i am a middle school student who is studying in olympiad training to hopefully participate in physics olympiads. i cannot stress how much you helped me and plenty of other students and people as well. Thank you so much! I aspire to be able to teach other people just like you do!
I'm in my second year doing Electrical Engineering, here in makerere university, Uganda... Africa, but your lectures have helped me a lot For me, you're the top/best professor in the world Stay blessed.
1:27, 1:28, and 1:29 u said let's assume the pulley has no mass and no friction Pls I can't picture that Is there any video solution to a problem where the pulley has a mass How does it affect the acceleration and the tension in the system
When the pulley has mass, you have to take into account the moment of inertia, and the problem is solved very differently. You can find those videos in this playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 Starting with this video: ruclips.net/video/XHkDJwQ4Xng/видео.html
Glad it helped! Yes this type of example encompasses the general principle of how to solve this type of problem taking the whole system into consideration.
Hi , I'm doing parachute auto steering for my project and I'm using motor to steer the parachute to the direction I need it to. However I need to find the tension and the force that is required for the strings and the motor to change direction. Could u please help me with this?
Sir i will be thankful of yours if you will upload a video for pulley system in which two pulleys are hanging on a ceiling( also have inertia) and a string goes over them and blocks are hanging over free ends
The only force M1 applies is its weight and it doesn't act in the direction of the acceleration. The normal force counteracts it, again not in the direction of the acceleration.
Isn't that shown in the video? The magnitude of the acceleration is given and the direction can usually be surmized by which of the two hanging weights is the heaviest. In this case it makes more sense to talk about clockwise or counterclockwise direction of the system.
S Joyce There are a number of examples in the video series with friction as well. For any hanging object (in an inertial system) it will always be: mg +/- ma
Michel van Biezen is there any way I can show you what I have calculated and you can look at my workings. Im just not sure I am calculating the tension correctly
S Joyce Feel free to show me. (there are a number of videos in this playlist; that show you how to calculate the tension) PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
It seems from the final equation that, any differential in weights attached to two sides of the pulleys shall accelerate the system, slow or fast. But in reality, the system won't accelerate at all till a certain threshold of weight imbalance. That differential in my opinion shall vary based on friction between pulley & rope and pulley inertia but more importantly based on "weights". More the weight, more the differential needed to start acceleration. Can you quantify it with equations in above case? Thanks
+Raj Learn Yes you can. Make one of the masses unknown, and then set the sum of all the forces equal to zero. You will find the minimum mass unbalance required to start the acceleration.
+Michel van Biezen Is it possible to elaborate it further? In my experiment with single pulley, I attached wts of 50g, 500g, 2400g and 4000g on one side. I noticed acceleration only after I put 400g, 1100g, 3100g and 6500 g respectively on other side. Then I took 2 pulleys, separated by 0.5 meters and attached same weights like before on one side. On other side I had to put wts of 500g, 1500g, 4600g and 8200g respectively to start acceleration. The pulley radius is 4 inch. I am unable to quantify these numbers with any theory. What this force is called in physics to overcome this inertia(?)?
+Raj Learn You are dealing with the difference between static and kinetic friction. Static friction is greater than kinetic friction, so it requires additional force to get the system moving than to keep it moving. Also if you are setting up an actual physical system, you are dealing with other factors such as play in the pulleys that could misalign, etc.
In this example the table is "frictionless" as indicated in the title. But if there was friction, we would need to take it into account as you indicated.
@@MichelvanBiezen Спасибо, у меня плохо с английским языком, поэтому я ориентировался на язык "математический и физический". Привет от учителей России!
thank you very much l am very happy with you and my question it (if the masses are equal each is 10 kg and instead of the body on the table we put Helical captivity whow much the force reading thank you again
That is usually indicated in the problem. If the coefficient of friction is zero, then there won't be any friction forces. If the coefficient of friction is not zero, then you have to take the friction forces into account.
When there is static friction it means the block is not moving, because there is a force trying to avoid its movement. If the block is about to move we call it the maximun static friction and it means that, if you apply just a little bit more force to the block, it starts moving. Therefore if the block is moving, then there would be kinetic friction and it means that there is a force trying to interrumpt the block´s movement while you are applying a force to it (while the block is moving), so the kinetic friction must be indeed, less tan the force you are applying to the block.
Why is the formula to find the force of the objects mg? I thought it was f = ma or do I have to use mg when dealing with surface and pulley problems? If mass 3 was nonexistent, and mass 1 was being pulled along the surface with mass 2 going downward, how can you find the acceleration? I remember being told that the acceleration between those 2 masses are equal but i do not remember why. If they are, can i find the acceleration by taking mg and diving it by the total mass? Im guessing the total forces would be mg of the 2nd mass or would i have to add the tension force of the first mass as well? Thank you
Think of the units on g as Newtons per kilogram, rather than meters/second^2. What it really is, is a force field strength. Just as electric field is force per unit charge and has the units Newtons/Coulomb, gravitational field is also a force per unit property of matter, in this case mass. It tells you what force gravity will apply to a given object at a given position, based on its mass. It is a "coincidence" that it also has the same units as acceleration, and ends up being equal to the acceleration of a free falling body when the only force is gravity by definition. I put "coincidence" in quotes, because in general relativity, you learn why this is the case, and it is not just a coincidence. What "F=m*a" refers to, is the NET FORCE equaling mass * acceleration. Not just any force. It is the vector sum of the forces, which we call net force. "Net" means total, in that you add up quantities taking in to consideration sign and direction where applicable. In this case, your "F=m*a" equation that involves gravity has to consider both forces acting on either hanging mass. Take m3 as an example. There is the tension in the string above it (T3) and its weight acting on it (m3*g). The F=m*a equation on this individual mass is T3 - m3*g = m3*a, considering both the tension in the string above it, and its own weight.
When solving problems using the equations of kinematics, direction of the vectors (displacement, velocity, and acceleration) are important, therefore you need to use -9.8 m/sec^2. In this example we use the magnitude of g, and therefore we can ignore the sign. (Magnitudes of vectors cannot be negative).
If you don't know which mass is larger, or which way it will accelerate, you can pick a direction arbitrarily for its acceleration and set up the force addition on each object accordingly. If it turns out you selected incorrectly, you'll simply get a negative sign in your answer for acceleration, which means acceleration is in the opposite direction you assumed. Since it is ultimately a system of linear equations, the magnitude of acceleration is still the same as you got, with your assumption that acceleration was in the other direction. If the hanging masses are both equal, the tension on the supported mass M1 will add up to zero, and there will be no acceleration. Using the same equations you solved assuming there would be acceleration, you will end up seeing that the acceleration is zero.
It is multiplied by the mass to give the force exerted on the object via gravity. The force exerted by gravity is required to determine the net force the cord is exerting on the object to accelerate it at the value determined earlier: 2.8m/s^2 F=ma We have acceleration, the mass, and we can easily determine one of the forces (Force of gravity), these are used to determine the unknown tension force. T = Fg + Fnet T = mg+ma T = m(g+a) I just realized your comments are 4 months old. Ohwell.
The tension is a force that is internal to the system and doesn't cause the system to accelerate. Only if you draw free body diagrams do you need to take the tension into account.
I have a problem just like this, only I am given the acceleration of the system and told to solve for the mass which the system acceleration is towards. I cannot figure out how to solve for mass, with mass being on the bottom of the equation as well.
Essentially you are solving for m3. Multiply a (on the left side of the equation) with the sum of the three masses (in the denominator on the right side of the equation) and then solve for m3.
You only need to consider external forces acting on the system, but you must include ALL the mass of the system since the WHOLE system is being accelerated.
Mass M1 is supported by the table below it, unlike the two hanging masses. As a result, the table below it applies a normal force (N) that will be as large as necessary to prevent M1 from sinking through the table, as long as the table is strong enough to provide this force. The forces in the vertical direction on mass M1 will therefore add up to zero, and not cause any acceleration on the system. No component of the weight of mass M1 will act in the direction of motion. If there were friction in this problem, then the weight of mass M1 would indirectly come in to play. The normal force is directly determined by the weight it supports, and the friction on M1 would be determined by the normal force and friction coefficient (mu_s and mu_k). At static conditions, the value of mu_s*N an upper limit to how large the friction can be. When there is motion of M1 moving sideways along the table, then the friction force will be equal to mu_k*N, in the opposite direction of the existing motion.
If you use "up" and "down" directions in the equations, you must set up independent free body diagrams, and then link them together. It is much easier to consider aiding and opposing the acceleration of the whole system as depicted in the video.
You can find you answer in this playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 ruclips.net/user/ilectureonlineplaylists?shelf_id=4&view=50&sort=dd
Took my online tutors hours to explain this using vague answers and not showing what to do. Took me a minute to search up the problem and 6 minutes to skim through the video and figure out what to do
+rashed a564 Respectfully, that is incorrect. When a force "reacts" to the application of an initial force, it is called a "reaction force". The box pushes against the table and the table pushes back.
@@mrrashedali "Reaction force" doesn't exclusively refer to the Newton's Third Law reaction pair. It also refers to constraint forces that are directly determined by the forces they oppose.
just a note: the weight of m1 and the normal force on it from the table happen to be equal and opposite in this case, but they are NOT a Newton's 3rd Law pair! My physics students get that wrong all the time.
Let's simplify this. Just take a table and place an block on top of the table with mass m. The weight of the block = mg The weight pushes against the table. What does the table do against the block? What is that principle called? Is that different from the situation when I push against a wall with a force F? What does the wall do back to me?
Michel van Biezen ah the book pushing on the table and the table pushing the book are indeed a pair. But “weight” is the force of gravity from Earth pulling the book. It’s equal+opposite force is the book pulling on the earth!
Michel van Biezen the table holding up the book has nothing to do with Newton’s 3rd law. What if the table is slanted? Suddenly the book slides, and the normal force is not equal and opposite to the weight. Has the book’s weight lost its equal and opposite pair? Does Newton’s 3rd law not apply to inclined planes?
@@willtanzola6629 Newton's third law DOES still apply, but the third law pairs to the forces involved in this problem are "out of the picture" of the problem we are trying to solve. The tensions are the only forces for which we need to consider Newton's third law pairs, and that is a simple matter of tension on any given cord segment being the same on both sides, and obviously opposite in direction. The Newton's third law pair to any given object's weight, is that it pulls up on the Earth with an equal and opposite force of gravity. The Newton's third law pair to the normal force on mass M1, is that mass M1 also applies a normal force that pushes down on the table, which is supported by the compression in the posts that support the table. Ultimately, there is a support force from the Earth, that passes through numerous structures to support the table, to support the object on the table, and to support the axles that support the pulleys. Because Earth's mass is significantly greater than the three masses involved in this problem, we can neglect the third law pairs to these forces that act upon the Earth, when solving this problem. The Earth will recoil by a negligible amount of upward acceleration, due to this system undergoing the situation described.
This is great work. But there is a problem of concept during explaination. The reaction force of m1 is not the normal force. The weight is a gravitational force and it's reaction force is the force with which it pulls the earth. The reason why the weight is equal to the normal reaction is due to the second law of motion because there is no reaction in the vertical direction.
+PrankingM@zeter You probably want to start with the basics regarding tension and Newton's second law. Take a look at the playlists: PHYSICS 4.5 - STATICS & EQUILIBRIUM and PHYSICS 4 NEWTON'S LAWS OF MOTION
BRUHH WTTFFFFF. I JUST GOT BACK FROM MY PHYSICS EXAM AND IT HAD A PROBLEM LIKE THIS, AM IM SO MAD NOW, BECAUSE THIS SHIT ISN'T TAHT HARD. all my practice problems always had two objects, which meant i only ever had to deal with one tension. i didn't know tension just added up masses like that. wtf . fucc.
i am a middle school student who is studying in olympiad training to hopefully participate in physics olympiads. i cannot stress how much you helped me and plenty of other students and people as well. Thank you so much! I aspire to be able to teach other people just like you do!
You are learning this while in middle school? That is amazing. Keep that up and you will do well in life.
I just wanted to say thank you. Your videos are super helpful, keep up with the great work professor.
My left ear is gonna get a 5 on the AP test
@Elijah Henrik Didnt work for me and the police have been called oh no
that's 2.5
I'm in my second year doing Electrical Engineering, here in makerere university, Uganda... Africa, but your lectures have helped me a lot
For me, you're the top/best professor in the world
Stay blessed.
Thank you. Welcome to the channel. Glad you found our videos. 🙂
This video helped me get a 100 percent !! Thank you!!
Wow! Thank you. You are the main physic online teacher that usually help me out.💞
We are glad to be able to help
Sal Khan at 1:33 would be like "As always I encourage you to pause this video and see if you can figure it out on your own!"
Thank you so much sir......I am totally doubtless after seeing this solution. Again thank you sir.
1:27, 1:28, and 1:29 u said let's assume the pulley has no mass and no friction
Pls I can't picture that
Is there any video solution to a problem where the pulley has a mass
How does it affect the acceleration and the tension in the system
When the pulley has mass, you have to take into account the moment of inertia, and the problem is solved very differently. You can find those videos in this playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 Starting with this video: ruclips.net/video/XHkDJwQ4Xng/видео.html
@@MichelvanBiezen thank professor you saved my life..even my lecturer couldn't explain this better 🙏🙏🙏🙏
Thank you, professor, this was extremely helpful in clearing up confusion and panic :3
Glad it helped! Yes this type of example encompasses the general principle of how to solve this type of problem taking the whole system into consideration.
thx a lot!
you solved my exam problems and i am so thankful that i know the concept of spring balance thoroughly!
thanks I finally understand and at the right timing my exams tomorrow 😊
Hi , I'm doing parachute auto steering for my project and I'm using motor to steer the parachute to the direction I need it to. However I need to find the tension and the force that is required for the strings and the motor to change direction. Could u please help me with this?
Sir i will be thankful of yours if you will upload a video for pulley system in which two pulleys are hanging on a ceiling( also have inertia) and a string goes over them and blocks are hanging over free ends
Thank you so much for your help, this really solidified information I was confused on for quite some time.
Glad it was helpful!
what about the force of M1? it also has a reaction force. shouldn't it also be part of calculating netforce?
The only force M1 applies is its weight and it doesn't act in the direction of the acceleration. The normal force counteracts it, again not in the direction of the acceleration.
M1 would only have a force if it had friction
Hi sir! Can you help me out? With a the same given, how do you determine the magnitude of acceleration and direction of a box?
Isn't that shown in the video? The magnitude of the acceleration is given and the direction can usually be surmized by which of the two hanging weights is the heaviest. In this case it makes more sense to talk about clockwise or counterclockwise direction of the system.
how is T2=(m2*g)-(m*a)?. Shouldn't it be (m2*g)+(m*a) ??. since you assumed positive direction in anti clockwise direction. Thank you
If you draw a free body diagram around m2 use F = ma you will find that T2 = m2 * g - m2 * a
yes sir. thanks it is clear now
thank you so much that was so helpful to me learnt something that i was struggling to know for the pas 2 days
would the tension be m3g - m3g?
Could you please show how to calculate the tension in the cables with friction force please
S Joyce
There are a number of examples in the video series with friction as well.
For any hanging object (in an inertial system) it will always be: mg +/- ma
Michel van Biezen is there any way I can show you what I have calculated and you can look at my workings. Im just not sure I am calculating the tension correctly
S Joyce
Feel free to show me.
(there are a number of videos in this playlist; that show you how to calculate the tension)
PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
Michel van Biezen do you have an email address?
S Joyce
Just go ahead and type in the equation you have here so I can take a look at it.
It seems from the final equation that, any differential in weights attached to two sides of the pulleys shall accelerate the system, slow or fast. But in reality, the system won't accelerate at all till a certain threshold of weight imbalance. That differential in my opinion shall vary based on friction between pulley & rope and pulley inertia but more importantly based on "weights". More the weight, more the differential needed to start acceleration. Can you quantify it with equations in above case? Thanks
+Raj Learn
Yes you can. Make one of the masses unknown, and then set the sum of all the forces equal to zero. You will find the minimum mass unbalance required to start the acceleration.
+Michel van Biezen Is it possible to elaborate it further? In my experiment with single pulley, I attached wts of 50g, 500g, 2400g and 4000g on one side. I noticed acceleration only after I put 400g, 1100g, 3100g and 6500 g respectively on other side. Then I took 2 pulleys, separated by 0.5 meters and attached same weights like before on one side. On other side I had to put wts of 500g, 1500g, 4600g and 8200g respectively to start acceleration. The pulley radius is 4 inch. I am unable to quantify these numbers with any theory. What this force is called in physics to overcome this inertia(?)?
+Raj Learn You are dealing with the difference between static and kinetic friction. Static friction is greater than kinetic friction, so it requires additional force to get the system moving than to keep it moving. Also if you are setting up an actual physical system, you are dealing with other factors such as play in the pulleys that could misalign, etc.
How would you approach a problem where the table was tilted at an angle? (say 20 degrees)... Thanks in Advance!
We have lots of examples in the following playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
my brain was blowing up at this problem, but you saved me
Сила трения груза М1 не должна препятствовать ускорению? Почему мы её не учитываем в системе?
In this example the table is "frictionless" as indicated in the title. But if there was friction, we would need to take it into account as you indicated.
@@MichelvanBiezen Спасибо, у меня плохо с английским языком, поэтому я ориентировался на язык "математический и физический". Привет от учителей России!
thank you very much l am very happy with you and my question it (if the masses are equal each is 10 kg and instead of the body on the table we put Helical captivity whow much the force reading thank you again
How do you determinate when friction matters or not?
That is usually indicated in the problem. If the coefficient of friction is zero, then there won't be any friction forces. If the coefficient of friction is not zero, then you have to take the friction forces into account.
?Dr what about if they say that there is a kinetic friction or a static friction
When there is static friction it means the block is not moving, because there is a force trying to avoid its movement. If the block is about to move we call it the maximun static friction and it means that, if you apply just a little bit more force to the block, it starts moving. Therefore if the block is moving, then there would be kinetic friction and it means that there is a force trying to interrumpt the block´s movement while you are applying a force to it (while the block is moving), so the kinetic friction must be indeed, less tan the force you are applying to the block.
Why is the formula to find the force of the objects mg? I thought it was f = ma or do I have to use mg when dealing with surface and pulley problems? If mass 3 was nonexistent, and mass 1 was being pulled along the surface with mass 2 going downward, how can you find the acceleration? I remember being told that the acceleration between those 2 masses are equal but i do not remember why. If they are, can i find the acceleration by taking mg and diving it by the total mass? Im guessing the total forces would be mg of the 2nd mass or would i have to add the tension force of the first mass as well? Thank you
Gravity is an acceleration so A=G therefore F=MA can be rewritten as F=MG which is what he uses.
Think of the units on g as Newtons per kilogram, rather than meters/second^2. What it really is, is a force field strength. Just as electric field is force per unit charge and has the units Newtons/Coulomb, gravitational field is also a force per unit property of matter, in this case mass. It tells you what force gravity will apply to a given object at a given position, based on its mass. It is a "coincidence" that it also has the same units as acceleration, and ends up being equal to the acceleration of a free falling body when the only force is gravity by definition. I put "coincidence" in quotes, because in general relativity, you learn why this is the case, and it is not just a coincidence.
What "F=m*a" refers to, is the NET FORCE equaling mass * acceleration. Not just any force. It is the vector sum of the forces, which we call net force. "Net" means total, in that you add up quantities taking in to consideration sign and direction where applicable. In this case, your "F=m*a" equation that involves gravity has to consider both forces acting on either hanging mass. Take m3 as an example. There is the tension in the string above it (T3) and its weight acting on it (m3*g). The F=m*a equation on this individual mass is T3 - m3*g = m3*a, considering both the tension in the string above it, and its own weight.
thank you so much i wasnt getting this concept and ur vdo really helped.... thanks a lot
Happy to help
Home come in some of your videos you used 9.8m/s^2 for gravity and sometimes you use -9.8m/s^2..? Thank you
When solving problems using the equations of kinematics, direction of the vectors (displacement, velocity, and acceleration) are important, therefore you need to use -9.8 m/sec^2. In this example we use the magnitude of g, and therefore we can ignore the sign. (Magnitudes of vectors cannot be negative).
You are a great teacher ❤
Thank you. We appreciate the kind words. 🙂
What if the two m's that were hanging have the same weight? What should I do??
Then the net force would be zero and therefore the acceleration would be zero as well.
If you don't know which mass is larger, or which way it will accelerate, you can pick a direction arbitrarily for its acceleration and set up the force addition on each object accordingly. If it turns out you selected incorrectly, you'll simply get a negative sign in your answer for acceleration, which means acceleration is in the opposite direction you assumed. Since it is ultimately a system of linear equations, the magnitude of acceleration is still the same as you got, with your assumption that acceleration was in the other direction.
If the hanging masses are both equal, the tension on the supported mass M1 will add up to zero, and there will be no acceleration. Using the same equations you solved assuming there would be acceleration, you will end up seeing that the acceleration is zero.
Thank you for this video, it helped a lot!
When calculating the tension for T3 why did you include 9.8m/s*2 in it, is that to account for the acceleration due to gravity?
It is multiplied by the mass to give the force exerted on the object via gravity. The force exerted by gravity is required to determine the net force the cord is exerting on the object to accelerate it at the value determined earlier: 2.8m/s^2
F=ma
We have acceleration, the mass, and we can easily determine one of the forces (Force of gravity), these are used to determine the unknown tension force.
T = Fg + Fnet
T = mg+ma
T = m(g+a)
I just realized your comments are 4 months old. Ohwell.
+Jewish Nazi why is your name "Jewish Nazi?"
@@ryanveyr9195 Probably some kind of an inside joke.
Superb explanation
hi great helpful video, could someone explain why T1G wasn't included in the equation
The tension is a force that is internal to the system and doesn't cause the system to accelerate. Only if you draw free body diagrams do you need to take the tension into account.
Can anyone explain what would happen if m2 and m3 have equal masses? Thank you:)
Then the system would not move.
this video really hepls me. Thank you
Thank you! This was very helpful!
I have a problem just like this, only I am given the acceleration of the system and told to solve for the mass which the system acceleration is towards. I cannot figure out how to solve for mass, with mass being on the bottom of the equation as well.
Essentially you are solving for m3. Multiply a (on the left side of the equation) with the sum of the three masses (in the denominator on the right side of the equation) and then solve for m3.
Nice teaching
If you are not taking M1 forces as part of the forces that act on the system, why is M1 mass used in the calculation of total masses?
You only need to consider external forces acting on the system, but you must include ALL the mass of the system since the WHOLE system is being accelerated.
Mass M1 is supported by the table below it, unlike the two hanging masses. As a result, the table below it applies a normal force (N) that will be as large as necessary to prevent M1 from sinking through the table, as long as the table is strong enough to provide this force. The forces in the vertical direction on mass M1 will therefore add up to zero, and not cause any acceleration on the system. No component of the weight of mass M1 will act in the direction of motion.
If there were friction in this problem, then the weight of mass M1 would indirectly come in to play. The normal force is directly determined by the weight it supports, and the friction on M1 would be determined by the normal force and friction coefficient (mu_s and mu_k). At static conditions, the value of mu_s*N an upper limit to how large the friction can be. When there is motion of M1 moving sideways along the table, then the friction force will be equal to mu_k*N, in the opposite direction of the existing motion.
THANK YOU SO MUCH THIS WAS THE MOST HELPFUL VIDEO
do you have any other form for grade9
good job, Mike.
Hello, thank you very much. Physics is fun with you. Thank you again
Glad you enjoy it! 🙂
Sir how T3=m3g+m3a ?
As g is in downward direction I think
T3=m3a-m3g?
Thanks (in advance)
If you use "up" and "down" directions in the equations, you must set up independent free body diagrams, and then link them together. It is much easier to consider aiding and opposing the acceleration of the whole system as depicted in the video.
I think we should not take total mass as you do but total is 3kg
Take the total mass of the system, (as shown in the video).
youre such a king
why do we consider the system as a single system? is it because there is no friciton?
+MrDoYouWannaBeOnTop With or without friction.Friction is an external force to the system.
how would i do this problem when the pulley has a mass?
You can find you answer in this playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 ruclips.net/user/ilectureonlineplaylists?shelf_id=4&view=50&sort=dd
thank you so muchhhh very very helpful
Glad it helped!
sir. if the table has friction , then how to find acceleration
We have a number of examples like that in the playlist.
Took my online tutors hours to explain this using vague answers and not showing what to do. Took me a minute to search up the problem and 6 minutes to skim through the video and figure out what to do
Fn isn't a reaction for Fg
because Fg is from the earth to the box and the reaction for this is a force from box to the earth . and Fn isn't that
+rashed a564
Respectfully, that is incorrect. When a force "reacts" to the application of an initial force, it is called a "reaction force". The box pushes against the table and the table pushes back.
+Michel van Biezen action reaction pair shouldn't act on the same object
Michel is right. Natural force pushes the object back up or else it would just fall through.
@@mrrashedali "Reaction force" doesn't exclusively refer to the Newton's Third Law reaction pair. It also refers to constraint forces that are directly determined by the forces they oppose.
Watching from India
Welcome to the channel! Glad you found our videos. 🙂
@@MichelvanBiezen 🙂 This type questions are asked in Medical & Engeering entrance exam in India.
thank you for the video
just a note: the weight of m1 and the normal force on it from the table happen to be equal and opposite in this case, but they are NOT a Newton's 3rd Law pair! My physics students get that wrong all the time.
Let's simplify this. Just take a table and place an block on top of the table with mass m. The weight of the block = mg The weight pushes against the table. What does the table do against the block? What is that principle called? Is that different from the situation when I push against a wall with a force F? What does the wall do back to me?
Michel van Biezen ah the book pushing on the table and the table pushing the book are indeed a pair. But “weight” is the force of gravity from Earth pulling the book. It’s equal+opposite force is the book pulling on the earth!
It is the same principle.
Michel van Biezen the table holding up the book has nothing to do with Newton’s 3rd law. What if the table is slanted? Suddenly the book slides, and the normal force is not equal and opposite to the weight. Has the book’s weight lost its equal and opposite pair? Does Newton’s 3rd law not apply to inclined planes?
@@willtanzola6629 Newton's third law DOES still apply, but the third law pairs to the forces involved in this problem are "out of the picture" of the problem we are trying to solve. The tensions are the only forces for which we need to consider Newton's third law pairs, and that is a simple matter of tension on any given cord segment being the same on both sides, and obviously opposite in direction.
The Newton's third law pair to any given object's weight, is that it pulls up on the Earth with an equal and opposite force of gravity. The Newton's third law pair to the normal force on mass M1, is that mass M1 also applies a normal force that pushes down on the table, which is supported by the compression in the posts that support the table. Ultimately, there is a support force from the Earth, that passes through numerous structures to support the table, to support the object on the table, and to support the axles that support the pulleys.
Because Earth's mass is significantly greater than the three masses involved in this problem, we can neglect the third law pairs to these forces that act upon the Earth, when solving this problem. The Earth will recoil by a negligible amount of upward acceleration, due to this system undergoing the situation described.
I like the Lagrangian mechanic method more for these kind of problems.
It is a very useful method indeed.
Really very helpful thank you sir
freaking helpful. Thanks a ton :D
awsm teaching # thanx
SIR THANK U😭
Thank you for the video!
Thank you very much
You da GOAT
If T3 = T4 is T1 = T2?
Yes, he also explicitly states T1=T2 at 6:33
He then states T3=T4 at 8:33
But T4 does not equal T1
So helpful
Glad it was helpful!
wow thanks sir you really helped me to understand i am in class 9 preparing for jee
That is a good time to start. Keep up the great effort. 🙂
Thanks sir for your blessing
Your video has no sound
Yes, it does, but since it is an older video, our older videos were recorded in mono, not stereo.
tnx alot sr
Ohhh fuck nevermind, i was reading your calculation wrong...
Keep up the great content ;)
Thank you!!!!!! Now I understand! :)
🔥🔥🔥
🙂
Great
I hear no sound 😥😥
Nathaniel John Abilar right speaker is blocked for some reason
100000thanks
thank youuuuuuuu
hello there
its me
lol
@@nachoproductions5984 hi dear
Heyowwwww
My right ear is not working 😂
Our older videos were filmed in mono sound :(
❤❤❤❤
Sir, I think you forgot to subtract friction from (m2g -m3g) to find acceleration.
The problem states that the table is frictionless
you can never have to many markers (;
We go through a lot of them every month.
W
This is great work. But there is a problem of concept during explaination. The reaction force of m1 is not the normal force. The weight is a gravitational force and it's reaction force is the force with which it pulls the earth. The reason why the weight is equal to the normal reaction is due to the second law of motion because there is no reaction in the vertical direction.
you sound like gru
You are not the first viewer to mention that. 🙂
hi klasmeyt
hi krast uwu
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@@abbydavid3133 hi dearrr
ruclips.net/video/mhE58_g6bv8/видео.html
HELLOOOO
sir i didn't understand a thing
+PrankingM@zeter
You probably want to start with the basics regarding tension and Newton's second law. Take a look at the playlists: PHYSICS 4.5 - STATICS & EQUILIBRIUM and PHYSICS 4 NEWTON'S LAWS OF MOTION
BRUHH WTTFFFFF. I JUST GOT BACK FROM MY PHYSICS EXAM AND IT HAD A PROBLEM LIKE THIS, AM IM SO MAD NOW, BECAUSE THIS SHIT ISN'T TAHT HARD. all my practice problems always had two objects, which meant i only ever had to deal with one tension. i didn't know tension just added up masses like that. wtf . fucc.
Thank you for this video!
thank you a lot