LOL at 2:30.... with the Inserted BUBBLE comment correcting your "m1" slip ..... your editing has evolved nicely with these added features..... :D ... lol thanks to your better half no doubt! Great video.... this series of videos are excellent... you truly drive home the concepts involved in learning how Tension can be determined..... EXCELLENT... !!
Why is the external force (NORMAL REACTION) at the the PULLEY, which provides the horizontal component of the combined system's (if we treat m1 m2 and string to be a single system) acceleration completely ignored ? If we only consider the external forces mentioned in the lecture, the "combined system" would not have any acceleration in the horizontal direction (as only m1g, m2g and N at m1 are mentioned, which are ALL in vertical direction), otherwise violating Newton's laws. However the combined system HAS an acceleration component in the horizontal direction (due to acceleration of m1) If we take the system to be combined m1 and m2, then the external forces acting on the system are not only the forces of gravity acting on the masses (m1g and m2g) , the normal reaction on m1 (which is nullified by weight of m1) but ALSO the NORMAL REACTION at pulley, point outwards (towards top right corner of screen) at an angle of 45 degrees with horizontal (for this specific problem), which is not at all mentioned. It is THIS force that provides the horizontal component of the system's acceleration. It is the vector sum of these TWO forces that drives the system (and not the single force m2g) and the acceleration vector of the CM of the system, is the vector sum of this Reaction (at pulley) and m2g divided by the total mass of the system (m1 + m2). In this specific problem, since the strings forms an angle of 90 degree at the pulley, the horizontal and vertical components of the Reaction force are equal in this case ( and makes an angle of 45 degree with horizontal) QUESTION: Why is the external Normal Reaction at pulley, that provides the horizontal component of the acceleration of the combined system(i.e. when treating m1 m2 and string as the system), NOT at all considered ?
Since the force of the pulley acting on the string, acts in the direction perpendicular to the string, it cannot do any work and thus it does not aid in the acceleration of the system.
@@MichelvanBiezen Thanks for your reply professor ! External Forces (responsible for acceleration in the desired directions) shown in the diagram are all in vertical (y) direction (e.g. mg). QUESTION: Which External Force gives the system (system = m1 plus m2 plus string combined) an acceleration in the horizontal (positive x) direction ? [When we take the combined system (m1 plus m2 plus string) , tension in the string is an internal force, and internal forces are not considered when evaluating F=ma of a system. As per Newton's Laws, there has to be an external force (or component of an external force) in the horizontal direction for the CM (Centre of Mass) of the combined system (i.e. m1 + m2+string) to have an acceleration component in the horizontal (positive x) direction.]
There are two ways to solve a problem like this. If you want to solve this problem and separate the horizontal from the vertical direction, then you must separate each mass and use free body diagrams. If you take the whole system as a single unit, then up/down, and left/right no longer has meaning, but in the direction of the acceleration or opposite to the acceleration does.
@@MichelvanBiezen Thanks for your reply again! If I get you, you are trying to say that the whole system (combined m1 + m2 + string) has an acceleration in the horizontal +x direction but no component of External Force in that direction ! And the total External Force is in one direction and the acceleration (rate of change of momentum) of the system (m1 + m2 + string) is in another direction!! Is this not clearly contradicting Newton's second law "Rate of change of momentum of body is directly proportional to the force applied and takes place in the DIRECTION of the APPLIED FORCE" ?
Hey, Michel wonderful video! At. 3:3, you ignored the Tension. 2 in the free body diagram which still perplexes me, is it because T2 would be canceled out by T1? Cause you said T2 is in internal of the system and can be ignored but when we calculate m2 we are not looking at the whole system but only m2 so I am kinda confused here.
Indeed if you want to isolate one of the masses, you will need a free body diagram and you will need to take T1 and/or T2 into consideration. But if you are calculating the acceleration of the whole system, then you don't need the internal forces of the system and you can ignore T1 and T2. Take a look at a few more examples and you'll see how that method works in every case.
When labelling the loads, does the hanging weight have to be labeled M2 or Mb? And the mass on the horizontal flatform has to be labeled M1 or Ma? Please reply...thnks
Once it is moving, it doesn't need a net force to keep moving. (Newton's first law). If it is not moving then yes, it will require a momentary force to get it going.
If two people pull on a rope, no matter how hard each one of them pulls and in what direction the rope is pulled the tension in the rope will be the same along the entire length of the rope. In other words the tension must be the same along the entire rope. The same here UNLESS the pulley has mass or has friction.
Work the problem (both for acceleration and tension) without plugging in numbers. Then in the end you'll see if you have enough information to determine mass or acceleration.
I can't express it enough...You make physics seem easy...THANK YOU!
You're a blessing, sir. God bless you even more!
God bless him double
LOL at 2:30.... with the Inserted BUBBLE comment correcting your "m1" slip ..... your editing has evolved nicely with these added features..... :D ... lol thanks to your better half no doubt! Great video.... this series of videos are excellent... you truly drive home the concepts involved in learning how Tension can be determined..... EXCELLENT... !!
+Philip Y Better Half thanks you.
Why is the external force (NORMAL REACTION) at the the PULLEY, which provides the horizontal component of the combined system's (if we treat m1 m2 and string to be a single system) acceleration completely ignored ?
If we only consider the external forces mentioned in the lecture, the "combined system" would not have any acceleration in the horizontal direction (as only m1g, m2g and N at m1 are mentioned, which are ALL in vertical direction), otherwise violating Newton's laws. However the combined system HAS an acceleration component in the horizontal direction (due to acceleration of m1)
If we take the system to be combined m1 and m2, then the external forces acting on the system are not only the forces of gravity acting on the masses (m1g and m2g) , the normal reaction on m1 (which is nullified by weight of m1) but ALSO the NORMAL REACTION at pulley, point outwards (towards top right corner of screen) at an angle of 45 degrees with horizontal (for this specific problem), which is not at all mentioned. It is THIS force that provides the horizontal component of the system's acceleration.
It is the vector sum of these TWO forces that drives the system (and not the single force m2g) and the acceleration vector of the CM of the system, is the vector sum of this Reaction (at pulley) and m2g divided by the total mass of the system (m1 + m2).
In this specific problem, since the strings forms an angle of 90 degree at the pulley, the horizontal and vertical components of the Reaction force are equal in this case ( and makes an angle of 45 degree with horizontal)
QUESTION: Why is the external Normal Reaction at pulley, that provides the horizontal component of the acceleration of the combined system(i.e. when treating m1 m2 and string as the system), NOT at all considered ?
Since the force of the pulley acting on the string, acts in the direction perpendicular to the string, it cannot do any work and thus it does not aid in the acceleration of the system.
@@MichelvanBiezen Thanks for your reply professor ! External Forces (responsible for acceleration in the desired directions) shown in the diagram are all in vertical (y) direction (e.g. mg).
QUESTION: Which External Force gives the system (system = m1 plus m2 plus string combined) an acceleration in the horizontal (positive x) direction ? [When we take the combined system (m1 plus m2 plus string) , tension in the string is an internal force, and internal forces are not considered when evaluating F=ma of a system. As per Newton's Laws, there has to be an external force (or component of an external force) in the horizontal direction for the CM (Centre of Mass) of the combined system (i.e. m1 + m2+string) to have an acceleration component in the horizontal (positive x) direction.]
There are two ways to solve a problem like this. If you want to solve this problem and separate the horizontal from the vertical direction, then you must separate each mass and use free body diagrams. If you take the whole system as a single unit, then up/down, and left/right no longer has meaning, but in the direction of the acceleration or opposite to the acceleration does.
@@MichelvanBiezen Thanks for your reply again! If I get you, you are trying to say that the whole system (combined m1 + m2 + string) has an acceleration in the horizontal +x direction but no component of External Force in that direction ! And the total External Force is in one direction and the acceleration (rate of change of momentum) of the system (m1 + m2 + string) is in another direction!! Is this not clearly contradicting Newton's second law "Rate of change of momentum of body is directly proportional to the force applied and takes place in the DIRECTION of the APPLIED FORCE" ?
Hey, Michel wonderful video! At. 3:3, you ignored the Tension. 2 in the free body diagram which still perplexes me, is it because T2 would be canceled out by T1? Cause you said T2 is in internal of the system and can be ignored but when we calculate m2 we are not looking at the whole system but only m2 so I am kinda confused here.
At 3:33
Indeed if you want to isolate one of the masses, you will need a free body diagram and you will need to take T1 and/or T2 into consideration. But if you are calculating the acceleration of the whole system, then you don't need the internal forces of the system and you can ignore T1 and T2. Take a look at a few more examples and you'll see how that method works in every case.
@@MichelvanBiezen Thanks a lot
Appreciate your effort 👍
Aliyan Tahir '
You'r videos are amazing! They helped me tremendously! :)
Prof von Beizen, Have you shown any of this type of problem with friction and with numbers?
Yes, there are lots of examples in this (and other playlists): PHYSICS 4.1 NEWTON'S LAWS EXAMPLES
Michel van Biezen they were VERY helpful. thank you so much. circular motion up next. i'm really enjoying physics and your videos.
When labelling the loads, does the hanging weight have to be labeled M2 or Mb? And the mass on the horizontal flatform has to be labeled M1 or Ma? Please reply...thnks
It doesn't matter. I suggest you labeled it such that you are most comfortable with it.
@@MichelvanBiezen thank you once again professor.. you are one of a kind..
Thank you! Your effort is highly appreciated.
Excellently explained. Thank you. :)
what if the string is suddenly broken what will be the acceleration of mass 1?
a would be zero
amazing
video
Glad you think so!
Does this apply when there is no net force?
If there is no net force, then once the system is moving, the acceleration would be zero and the objects would move at a constant speed.
Michel van Biezen Thank you, but what if the system is in equilibrium and there is no movement?
Once it is moving, it doesn't need a net force to keep moving. (Newton's first law). If it is not moving then yes, it will require a momentary force to get it going.
Y is t1 n t2 the same when m1 n m2 is different
If two people pull on a rope, no matter how hard each one of them pulls and in what direction the rope is pulled the tension in the rope will be the same along the entire length of the rope. In other words the tension must be the same along the entire rope. The same here UNLESS the pulley has mass or has friction.
Thanks very much
so helpful thank you !
What if the tension is given, you have to find mass but there is no acceleration given.
Work the problem (both for acceleration and tension) without plugging in numbers. Then in the end you'll see if you have enough information to determine mass or acceleration.
i love u sir...i reallyy love u
What if we are given a coefficient of kinetic friction?
Lara.c we need to substract frictional force from total force
what if miu is equal to something
We have a lot of example videos like that with and without friction in the Newton's Laws application playlists.
thank you@@MichelvanBiezen
Change playback speed to 1.5. You're welcome.
Pdf
Thank you sir
Danke Michel :)
+MrShanqwert Bitte
+Michel van Biezen Het is mijn plezier