Physics - Mechanics: The Pulley (2 of 2)

[@aidedominguez7141]

You are the best! Seriously, my teacher goes too fast and I pay attention but don't understand much. It is nice to know that I can count on you as a support. Thank you so much!!!

[@ggg9gg]

I play this at 2x the speed

[@eadgbeskate]

these videos are undisputedly the best mathematics and physics videos on youtube

[@mcamaro7]

Thank you, this really helped me understand Atwood machine and pulley problems in general.

[@talashka88]

first technique is more comfortable, Thank you Michel!

[@kendronkoay5982]

talashka88 iiiii

[@superspike769]

Ye i agree. Second one is far too long and confusing

[@Hemantasati]

Likes should be in millions ... Nice Explanation .. 👍

[@wint1964]

My left ear learnt a lot! ty

[@MichelvanBiezen]

😂

[@flip2shoot567]

Finally it makes sense, great video :D

[@harithakassireddy6412]

I like all of your explanations they are knowledgeable and interesting as well.

[@andsn537]

Why doesn't 'g', acceleration due to gravity not a negative value i.e. -9.8m/s2. In this case, it is positive, why?

[@MichelvanBiezen]

Here we are calculating the magnitude of the tension and we are using the magnitude of the vector g. The magnitude of a vector cannot be negative.

[@canibotlin8302]

This deserves to be at the top, I was confused on this exact concept.

[@peterkovinski8476]

Lost when T1 = T2 and a1 = -a2

[@MichelvanBiezen]

The direction of the tensions are equal in both free body diagrams and the accelerations are in different directions for both free body diagrams

[@hammadforever1185]

sir i am watching this amazing video in 2024 which is after 11yrs of release and this is amazing

[@MichelvanBiezen]

Thank you!

[@JacksonChishimba-n5u]

Wonderful teaching

[@MichelvanBiezen]

Glad you think so!

[@sreelatharaghu1748]

this, i find is the better method for solving the problem

[@DonatoGreco]

nice crystal clear explanation, a pity the single channel audio while hearing on headphones ...

[@MichelvanBiezen]

Thank you. Yes, our older videos were filmed in mono. We fixed that in our newer videos.

[@DonatoGreco]

indeed, I found you corrected it, a great improvement for these juicy lessons ;-)

[@originalpost5465]

8g-T=8a , and T-5g=5a, so adding both 3g =13a so a=3g/13 simple problem dont make it compicated teacher

[@varunkumar7237]

dont watch then if you are PHD yourself

[@SilaTemane]

You are the best

[@arnavdinker393]

Thank you sir u r the best physics teacher

[@RedmiNote-xc9ls]

Dear professor Michel. I'd like to ask you that in the above solved question instead of T1 - m1g = m1a it should be m1g - T1 = m1a as the 8kg block is moving downwards and hence m1g will be greater than T1??

[@MichelvanBiezen]

When using free body diagrams we take into account direction and thus T1 - m1g = m1a will result in a negative a. (The video is correct)

[@RedmiNote-xc9ls]

@@MichelvanBiezen okay sir thank you for guiding me.

[@ChayaWagschal]

In the problem "Physics-Pulley System on a Table" the heavier mass was considered positive (even though it was going down), since it was aiding acceleration, and the lighter mass was negative because it was opposing acceleration. Why in this problem is positive acceleration the mass going up, and negative acceleration, the mass going down?

[@MichelvanBiezen]

This problem is solved using the free body diagram technique. In this case you strictly us up as positive and down as negative. Then you solve the problem accordingly. However note that the pulley redirects the direction of the tension of the string connecting the two masses. (That is why I like the other method better).

[@ChayaWagschal]

thank you!!

[@vitusbermundo-diva5578]

Thank you for this. I understood the topic much more clearly.

[@MichelvanBiezen]

Glad it was helpful!

[@XxWH0M3xX]

Since we already know that m1>m2 and therefore a1 is negative and a2 positive, why is this not considered in the equations? Shouldn't the equation then be T1-m1*g=m1*(-a1) ?

[@MichelvanBiezen]

There are several different ways in which you can solve a problem like this. One way is the way you describe. You draw a free body diagram on each object, and set up an equation for each. Then solve them simultaneously. Or you can ignore all internal forces and solve the whole system as one body as shown in the video.

[@nathanielreyes727]

Thank you very much. ❤ It was a really good explanation. 😊

[@MichelvanBiezen]

You're welcome 😊

[@_nanaamakwansima_]

Thanks!! I think i prefer this method to the previous one.

[@aljoschakainz3365]

great video thank you

[@MichelvanBiezen]

Glad you enjoyed it

[@RK-jb6rq]

Thanks..... This helped me a lot

[@dramram3653]

Nice solution

[@MichelvanBiezen]

Glad you liked it.

[@varunkumar7237]

You sir earned another sub , thanks for your explanation

[@johnhippisley9106]

Why is T1 = T2? Aren't T1 = m2(g) and T2 = m1(g)? m2 /= m1 therefore they are cannot be equivalent. I know this is incorrect, however, I just can't wrap my head around why this is so.

[@MichelvanBiezen]

Watch some of the other pulley problems where the tensions are calculated and shown to be equal.

[@Leonardocn98]

Prof. I think the correct formula is a1-a2=0, therefore a1=a2. Isn’t it?

[@MichelvanBiezen]

If we are using the magnitude of vectors, you are correct. But when we us free body diagrams to solve the problem, we must use the directions as well and then a1 = - a2

[@BalaCliffDezjnr]

So in this case, without friction, T1 = T2 also a1 = -(a2)…

[@MichelvanBiezen]

That is correct.

[@Akki420ish]

Excellent sir!

[@lalalia7724]

this man saved me lmao

[@naeemghafori5046]

Sir, what is the difference between this problem and the problem you solved before ??

[@MichelvanBiezen]

+Naeem Hakimi Similar problem, but different method of solving the problem.

[@chandrakalasathivada111]

Superb explanation sir

[@MichelvanBiezen]

Thank you

[@mwont]

Your are true master! Thank you.

[@MichelvanBiezen]

You are very welcome

[@boubinijones-wonni3865]

Thank you for the video. I have just one question: why is T1 = T2 when there are different masses attached to each end of the string. Shouldn't the string carrying the heavier mass have more tension to balance it out?

[@MichelvanBiezen]

Unless the pulley has mass or friction, the tension must be the same throughout the whole string. (There must be some force that acts on one part and not the other for the tension to be different) See the other examples.

[@TheDRUTTA]

what ih the pulley has its own weight?

[@carultch]

@@TheDRUTTA If the pulley has its own mass that is significant, then you would need to know its radius and its distribution of mass to find its rotational inertia (aka moment of inertia). You then would have two different tensions, and the difference between the two tensions, multiplied by the radius, would be the torque on the pulley. Assuming negligible bearing friction, this torque is also the net torque on the pulley, that when divided by the moment of inertia, will give you the angular acceleration. Relate angular acceleration to tangential linear acceleration to find how the pulley's kinematics relate to the kinematics of the hanging masses. Wikipedia has a derivation of this problem for a pulley with inertia and friction: en.wikipedia.org/wiki/Atwood_machine#Equations_for_a_pulley_with_inertia_and_friction

[@symbiotezilla12345]

Jazakullah Kair!

[@beesalazar2415]

Never thought I'd say this but math is cool. And so are pulleys. Is there an equation yo use for a 5 to 1 pulley. This is for 1 to 1 pulley with a change in direction?

[@MichelvanBiezen]

Look in this playlist for multi-pulley systems: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS

[@johnfei434]

KZbin helps me more than the school does.

[@didiershema9833]

Thanks alot

[@MichelvanBiezen]

Most welcome

[@metturamadevi3459]

Why would we equate both tension

[@MichelvanBiezen]

To show that they are equal.

[@vyoma1998]

this video was sooo helpful!!!! Thank you so much!!!!

[@lakshanar.h.7647]

Extremely useful thank you!

[@MichelvanBiezen]

Glad it was helpful!

[@tashfeenahmed6255]

Sir can you please explain the relationship of height and force of a mass suspended by a pulley...Meaning if two different a weights are hung the larger weight will be closer to ground but final they both stop moving.why is that ? Should the larger mass just go down and down until it reaches ground?

[@MichelvanBiezen]

The larger mass will keep moving until it reaches the ground (or until the small masses crashes into the pulley).

[@bekuma6609]

Nice

[@MichelvanBiezen]

Thanks

[@stevekim5233]

In your personal opinion, do you prefer force diagrams looking at the forces as a whole, or free body diagrams looking at forces individually? My ap physics teacher likes the free body diagrams, but I was wondering what you prefer

[@MichelvanBiezen]

They are both good techniques, but if all you are looking for is the acceleration of the system, looking at the system as a whole is a lot easier.

[@stevekim5233]

Then when would you say is a "better" time to use the free body diagram technique?

[@MichelvanBiezen]

When you are required to find the tension between 2 of the masses, that can only be done by drawing the free body diagrams.

[@beesalazar2415]

Thank you will do!

[@LincolnLian]

how to calculate the length of mass 1, 2 or total length of the string

[@brian12463]

Why can you assume that T1 = T2 ?

[@MichelvanBiezen]

If the pulley does not have mass, then it requires no work to make it turn, and then no energy is lost, therefore the torque on both sides must be the same and therefore the tension on both sides must be the same.

[@lyphing2482]

But what if they didn't tell us the mass of both m1 and m2??

[@lyphing2482]

But they tell us that m2 us bigger than m1

[@MichelvanBiezen]

Work the problem the exact same way, but use m1 and m2 instead of the numbers given. Also know that the heavier object will move downward and the lighter object will move upward.

[@ch0vits]

Thanks

[@najeemnafla2989]

Why T 1= T 2 pls tell the reason using taw of pully

[@MichelvanBiezen]

If the pulley doesn't have any friction, or any mass, then the pulley cannot produce any force to cause the tension to be greater on one side compared to the other side.

[@ashkelsproductions1]

Hi I am taking a physics class and we are trying to find the acceleration of an object as it falls from a pully system. There is only one mass, we are supposed to be holding the other side and then we let go of it. I was wondering if you had any hits how to solve this problem? Please let me know! I think this video is close to what we are supposed to be doing but also different since we only have one mass. A thin thread is wrapped around a pulley of moment of inertia I and radius R, and is connected to an object of mass M. You release the mass from rest, and I would like to know the acceleration of the object as it falls, in terms of M, I, R, g.

[@MichelvanBiezen]

Natalie Clarke Take a look in the playlist: PHYSICS 11 ROTATIONAL MOTION The last problem is like the one you describe (a yo-yo)

[@Trancecend]

+Natalie Clarke If there is no counter weight, the only acting force is gravity... if I understand your problem correctly.

[@gomezalejandrog]

was able to get the same answer not including T? I suppose that the only reason you'd use this longer method is only if you need to solve for the tension?

[@MichelvanBiezen]

+Alejandro Gomez There are different methods and it is a good idea to learn all of them. Here we use the method of free body diagrams. Compare that to the other method we use in the other videos.

[@davide7931]

thank you !!!

[@vansf3433]

The correct name is force diagram, but not free body diagram

[@MichelvanBiezen]

A free body diagram and a force diagram are one and the same, except that a free body diagram specifically isolates a single member or a section of the whole system.

[@vansf3433]

The word "free body" are inappropriate for such an expression because all are about interactions among forces. and a system can be a single particle or object or many of them What is the most important in science is how to use the language clearly and unambiguously and logical reasoning based on objective and undeniable facts which are the closest to being parallel to objective reality, The phrase"free body" can also means other different things than the meanings in the context of physics which is ambitious or vague So when using the word " force " it gives you a direct meaning without any ambiguity

[@Jonathan-vx7xi]

Hi, just one question...at 5:58 you started replacing some of the units from the other side..how can we decide which ones to change?

[@MichelvanBiezen]

I am not sure i understand your question. But if a1 = -a2 then a1 can be replaced by -a2 and if T1 = T2 then T1 can be replaced by T2

[@Jonathan-vx7xi]

@@MichelvanBiezen Thank you

[@prathamkundan6473]

Thanks a lot.

[@arunfz]

sir , i have one doubt . in this video u said T1=T2 as pulley do not have mass . i am confused with this . how this pulley will rotate if tension in both sides are equal . what gives its ability to rotate ? do u mean the string is just sliding over the pulley(no rotation ) as there is no friction between string and pulley ? i am confused with the same tension concept .

[@MichelvanBiezen]

The pulley will rotate. But since the pulley "has no mass" the pulley does not offer any resistance to the motion or the acceleration, and therefore the tension in the string must be the same on both sides of the pulley. If you look at each free body diagram, since m1g is greater than T1, block m1 will accelerate downward. Since T2 is greater than m2g, mass m2 will accelerate upward and yes, T1 = T2.

[@arunfz]

then what causes the pulley to rotate ? from where it gets the turning effect ?

[@MichelvanBiezen]

The friction between the string and the pulley.

[@DeniseP]

Why is a2 negative if it's going up? at 5:08 (im assuming up is +ve and down is -ve

[@MichelvanBiezen]

It depends on how you want to solve the problem, especially with systems were the components are linked together and accelerate in different directions. So with systems like the Atwood machine, you can call the direction of the acceleration positive which means up for one mass and down for the other mass.

[@DeniseP]

@@MichelvanBiezen Thank you for your prompt response!

[@trevordfan4954]

is there a video explaining how i would find the weight of the system on a spring scale when the system is accelerating

[@MichelvanBiezen]

Take a look at this playlist: PHYSICS 17 WEIGHT AND TENSION

[@trevordfan4954]

Michel van Biezen thank you!

[@superspike769]

I dont see the point of this method. Its much more complicated and takes alot longer. The first method in your previous video works alot better and is easy to understand and takes 10x less time to do.

[@MichelvanBiezen]

It is a technique we should learn how to do as well, but yes the other method is much easier and faster (which is why we teach that method).

[@mbaizisilas6937]

Great

[@victoriacorcimaru1731]

Can anybody please explain why T1 is equal to T2?

[@MichelvanBiezen]

If the pulley has no friction and the pulley has no mass (eliminating the need to take into account the moment of inertia of the pulley), then the pulley only applies a force perpendicular to the string, and therefore cannot add tension to the pulley and therefor the tension must be the same on both sides.

[@victoriacorcimaru1731]

@@MichelvanBiezen Thank you! I noticed this issue of mass influencing the tensions in the rope in another example too: two accelerating blocks (m1 and m2) on a flat surface being connected by a non-massless string. The textbook says that T1 and T2 for the blocks are different. Any ideas why? And does this somehow relate to your previous reply? Thank you! Your videos are the BEST!

[@MichelvanBiezen]

If the string is non-massless, the the tension will gradually increase as there is more mass to pull with each increase in string length. Most problems will use "massless" strings so we don't have to account for that.

[@victoriacorcimaru1731]

@@MichelvanBiezen Thank you, professor!!!

[@juan125873a]

is it just me or is the technique different than statics... this had me very confused

[@carultch]

Since acceleration is involved, you cannot solve this problem with statics alone. You could in concept apply inertial forces (aka D'Alembert Forces) to each object, equal to the negative of their masses * their accelerations, if you choose to do so, and then apply statics principles. But this is a confusing thing to set up, and distracts from the Physical principles that are really involved

[@yomama78ify]

shouldnt m1g be greater than the tension?

[@MichelvanBiezen]

+MunkresdoesMC It is. (a1 is negative)

[@yomama78ify]

oh, i see now, thank you

[@SadSocks]

Dear professor Biezen, I'd like to ask you a question. I discovered that the tension T3 that sustains the whole system is 2T, correct me if I'm wrong. Why is it so? Why is it equal to m1g + m2g only when the system is at equilibrium and not always?

[@MichelvanBiezen]

There is only T1 and T2. (There is no T3).

[@alvmech]

я вижу одну ошибку - а2 равен «минус» 2,26 не «плюс» 2,26 м в секунду в квадрате

[@antonandreev8543]

Hello. Could you please explain why a1 is -2.26 and not 2.26? Thnx.

[@MichelvanBiezen]

Yes, it is confusing with the red a2 arrow drawn the way it is. The "negative" for the acceleration of m2 comes from letting a1 be positive relative to the red a1 arrow drawn downward. If a1 is "positive" in the down direction then a2 is "negative" in the up direction. (I should not have drawn the a2 arrow)

[@Mcthindi]

#100comment Sir my question is if weight on both side is same & what will be T ? Is it m1+m2 ?

[@MichelvanBiezen]

If the weight is the same on both sides, then there will not be any acceleration and then the tension = mg (m1 = m2 = m)

[@lukemassa3526]

thanks :)

[@MichelvanBiezen]

Welcome!

[@MrDoYouWannaBeOnTop]

what if the pulley has a mass?

[@MichelvanBiezen]

+MrDoYouWannaBeOnTop Then you work the problem as explained in this video Physics - Mechanics: Application of Moment of Inertia and Angular Acceleration (2 of 2)

[@MrDoYouWannaBeOnTop]

ok. thanks!

[@supere872]

why is a1 & a2 assumed to be a vector while g is not?

[@MichelvanBiezen]

When working with free body diagrams, it is important to keep track of the directions as was done in this example. g is just a constant (like the magnitude of a vector) and thus must be positive. When used in certain contexts, (like in the equations of kinematics) it can be given a negative direction.

[@andrewfranklin3242]

I am confused as to why a1 and g have opposite signs if they are in the same direction. Should you have used -9.8 when solving for a2 since a2 and g are in opposite directions?

[@carloschuecos7014]

How would you solve this problem if the whole actual system is affected by another mass "m" at the extreme of another string wrapped to another fix pulley? Thanks

[@myname-pe2pe]

what if your solving for a pulley with mass what should you do then

[@myname-pe2pe]

would I have to use moment of inertia I didn't know if that would work or not since no pulley is actually a cylinder because of the room on the side to hold the rope

[@MichelvanBiezen]

+chuck Ramsey Take a look at the following playlist. You'll find those types of videos that you are looking for: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2

[@myname-pe2pe]

+Michel van Biezen thanks, your videos are very helpful

[@sebastianmse]

Hello! Thank you very much for your videos! they really help me in many ways, but i have this question, how to calculate the Tension in the horizontal line of this pulley system, (sorry about the diagram, didn't know how to attach a pic) Many thanks!!!! T=? fix point =? O-------------------------0 fix point =? I I I I T=98N T=98N I I I mass= 10kg Thank you very much!

[@MichelvanBiezen]

+Sebastian Amerise It turns out, that if the pulleys are "ideal" pulleys and don't have mass or friction, the tension is the same everywhere in the string and thus the horizontal part also has a tension of 98 N

[@sebastianmse]

Thank you very much! so now learn how to calculate the friction and check the mass of my pulleys! Thanks!!!!!!

[@ny6u]

freebody diagrams are easier

[@MichelvanBiezen]

I prefer the other method. :)

[@priyanshupriyam5024]

Wow

[@tomhoward6609]

OMG IM FAMOUS

[@tyriquepillay4226]

thanks

[@MichelvanBiezen]

You're welcome!

[@SivasakthiTraders]

Thanks a lot

[@MichelvanBiezen]

You are welcome.