Physics - Mechanics: Applications of Newton's Second Law (13 of 20) table and pulley

[@proudmuslim_29]

Greatest physics teacher in my life

[@MichelvanBiezen]

Glad you enjoying the videos. 🙂

[@MichelvanBiezen]

Ravi, There are lots of examples in the playlist with and without friction.

[@TheBuzz187]

Another stellar video from Professor Biezen!!!! Awesome! Keep em coming!

[@unknown-iu5wo]

Hi prof! Ive been watching your physics mechanics videos one by one but theres something I dont understand. In some cases we go ahead and find the components and in some other ones (like in this video) we go straight for Fnet=mtota. Why so? When are we meant to find the components (mgsin/cos) and when should we just go for Fnet?

[@MichelvanBiezen]

You can indeed break this type of problem down into block diagrams while keeping track of all the forces in the x and y directions. But it is a lot easier to look at a problem like this in terms of the whole system moving in unison as shown in the video.

[@MichelvanBiezen]

Also this way we ignore all internal forces within the system and only consider external forces acting on the system as a whole.

[@ptyptypty3]

Hello Michel and WIFE :) This is PHIL ..long time... all is well in Michigan.. Hope you're keeping COOL and enjoying Summer....I can't believe I came back to one of your videos that is TEN YEARS OLD!! I do have a question... what would the Acceleration of the two masses be when the Block on the Table went over the edge and both masses were subject to Gravity? that is, M1g and M2g ?? I may have just figured that out... they BOTH would be accelerating at "g" .. so the Net Accerlation would change from a = (m1g/(m1 +m2)) g to a = g so the accelration would Increase., RIGHT? thank you... congrats on having OVER 1 million Subscribers!!

[@MichelvanBiezen]

Hi Phil. Good to hear from you. Hope all is well. Where we live it is rather comfortable. We have that large Pacific Ocean keeping things cool. You are indeed correct with the acceleration. a = g in that case.

[@James189hshwh]

Master please i have got a question in my mind for a long time i am a highschool senior student my teachers dont explain this and i dont know if the rule like we say if the direction of the matter moves just like the direction of clock ticking then the force is positive if its opposite then negative but i want to say if this rule also apply for this or not if yes how? thanks so much

[@MichelvanBiezen]

I think that trying to enforce that rule in this type of problem will just confuse the student. If you look at this setup from the other side you would get the very same answer and one object would be moving to the left and one object would be moving down, and then if you add a third mass on the other side which is bigger, then one object would be moving up, the second would be moving to the right, and a third object would be moving down. BUT if that is what you teacher wants, then I suggest you follow your teacher's rules.

[@James189hshwh]

@@MichelvanBiezen thanks so much no i didn't want it for my school test that teacher ask us i wanted this for the national examination and this answer was totally clear thanks master

[@MichelvanBiezen]

You are welcome. All the best on your exam.

[@pranavamali05]

Just enjoying diff question throughout the day thanks😊

[@MichelvanBiezen]

Good way to keep up your knowledge. 🙂

[@alixavi7933]

Plz pro i want example about impulse and momentum to disk G.P.M by belt to another disk rotatin !

[@nedimkorcic1607]

Why we did not calculate tension in the system?

[@drdeepika703]

What if we were to find out acceleration of both blocks individually?

[@MichelvanBiezen]

Since the blocks are connected they both have the same acceleration (magnitude).

[@nouraa4874]

I thought that the system would be in equilibrium ?! because mass 2 is heavier so mass 1 would be just suspended

[@MichelvanBiezen]

Our intuition sometimes misleads us. When we calculate the net force on the whole system, we find that the net force is NOT zero and thus there MUST be an acceleration. That is how we know the system is in equilibrium or not.

[@nouraa4874]

yes you are right .. that make sense Thank you ()"

[@delta1404]

Would someone care to explain why is m1 the mass of Fnet and not mT? :)

[@MichelvanBiezen]

The force of gravity acting on m2 cannot accelerate m2, since it acts in the wrong direction.

[@moezzatehseen5831]

Sir,why are we not taking in account the tension in he string?

[@MichelvanBiezen]

Because that is a force that is internal to the system and does not contribute to the acceleration. Only if you look at each body separately using free body diagrams to you need to take the tension into account.

[@kierantaylor2001]

@@MichelvanBiezen Sorry for such a late comment on this video but do you have any videos doing this as currently trying to get some work done that specifically requests to find the acceraration of each block tension in each of the cables. Any help would be appreciated

[@abre.ham121]

why not used Fnet equal to T - m1g ? Thanks

[@MichelvanBiezen]

Fnet is always equal to : total mass of the system x acceleration of the system.

[@ravivignesh3505]

How to problem when there is friction coming into play?

[@HirokoNihon]

Sir, why is it m1g over the total mass? Isnt it the weight of the total mass, so mtotal*9.8?

[@MichelvanBiezen]

Depends on what the "it" is. In the case we start with a correct equation: F = m a and then when we substitute the correct values we algebraically find the equation in the video.

[@khodadadvahedi2457]

great teacher

[@MichelvanBiezen]

Thank you! 😃

[@charlynesmith2097]

Why was the tension force not included in equation?

[@MichelvanBiezen]

+Charlyne Smith Tension is an internal force which does not have any effect on the system as a whole. We only want to consider outside forces acting on the system. That said, there is another method to solve for the acceleration of the system by drawing free body diagrams of each part of the system and solve those equations simultaneously, and then you have to take the tension into consideration. That method is much harder.

[@charlynesmith2097]

Yes, I am more familiar with that method because I practice it a lot. I prefer yours though because it saves time. I was just curious as to the reasoning behind neglecting the tension force. Thank You!

[@surya8891]

+Charlyne Smith In that case m1a = T m2a= m2g - T (m1+m2)a=m2g a=m2g/(m1+m2) Tensions cancel out each other and hence have no effect on the system

[@faizanaz1092]

i know this comment was posted a while a go but why is "T" negative? is not suppose to be positive, since it is going in +y and +x direction.

[@Amirali1amir]

Excellent

[@i.apilado]

I doubt why the acceleration of it is positive? Why shouldn't be negative since the magnitude is downwards?

[@MichelvanBiezen]

Since we are finding the acceleration of the whole system, not just the block moving downward, you cannot call the acceleration negative. Note that the block on top of the table accelerates in the positive x-direction.

[@i.apilado]

@@MichelvanBiezen oh thank you I understand

[@doodelay]

Darn, I saw that the net force would not equal zero but assumed no movement anyway for some reason 😑

[@alixavi7933]

Hi prof. I,m from iraq and am like your lecture .. my qustion. is ax = ay in this proplem .? Or a total = ax+ay (vector) plz i want answer quicky Thank you 🙏🏻

[@MichelvanBiezen]

First, welcome to the channel. No, the total is NOT ax + ay. Note that with vectors, you cannot add them unless they act in the same direction. You can however say: the magnitude of the acceleration of the block on the table moving in the x-direction is equal to the magnitude of the acceleration of the hanging block moving in the y-direction. (Also note that only the magnitude of ax = the magnitude of ay)

[@alixavi7933]

Michel van Biezen thank you very much prof.

[@livelife623]

How does the Normal force on mass #2 cancel?

[@livelife623]

Oh I see. The normal force on mass 2 cancels out with the weight of the object but for mass 1 the Normal force is not equal to the object's weight since it is suspended vertically on a rope.

[@suvamlenka2104]

thank u sirrr

[@Broccoli_kb]

Thankkk uuuu

[@MichelvanBiezen]

You are welcome. Glad it was helpful. 🙂