@@aleksandramichalak7826 Except his shirt pocket is on the wrong side (for men's shirts), his shirt buttons are reversed from normal (should be left over right), and it's statistically unlikely that he and every other instructor who uses this technique are left handed. The image is almost certainly being mirrored (flipped) in post-production.
I got everything, except one part: How can mass 2 have a non-zero acceleration, but the free-moving pulley it's directly attached to, pulley B, has NO acceleration?
This has been such a good explanation. But now i have just a little problem with pulley B. We assume that the pulleys are approximately massless and when we get that T3=2T does it mean that the pulley is stationary or that it is moving with constant velocity because i think it can not be stationary given the fact that m2 is pulling it downwards and if it is moving then it is accelerating with thesame acceleration as m2 but will be in disagreement with the equation established.I kind of understand here that it is the mass been approximated to zero that sets all this but what is physical interpretation of the fact that we approximate the mass to be zero and it leads to forces acting on it being balanced So actually I am lost at that point someone please help me with an explanation
I'm sure you don't care about it 3 years later, but this is how I understand it. First, think of the pulley having some mass, m. In the case of the tensions on one of the pulleys, so for pulley B, 2*T - T3 would be the net force on the pulley, and Net force = ma, so 2*T - T3 = ma. If you solved for T, you would get T = (ma - T3)/2. Maybe if you look at this equation, you can see that no matter the acceleration, as m approaches zero, T will approach (T3)/2, so even if there is acceleration, the forces have to balance out.
Some of them pulleys are pretty dog gone heavy. It is nice to just assume things, you know it is so much easier. Hey lets all go home now and have a beer.
@@neoneon6076 What about the equation of m1? Since pulley B and block 2 are pulling downward, the block 1 will move upward. As we assume that taking down is positive, the acceleration a1 of m1 would be negative but not positive. He wrote it as positive. Is it wrong? Please, answer. Thank You.
3:21 - "Our Rope is not slipping"
3:25 - "Rope is actually just slipping"
Our rope is not stretching (nor compressing)
Our rope is just slipping (if not, either it's fixed to the pulley or it's extensible)
He knows how to write backwards very well :/
He's not writing backward. Computer program flipping it!!
Lol
technology on attack..HELP!
It's called a learning board. No computer flipping involved 😊
@@aleksandramichalak7826 Except his shirt pocket is on the wrong side (for men's shirts), his shirt buttons are reversed from normal (should be left over right), and it's statistically unlikely that he and every other instructor who uses this technique are left handed. The image is almost certainly being mirrored (flipped) in post-production.
Thank you very much, I must say this level of understanding is just superb
I got everything, except one part: How can mass 2 have a non-zero acceleration, but the free-moving pulley it's directly attached to, pulley B, has NO acceleration?
This is the worlds most complicated explanation on calculating pulleys! Lol! This is the difference between engineering and science.
But which one is easier
you are making this way more complicated than it needs to be, just watch van Biezen
Great sir.......awesome.........
Amazing explanation...
This has been such a good explanation. But now i have just a little problem with pulley B. We assume that the pulleys are approximately massless and when we get that T3=2T does it mean that the pulley is stationary or that it is moving with constant velocity because i think it can not be stationary given the fact that m2 is pulling it downwards and if it is moving then it is accelerating with thesame acceleration as m2 but will be in disagreement with the equation established.I kind of understand here that it is the mass been approximated to zero that sets all this but what is physical interpretation of the fact that we approximate the mass to be zero and it leads to forces acting on it being balanced So actually I am lost at that point someone please help me with an explanation
I'm sure you don't care about it 3 years later, but this is how I understand it. First, think of the pulley having some mass, m. In the case of the tensions on one of the pulleys, so for pulley B, 2*T - T3 would be the net force on the pulley, and Net force = ma, so 2*T - T3 = ma. If you solved for T, you would get T = (ma - T3)/2. Maybe if you look at this equation, you can see that no matter the acceleration, as m approaches zero, T will approach (T3)/2, so even if there is acceleration, the forces have to balance out.
Mind BLOWN
Thank you sir
How is he able to write in the reflected manner so easily..??
the video is flipped. Notice he's writing with his left hand when it's more than likely he's right handed
Gracias
nice video.
Some of them pulleys are pretty dog gone heavy. It is nice to just assume things, you know it is so much easier. Hey lets all go home now and have a beer.
tnx xx
Pulley B is not fixed it is coming downward so we cannot write it as T3-2T=0
It is a massless pulley so m_B = 0, so m_B*a = 0.
@@neoneon6076 What about the equation of m1? Since pulley B and block 2 are pulling downward, the block 1 will move upward. As we assume that taking down is positive, the acceleration a1 of m1 would be negative but not positive. He wrote it as positive. Is it wrong? Please, answer. Thank You.
nice one .helping one.....
Board is an abscurd.
Plz in hindi
Haha fuck you!
you hindi people can learn from a channel called physics wallah
bro every learning vid in hindi pisses me off