Physics - Mechanics: Applications of Newton's Second Law (11 of 20) accelerating pulley

[@novemberftblue6937]

In the video , professor used the method of Free body diagram, so the upward direction is always positive. For those who wondered : m2g-T=m2a2 ( you used the method which the direction of acceleration of the system follows the movement of (m2). So, here is your full answer: -System (m1): T - (m1)g =(m1)(a1). (1) -System (m2): (m2)g - T = (m2)(a2). (2) (1) +(2) we get: g[(m2) - (m1)] = (m1)(a1) + (m2)(a2). (3) Suppose (A) is the acceleration of (m1) and (m2) before applying the acceleration(a) ~(A) - (a) = (a2) {(m2) moves slower when the system is affected by (a). ( opposite direction)} ~(A) + (a) = (a1) {(m1) moves faster when the system is affected by (a). (same direction)} => (a2) = (a1) - 2(a) (4) substitute (a2) into equation (3) we get: g[(m2)-(m1)] = (m1)(a1) + (m2)(a1) - 2(m2)(a) Hence, (a1) = {1/[(m1)+(m2)} x { g[(m2) - (m1)]+2(a)(m2)} (PROVED) To find (a2), simply substitute (a1) to equation(4) which the result is positive. ( the result in the video is negative just to show the direction of (a2) relative to (a). Then you have it.

[@monkey123491]

Holy shit thank you so much I thought I was losing my mind. When 1 comment makes more sense than a whole 15 minute video.

[@aqilaafarwin7763]

Thank you😊🤗👍

[@naz-tl4uj]

allah razı olsun kardeşim🙏

[@QuyNguyen-pv1fg]

Nice

[@luisenriqueerciasueiro3573]

When i am looking for a topic with some extra complexity you are always there to save the day! Thanks sir!

[@BeMedic]

I think what has got a few people confused is that a1 and a2 are used in two different places. a2 = -a1 and then a2 = -a1 + 2a. Both of these cannot be correct, unless a = 0. A double subscript system might be used in this case. Great video nonetheless!

[@joeyborja423]

Personal note: why a2 = -a1 + 2a. We know a2 = -a1. With a (2m/s^2) added to both sides and considering opposite directions and signs (downwards positive, upwards negative), we get a2 - a = -a1 + a. Solving for a2, a2 = -a1 + a + a or a2 = -a1 + 2a.

[@canarat417]

Considering both accelerations like vector problems make it easy to understand. To set them equal Thanks

[@MichelvanBiezen]

I respectfully disagree with your assertion that the method is wrong.

[@dwightd3659]

+Michel van Biezen How did you get 5.9? If we did it the way we always has done it you would get: a1=5.26m/sec^2

[@sedaozkan3656]

How do you get the relationship between a1 and a2? I have a problem at this point. Please help me.

[@daniyalravandeh5223]

While I do appreciate your useful video, I'd like to put forward that the negative quantity for a1 while the assumed arrow is upward could not be correct and I believe that it should be positive by considering arrows direction on the FBD.

[@rishabhbhardwaj2873]

Ya you are right as the two co ordinate systems are moving in same direction there displacements should add up and if we take the 2nd derivation then we should get a1+a not a1-a

[@ironuranium3927]

Bezen I got the same result a1=5.9 & a2= 1.9 by solving these two equations m1a1+m2a2=m2g-m1g & a1-a2=4

[@anasmamdouh4050]

Im sorry but i cant wrap my head around the tension of the left mass. Been thinking about it for like 2 hours now. So you said we will put a negative sign in front of the m(2)g because the direction of gravity is negative, but isn't the direction of the acceleration also negative? Additionally, we should be doing vector addition, as in T+m(2)g=m(2)a(2), then applying the negative and positive signs depending on the directions of the vectors. What you did is that you began with the equation: m(2)g-T=m(2)a(2) then applied the negative signs to the acceleration and gravity then divided by -1. (i think anyways)

[@JoshuaJohnnC]

great work, but to save time from solving a2, the equation a2 = -a + 2a can be used. Since a1 was solved, then just substitute values. well, assuming that a1 is correct. :D

[@jeetendrasharma2283]

Sir, I solved this problem applying concept of pseudo force and got the same answer that is given in the video.

[@MichelvanBiezen]

Yes, that is an excellent method to use on this problem.

[@ryanshiu9289]

I was wondering how you would do this problem but instead of having a known acceleration of the whole system going up at 2 m/s2. Have a known applied force going up at say 250 newtons on the whole system

[@georgemtolo9098]

thanks prof your video was very helpful to me

[@obakengatom_sweetbrotherof1081]

sir how can a1 be negative since its going upward

[@sanskartiwari2996]

sir can you please recommend some books for mechanics. you are awesome

[@shanzidmahmood2243]

Hi Mr. Biezen, I was wondering whether applying (T=mg +/- ma) would be applicable to find the tension. If so, would we use + or -. If not, why can we not apply it? Thanks so much!

[@MichelvanBiezen]

In this case we can't use it, since the whole system is accelerating as well as the individual masses.

[@shanzidmahmood2243]

@@MichelvanBiezen thank you! That makes sense. On a seperate note, I just wanted to let you know that you’ve been a very big inspiration. I completed an Honours in Applied Accounting and have worked in that field for the last decade, but you’ve given me the courage to pursue something I was deathly afraid of all my life. I start my civil engineering degree this month. It’ll be a long 4 years, but atleast I know I have your videos to help me through it all. Thank you SO very much and expect a few more questions from me down the line, haha. Cheers.

[@MichelvanBiezen]

That is great! Keep it up and pursue what interests you.

[@ptyptypty3]

Great VIdeo... I'm just curious, what would the TENSION in the Cable be if a cable was accelerating the entire system up? I see that acceleration is equal to 2 m/s... I guess the Force would equal M (total) x a ... but since one of the masses is accelerating DOWN... it seems that the UPWARD Tension, in a cable, would be less than the TOTAL Mass?? yes? no?.. Thanks Michel ...

[@MichelvanBiezen]

Since each mass is accelerating at a different acceleration, you have to work out each mass separately and then add up the results: F1 = m1g + m1a and F2 = m2g + m2a The sign for the second term must be changed if the acceleration is downward.

[@ptyptypty3]

Thank you Michel!!!.. I truly appreciate the time you took to consider my question.. :) by the way... we had our FIRST day of 2018 where the temperature outside was 70 degrees !!!.. but that was at 2pm.. right now, at 10:33pm, it's 38 degrees... welcome to Michigan!! lol... thanks again Michel...

[@mahakgoyal2006]

as m2 is accelerating downward so T=m2g-m2a2??

[@MichelvanBiezen]

yes it is

[@mahakgoyal2006]

thn how T-m2g=m2a2??

[@alialoush9121]

@@mahakgoyal2006 hope it isnt too late. beacause the a2 going down which means its negative

[@hamzasaleem2550]

as we always take right direction positive in case of acceleration but the M2 is moving to the left so the acceleration of M2 was a negative quantity and due to the acceleration upwards positive acceleration is added to a2 so it became a smaller quantity opposite for a1 .... am i right sir? ................ my 2nd question is that is it due to the fact that we are watching the whole scenario as some kind of reference ? Thank you

[@MichelvanBiezen]

With pulleys, it is better not to use + and - for the direction, unless you use the technique of free body diagrams. The technique shown it this and the other videos works.

[@irfankhan_extca_0273]

sir, how did this happen a1-a=-(a2-a)

[@iyadindia862]

a1=--a2 Since directions of accelarations are opposite to each other

[@الاستاذحسينعلييعقوب]

Thank you sir ******Can we firstly neglect the acceleration (a = 2 ) then solving then determine acceleration for both ( 49/15) then + 2 and - 2.finally the results are ( 5.2. AND. 1.2 ) thank you again

[@MichelvanBiezen]

No, you cannot neglect the acceleration and work the problem as if there is no acceleration first.

[@الاستاذحسينعلييعقوب]

Michel van Biezen Thank you very much sir

[@gladnoah9426]

for the m2g which is accelerating downward the equation is supose to be m2g-T=ma

[@MichelvanBiezen]

Glad Noah, The equations are correct as stated in the video.

[@gladnoah9426]

Okay thank you

[@ratinislam4909]

Michel van Biezen But how? :/ i also thought it should be m2g-T=m2a2 ; isnt m2g>T ? as m2 is accelerating downwards.. could u kindly clarify?

[@MichelvanBiezen]

Ratin Islam This problem is different since the pulley is accelerating upward as well. So it is a non-inertial reference frame. The problem needs to be solved as I showed in the video.

[@ratinislam4909]

Michel van Biezen ooh i see..thanks a lot for clarifying this ! :)

[@reyna5443]

shouldn't be a1+a=-(a2-a) bc a1 and a are in the same direction?

[@MichelvanBiezen]

The equation in the video is correct. Thank you for checking.

[@TheBuzz187]

I tried another way of solving this problem after learning the way you used in the video. I solved for "a" using our normal method, m2g-m1g=MtA, and it didnt work. My logic was: if I could find the regular acceleration first, I could just add the given upward "a" to the right side and add "-a" to the left side. It turned out close but slower than the answer we got in the video. Can you explain why the upwards system acceleration increases the acceleration relative to the pulley please? I got 3.3m/s^2 as the acceleration relative to the pulley. But when you add the 2m/s^2 system acceleration it's a tiny bit slower as i mentioned above. So much for an easy way haha.

[@MichelvanBiezen]

I have only found 1 correct method to work this type of problem (and it is the method on the video).

[@surya8891]

+TheBuzz187 I did the same thing , the answer was close

[@sanskartiwari2996]

sir shouldn't the equation T-m2g=m2a2 be this T-m2g= -m2a2 (negative) because a2 is negative. or we could see it this way m2g-T=m2a2 (same as above) because I could subtrarct the other force(tension) from dominating force m2g taking the direction of m2g positive(because it is the dominating force).

[@MichelvanBiezen]

The rule used here is that any force that aids the acceleration is considered positive and any force that opposes the acceleration is considered negative.

[@sanskartiwari2996]

sir, but then according to it the equation shall be m2g-T=m2a2 (because m2g is aiding a2 and T is opposing a2) but then you are also right at one point because according to m2g-T=m2a2 we would get a positive value for a2 while it is negative. ps: your series on fluids helped me a lot

[@MustafaYlmaz-ri6qv]

Sanskar tiwari you are right, your equation is true but the teacher seems not to accept his error unfortunately altough he is very well at way of thinking and implementation.

[@novemberftblue6937]

In the video , professor used the method of Free body diagram, so the upward direction is always positive. For those who wondered : m2g-T=m2a2 ( you used the method which the direction of acceleration of the system follows the movement of (m2). So, here is your full answer: -System (m1): T - (m1)g =(m1)(a1). (1) -System (m2): (m2)g - T = (m2)(a2). (2) (1) +(2) we get: g[(m2) - (m1)] = (m1)(a1) + (m2)(a2). (3) Suppose (A) is the acceleration of (m1) and (m2) before applying the acceleration(a) ~(A) - (a) = (a2) {(m2) moves slower when the system is affected by (a). ( opposite direction)} ~(A) + (a) = (a1) {(m1) moves faster when the system is affected by (a). (same direction)} => (a2) = (a1) - 2(a) (4) substitute (a2) into equation (3) we get: g[(m2)-(m1)] = (m1)(a1) + (m2)(a1) - 2(m2)(a) Hence, (a1) = {1/[(m1)+(m2)} x { g[(m2) - (m1)]+2(a)(m2)} (PROVED) To find (a2), simply substitute (a1) to equation(4) which the result is positive. ( the result in the video is negative just to show the direction of (a2) relative to (a). Then you have it.

[@ALAL-jm1lr]

Sir, when acceleration is not same, doesn't this means smaller mass will run faster than the other big mass? and this lead to loose of connection(tension) between two masses ....until big mass catch same traveled distance as smaller mass, then now it will drag it ...and so on?? is there any real experiment to show us what will really happen for this system?

[@nurettinserdar8918]

thank you from Syria

[@MichelvanBiezen]

Welcome to the channel

[@wasafitvonline9248]

I got a1= 0.6 a2=3.4 in which when I add them and divide by 2 as you said .... I get 2 for a.... and I can see you mistaken from T2 for Mass 2. that T= m2g+m2a2 instead of T=m2g-m2a2 I'm correct pro???? please give me feed back for this question sir!! shalom

[@anirbandey6206]

Yeah bro you are right

[@rahultiwari9003]

when can we consider the tension on both sides of the rope to be equal in an Atwood machine

[@MichelvanBiezen]

+Rahul Tiwari Only if the pulley doesn't accelerate upwards

[@patrickmoloney672]

And no friction.

[@shubhamkukreja95]

Sir what if we consider the whole system at rest find a and then add or subtract by the acceleration of the system.wouldnt that save time?

[@MichelvanBiezen]

Try it and see if you get the same answer.

[@rahultiwari9003]

sir I did this question in the non inertial frame of the accelerating lift but I am not getting the correct answer.I set up the following two eqns 1.m2g+m2a(a of the lift) - T = m2a(a of the block) and 2.T-m1a(lift)-m1g-m1a(block)=0 sir can you can please tell me where am I doing wrong in these eqns please

[@recall660]

I have a problem understanding the acceleration part, this how I think about it ,since the mass 5kg moving up and the mass 10 kg moving down ,so mass 5kg moving in the same direction of the pulley ,so I add a1 to a (a1+a),and since mass 10kg moving down its in the opposite direction of the pulley (a2-a).can you explain prof and thanks

[@fernandosousa1542]

Very good teacher

[@YogiliciousP]

Note to self: I need to come back here! Woah

[@adharshsuresh997]

You r nice in teaching sir

[@thounaojamanilsinghluwang1464]

Sir even though m2 is greater than m1 you wrote the equation T -m2g because the whole system is going upward right?

[@MichelvanBiezen]

The assumption was made that m2 is accelerating upwards because the pulley is accelerating upwards.

[@thounaojamanilsinghluwang1464]

Michel van Biezen thanks sir for your time

[@allanthomas6489]

Since 10kg is accelerating downwards t=m2g-m2a2. Is it right or wrong. In the previous video you taught us this way

[@MichelvanBiezen]

This is a special case where the whole system is accelerating upwards. When the pulleys is stationary you are correct.

[@allanthomas6489]

@@MichelvanBiezen ok Sir. Thankyou😊

[@rustamshahverdiyev4164]

Professor hello, in this problem you did not apply aiding and opposing approach acceleration as oppose to other problems,why is that ? Thanks before.

[@MichelvanBiezen]

Because the whole system is being accelerated upwards. (This is an unusual situation), but has to be handled differently.

[@rustamshahverdiyev4164]

@@MichelvanBiezen Thank you very much

[@iyadindia862]

Is there any cases in which tension at the ends of string is not same to each other Professor Pls reply

[@MichelvanBiezen]

Yes, in the case that the pulley has friction or has mass the tension will not be the same on both sides.

[@iyadindia862]

@@MichelvanBiezen Pls post a video on this case when you get your free time I am eagerly waiting💕

[@MichelvanBiezen]

Those videos can be found in this playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2

[@garethm3171]

Please could you explain in more detail and slower than on the video how you arrived at a1 - a = -(a2 - a). It's just that the explanation on the video is very rushed and a bit incoherent.

[@spstcm4153]

a1=-a2 before 'a' is introduced. a1=-a2+2a * after 'a' is introduced which is logical as a1 and a2 both increase by 'a' making a net difference of 2a. Let a1=5 and a2=-5 then after 'a' introduced a1=7 and a2=-3 which satisfies * when 'a'=2.

[@radimollov4562]

I can't understand why a2=-a1. Can someone explain it to me?

[@MichelvanBiezen]

Since the two masses are connected with a string, the magnitude of their accelerations must be the same. But a1 is the acceleration of m1 upwards and a2 is the acceleration of m2 downwards, thus they have opposite directions.

[@radimollov4562]

@@MichelvanBiezen Thanks.

[@abre.ham121]

why a2= -a1? you said a1 and a2 are not equal? Thanks

[@MichelvanBiezen]

It depends on the reference point. Referenced to the pulley they will have the same magnitude. Referenced to a fixed point their magnitudes will not be the same.

[@obakengatom_sweetbrotherof1081]

answer i got 5.8 and -1.8 pls do it again sir

[@physicshacks6349]

Apply a(pulley) =( a1 + a2) /2

[@tsoojbaterdeneharvard3187]

a2=-a1 then why it cant be 5.9=-5.9 I really hope you will answer my question🙏🙏🙏

[@MichelvanBiezen]

a2 = - a1 simply compares the acceleration of 2 objects moving in opposite directions. The statement on the right is setting 2 numbers opposite in sign equal to one another which is not mathematically possible.

[@abre.ham121]

Can you explain please why not M2g-M2a2 = T . We used it like this in the privious examples?

[@MichelvanBiezen]

In this case the whole system, including the pulley, is accelerated upwards.

[@Reseme92033]

Great , thanks man

[@adharshsuresh997]

Tks

[@petereziagor4604]

Smiling at 6G LTE

[@MichelvanBiezen]

🙂

[@canarat417]

Using pseudo 10+2=12 12*10=120 12*5=60 120-60=60 60/15=4 a1= 4+2 a2= 4-2

[@anuragpaul7392]

Proffessor there is something wrong....it will be m2g-T = m2a2

[@MichelvanBiezen]

The video is correct. Thanks for checking.

[@vaibhawtiwari9084]

@@MichelvanBiezen a Yes sir it's moving upward resultant

[@obakengatom_sweetbrotherof1081]

YOU'RE WRONG SIR CAUSE I THINK FOR M2 IT WAS SUPPOSED TO BE M2g-T=M2a2 since it is accelerating is determined by gravity of block m2

[@sarfarazahmedsuri]

the method you used in acceleration ie a1=-a2+2a is wrong. mathematically its wrong method.if you have better knowledge then explain me plz.but i have seen you have not replied to any comment.i wish i will be replied.thank you.

[@recall660]

actually I got a1=a2-2a