393-95748-9

[@rakshitsohlot7461]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you very much

[@terra4837]

Hi Prof- If an object has net Forces = 0, net Torques = 0, but has a constant velocity, is that object in static equilibrium? Or does the object need to have v = 0 to be in static equilibrium? Thank you

[@lecturesbywalterlewin.they9259]

>>>is that object in static equilibrium?>>> yes it is in static equilibrium in the frame of the moving object with constant velocity

[@terra4837]

Thank you Professor!

[@michelm.6704]

I love these lectures ! But I have a question about the rotational equilibrium condition : it simply seems very artificial to me that the rotational equilibrium of an object is dictated by the fact that the sum of all the forces times their distances( taken perpendicularly to the forces) to a specific point has to be zero for every point in space . I understand it is an experimental fact , but is there any way of proving it only based on theory ? I understand that torque is proportional to angular acceleration and so if we make sure for an object , that the total Torque that is acting upon him is zero relative to any point in space , it means that the angular accelerations relative to all these points must also be zero , and so rotational equilibrium is reached .

[@lecturesbywalterlewin.they9259]

question unclear. How many minutes into the lecture? also simplify your question.

[@lecturesbywalterlewin.they9259]

What I wrote early in my lecture is correct. Static equilibrium means that the vectorial sum of all forces and all torques are zero. The object can then still have a constant velocity, v, along a straight line. But I can always choose a reference frame with that same velocity v. The object can then still rotate in this reference frame. To keep it rotating requires no torque and no force under ideal conditions. We call this therefore "static equilibrium"; we do not call it "rotational equilibrium". However, in the reference frame of the rotating object there would also be "rotational equilibrium".

[@michelm.6704]

@@lecturesbywalterlewin.they9259 Dank u wel , professor , it's kind of you ;)

[@govindaraj726]

@@lecturesbywalterlewin.they9259 Sir did you teach only 8.01,8.02,8.03 course in mit, if you taught more where can I find those precious lectures

[@rakshitsohlot7461]

Professor to which chapter did you refer to because I have 3rd edition not 2nd

[@lecturesbywalterlewin.they9259]

You should be able to figure this out - read the index!

[@khalednawaz8964]

professor i am civil engineer, pakistan. Apart from construction engineering , i also teach engineering mechanics. Once i was ssearching for topic on youtube and i came across your lecs. I must say it was astononishing and i am planning to do MS in physics.

[@prakharbhalla9461]

sir if large number of forces are being applied on the body. then is there any way to predict the point about which body will rotate assuming that there is resultant force not equal to zero. this type of problem will arise when man on ladder will climb just more than half of the length. then about which point ladder will rotate and will its center of mass also move in direction away perpendicular to wall?

[@lecturesbywalterlewin.they9259]

any object that you throw up in the sky will rotate about its center of mass and the center of mass will have a parabolic trajectory. Watch my lectures

[@prakharbhalla9461]

Sir in case of ladder against the wall question when man climbs distance just greater than half the length of ladder then normal reaction from wall becomes a infinitesimal larger than friction from ground. Hence a result force is perpendicular away from wall then centre of mass should move in the direction of force but it also moves down towards ground. How? Thanks in advance

[@lecturesbywalterlewin.they9259]

If the sum of all forces and the sum of all torques (relative to nay point) is zero. Their is static equilibrium. If that is not the case the object will move.

[@prakharbhalla9461]

i got that from your lecture sir but my question is suppose a rod is vertically aligned at an angle with ground as / then is there a reason why it rotate about point of contact of ground and rod. And not about any other point. because torque about any point on rod would be same for such situation.

[@lecturesbywalterlewin.they9259]

question unclear

[@d4dbeh]

which textbook that was sued for this course?

[@rakshitsohlot7461]

I referred to the index and chapter 'statics and elasticity' but didn't find the derivation

[@lecturesbywalterlewin.they9259]

perhaps they have deleted it

[@rakshitsohlot7461]

Lectures by Walter Lewin. They will make you ♥ Physics. Where should I refer for the derivation

[@nileshrathod3153]

what will happen if Fs is greater than Ff max? why the condition can't be Fs >= Ff max?

[@nanor8921]

nilesh rathod Fs cannot be greater than Fmax since Fmax is the full force of the friction. If Fs had to be greater than Fmax it would mean that Np is creating a large force, and Fs cannot compensate so the ladder will fall

[@SaritaYadav-ye4eu]

Thank u sir I am in High school (class 10th) and I am aspiring to get in at MIT ur lectures are really helpful

[@abhishekcherath2323]

Sir what's in that cool red ring?

[@lecturesbywalterlewin.they9259]

hidden power

[@thegreatest1948]

Where can I download the book of this course ?

[@lecturesbywalterlewin.they9259]

not possible

[@Raphael_NYC]

Thank you Dr. Lewin. raphael santore

[@trishitchandra1549]

Sir, I have a question about a toy of mine which is a good example of static equilibrium. But still I am not able to find the mystery of the physics of that toy. And I am trying through various social communication platforms to show you the video of that toy. But nothing happened with those contact information. So please help me to reach you and find the mystery and I'm pretty sure that you would like it too.

[@lecturesbywalterlewin.they9259]

this KZbin channel is the only way for us to communicate.

[@trishitchandra1549]

okay. Actually it is about a spinning top which I am talking about. So if I post a link of the video of that spinning top then will you be able to check that video?

[@trishitchandra1549]

kzbin.info/www/bejne/goq6qJqJoLmGpMk this is the link of the video of that top. Please kindly check the video sir and help me out.

[@lecturesbywalterlewin.they9259]

this behaviour is discussed in great detail on line. Use google

[@trishitchandra1549]

okay sir.

[@souvikmahato]

12:36 , i thought we have to consider the normal force applied by the man on the ladder, instead of its full weight ! thought it would become complex !😅

[@DemiHalf]

So do you guys just understand how the whole three equations give us a max angle even with wall friction in the textbook or just ignore the fact that idk how to prove this fully and move on?

[@ManojKumar-cj7oj]

That ropwalking thing drived me crezy like helllllll,oh my god🎩🎩🎩🎩

[@anuguarun911]

sir i had a doubt regarding statics A frictionless surface is in the shape of a function which has its endpoints at the same height but is otherwise arbitrary. A chain of uniform mass per unit Figure 1.10 length rests on this surface (from end to end; see Fig. 1.10). Show that the chain will not move. this is the question taken from the book david morin introduction to classical mechanics proble 3 chapter 1 could u pls solve it for me this is not my homework

[@lecturesbywalterlewin.they9259]

I do not solve problems for viewers. I teach Physics. Watch my 8.01 lectures and you will be able to do this easy problem

[@pierocahuanovera7899]

Hi proffesor Wlter Lewin, you use twitter?

[@timetochangetot4094]

Be healthy sir.

[@akramal-khazzar5450]

the guy @30:00 like a robot if mr lewin asked me something i ll jump of my chair running to do it

[@aman2426]

I love the vibe.

[@amologusmogusmogumogu2535]

Thank you sir

[@giantmiller1136]

Ik hoop dat u nog een beetje Nederlands kent ;-) ---> Waarom is er bij Ff max al geen rekening gehouden met de afstand waar de persoon zich op de trap bevind (d)? Immers aan het begin van de trap komt de massa van de persoon geheel op punt Q, dus dan kan de gehele massa wel worden meegenomen (M+m) maar waarom niet halverwege de trap een half (of zoiets afhankelijk van de hoek) maal m nemen? Zal misschien uiteindelijk niets uitmaken, maar toch....?

[@lecturesbywalterlewin.they9259]

Hoeveel minuten in de lecture" Mijn Nederlands is na 50 jaar VS nog perfect luister maar eens kzbin.info/www/bejne/b4C2hp2ep7Oaj7c

[@giantmiller1136]

U bent inderdaad nog goed te verstaan, je hoort 50 jaar afwezigheid echt niet :-) Is dat ook zo met het schrijven en lezen? In minuut 12/13 staat de vergelijking.

[@lecturesbywalterlewin.they9259]

Ff _max is uitsluitend bepaald door N_Q afstand van de persoon komt niet voor in N_Q F_f moet dus kleiner of gelijk zijn aan F_max Ik begrijp echt niet wat je bedoelt.

[@lols5761]

This is very similar to Irodov's PROBLEMS IN GENERAL PHYSICS 1.93..

[@PrinceKumar-ig5oc]

Thanks sir

[@PrinceKumar-ig5oc]

From india