It is good to see that these videos are peaking your interest in physics and science in general. It does appear that you are re-discovering the principles and understanding them. Good for you!
I wish that you teached me when I was in high school, I always loved physics but my grade are not that good, but now thanks to god that I know your channel...
Your teaching is excellent but I do have one question, why did you make the gravity positive? I would have made the gravity negative because its going downwards, but it would give me the wrong answer.
@@abdulnaqvi8284 Because gravity itself, fundamentally, is a DOWNWARD force. (As it *PULLS* objects toward earth) So the positives value will always be downwards, and negative value will be upwards because it is against gravity. Hope u get it, correct me if I'm wrong.
Suvrath Hegde Sure. It makes everything a lot easier. If you are just trying to understand the concepts or show that you know how to solve a problem 10m/sec^2 makes it much simpler. But when the "exact" answer is asked for you must use 9.8 m/sec^2
Faisal, At first only using variables (and no numbers) may look strange and confusing. But it is the best way to learn how to solve problems in physics. I recommend that you learn this technique, because in the long run, it will greatly improve your ability to solve physics problems. Just take it one step at a time and make sure you understand each step before going on to the next step. After giving it some time it will make sense
it is pretty nice to see that you are still replying to comments considering the fact that the video was uploaded 9 years ago! I am generally solving these questions by turning these two masses into one body and finding the acceleration from there.
There is a formula for this type of question which is actually a derivation. Here the mass of object 1 should be greatest than mass of object 2 (m1 > m2) and should be in Vertical motion For acceleration a = [(m1 - m2)÷(m1 + m2)]×g For tension T= [(2m1m2) ÷(m1 + m2) ]×g (I mentioned in magnitude) If m1 = m2 Then a=0 If i am wrong correct me
@MichelvanBiezen Sir, it's actually very helpful. Your way of teaching is great and beginner friendly too and you explain the concept crystal clear.I really appreciate it.All I wanted to convey is that if direct formula and tricks are given it will be helpful for students appearing in competitive exams that's it. And Sir, I have been watching your videos for a long time and it really helped me to ace my exams😊. Thank you so much. Wow it's been 11years and you still reply. Hats off And I am going to attend an entrance exam. Wish me luck please🙏
why did he write m1g minus m2g and not plus? if you look at the diagram both forces act in the same direction so the resultant force on the system is g(m1 + m2). what is the explanation for putting a minus?
To find the net force you must add all the forces that aid the acceleration (they act in the same direction as the acceleration) and SUBTRACT all the forces OPPOSING the acceleration (they act in the opposite direction of the acceleration).
this saved my life from a high school teacher that doesnt know how to teach properly, basicaly a self tought class. This helped me very much thank you!
i know that the m2g is slow down the systyem acceleration but the definition when we sum the force is force in the same vector need to add not subtract
Thank you, Sir! This was great. I had two physics teachers that tried to explain this to me and they made it so complicated by subbing equations into each other. You made it so easy and I have a test tomrrow.
We have examples of that when the pulley has mass. Physics - Mechanics: Application of Moment of Inertia and Angular Acceleration (2 of 2) th-cam.com/video/wrhT5xGS-f8/w-d-xo.html
The tension in the string is 60.3, the resultant force is 60.3*2 = 120.6(N), which is less than the sum of the two weights (5+8)*9.8 = 127.4 (N). Please explain why!
The tension in a string can only equal the weight of the object havning from it, IF the object is not moving or moving at a constant speed. If the object is accelerating, the tension will not be the same. (See the other videos in the Newton's application videos).
For the tension to change from one part of the string to another part of the string there must be a force (from outside the system) somewhere that pulls on the string other than at each end by the two masses. If there isn't (like in this example), then the tension must be the same on both ends. Newton's third law states that for every action, there must be an equal and opposite reaction. The pull by the one mass is therefore equal and in the opposite direction to the pull of the other mass.
@@MichelvanBiezen sir if there is a string where one end of the string is attached toa rigid support and other end is passed through a pulley and is attached some load now in between pulley and one end of the string that is attached to the rigid support there attached another load directly to the string, now will the tension be same over the whole string sir
@@aryasomayajularajeswarisra9794 Tension is a "stretching force" that opposes a stretched structural member from increasing its length. In this example with an idealized string that is massless and inextensible, the string length remains constant. Tension will be as large as necessary to prevent the distance along the string between the two masses from increasing. If one tries to move away from the other along the path between them, the string's tension will oppose this motion. The string length remains constant, and the tension is the same on both sides of the string. The tension is the same throughout the string because the string is idealized to have no mass of its own, and the pulley is idealized to be massless and frictionless. All the pulley does in this example, is redirect the tension so that it pulls both masses up. There is a bearing that supports the pulley, which applies a force of 2*T to allow the pulley to support the tension from below, and be supported from above.
The mg force is in the Same direction so we can’t calculate in one system because F =ma can use only F make the object has Same acceleration but we need to calculate in each object and add 2 equation so we can get your equation does it correct
There are 2 ways to solve a problem like this. You can take the whole system at once and calculate the net force on the whole system, or you can draw a free body diagram around each mass and calculate the total force on each block, and then solve the the two equations simultaneously. The first method is easier and faster, but they are both correct methods to solve the problem. We show the second method in a different video.
What do you mean by the "source". We are only dealing with mass and force, not sources. Also, what do you mean by "solution"? We are either looking for acceleration of the system, or the tension in the string.
For the tension to change from one part of the string to another part of the string there must be a force (from outside the system) somewhere that pulls on the string other than at each end by the two masses. If there isn't (like in this example), then the tension must be the same on both ends. Newton's third law states that for every action, there must be an equal and opposite reaction. The pull by the one mass is therefore equal and in the opposite direction to the pull of the other mass.
The size of the mass doesn't matter. The rope is pulling on each mass with the same force and hence the tension is the same everywhere in the rope. It is physically not possible for the tension to be different, unless there is an external force (not part of the system) exerting a force on the string along the direction of the string.
We usually ignore friction (since it tends to be insignificant), but if the pulley has mass you must take into account its moment of inertia: see these two chapters: PHYSICS 13 MOMENT OF INERTIA APPLICATIONS and PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 starting at: th-cam.com/video/dEAtTpIG8AM/w-d-xo.html and th-cam.com/video/XHkDJwQ4Xng/w-d-xo.html
Let us say I'm making a 2:1 ratio pulley system. 1st mapping is: (A) 10kg weight on ground rope tied to it, (B) Pulley anchored on ceiling, and (D) I am hand pulling from free end on ground over top pulley to move (A) to a balanced point and progress capture. Here I get 10kg to move, 20kg on pulley anchor, while I tugged at 10kg. Correct? OK... Map 2: What if I add another identical size pulley hanging same height @ 60cm away... then run atop both pulleys still in a 2:1 (no direction change). Even if angle of the two pulleys is more or less out of the equation... Am I reducing friction? Does this have potential for preventing potential vectoring or twist of my tugging position? Did I make any difference like changing acceleration or ease I might feel by possibly sort of creating a larger bullwheel so there is less bite over just atop the single pulley? I know that is a lot to ask. Thank you for considering my inquiries.
MY first questions get transformed into another practical or not reason to know where I am going with this. Map X: I am standing on the shore trying to free a pinned raft. I am fortunate enough in this scenario to have the pulley attached to the Raft. I make a 2:1 Point (A) = Shoreline anchor Point (B) = Pulley secured to Raft Point (C) = Me tugging on the free end of rope. Here I should get mechanical advantage of shoreline anchor receiving same load as me tugging, while the pulley to Raft is doubled. Correct? SO... what if I employ a second identical pulley inline with first, same 60cm apart (maybe/likely even more) still 2:1 as in no direction change for second pulley. Same inquiry: am I achieving anything other than a backup pulley? Thank you again.
We call it Atwood's machine because Mr Atwood came up with this machine for the purpose of slowing down the acceleration of gravity, so that it would be practical to measure it accurately when subject to human reaction time. He also used it as a way to experimentally verify the constant acceleration kinematic formulas.
A very interesting lecture. My question: How would you explain this by the old system? Since you only examine a static condition, the forces in the rope’ s sides while the pulley is not rotating. I spent 60 years in static engineering and I am glad that I got out of it in the right time. One of my daughter is a physicist from Cambridge.
Michael thanks for your reply. my phone is this year's ZTE . iPhones may allow changing resolution but as far as I know I can not change the resolution .
We just checked it out on one of my kid's phone and it looks fine on that one as well on the regular setting. And she can increase the resolution as needed. (that is part of the youtube settings and it should not depend on your phone).
How can the acceleration be the same if the mass is bigger on one end meaning the acceleration is bigger would have to be bigger for m2 to have the same force?
Unless the pulley has mass or friction, there is nothing between the left end of the string and the right end of the string to make the tension different. Draw a free body diagram around the pulley (and not around the 2 masses) and you'll see that the tension must be the same unless there is another force to make them different. Watch the other videos in the playlists to get a better understanding.
Wow this video was God sent! Someone recommended this video in an MCAT practice question thread and this just explained everything so well. Thank you :)
Thank you so much.If I were not so daft I should have realised that.However, as far as I can see, we always seem to deal with a light string.Well, supposing that under the same conditions, as in the the simple example of 1;2 we deal with a heavy rope of mass m.Could we answer the question by the proportionality within the rope? I have tried this but still fail to find the right answer.
Sir, I also did a problem like this on the test and I have rounding errors as well. I'm kind of curious about how those rounding errors are created because the tension from the two sides (same rope) are really really close but it's not exact.
@@MichelvanBiezen Okay thank you! I kept more decimal places the last time I calculated it, and it perfected the result although they're still missing by 0.02.
i have one doubt . Here its said that tension in the string is same in both side of the pulley .but a pulley will be in static equilibrium if the tension is equal on both sides as turning effect clockwise and anticlockwise cancels out the rotation chance . here in this question the pulley will rotate anyway . so how we can say tension in the string is same on both sides ?plz help
@@MichelvanBiezen but how a pulley can rotate if the tension in the string on both sides are equal ?turning effect will cancel out and pulley will be at stuck na ? my email id is arunkumar8c@gmail.com . if u can elaborate with any sketches u can mail me to it . will be helpful if ur helping me .
I wish my professor would teach these concepts straightforward like Biezan, he does it in like 10min too. My professor just explains how the equations are derived and has us do practice questions as HW, without actually doing at least one problem to show us how its done. Going though step by step of how to solve physics questions in these videos helps me understand more and gives me hope to pass my physics class
The tension on both sides is the same due to two conditions, both of which are necessary: (1) the pulley is massless/frictionless so that there is no friction between the string and the pulley and (2) the string is massless. ONLY under those conditions is the force of tension constant throughout the string. It can be shown by writing N2L for any piece of string. Because the quantity (ma) for any piece of string is zero (due to its mass being zero), the tensions pulling it in the opposite directions must be equal. There are problems in AP Physics C where the tension in a string is NOT constant throughout. THe video does mention, of course, that the tension forces are of equal magnitude - but I don't think it ever explains why. In fairness, most online videos and even many textbooks gloss over this as well - which is fine unless you encounter something less trivial. Also, check this out: en.wikipedia.org/wiki/Atwood_machine
@@MichelvanBiezen oh no. I mean if mass 1 or m1 is required to be the heavier mass in the equation. For example, on one end of the rope has 20kg of weight and the other has 25kg. Would the 25kg object take the place of mass 1? Or is it interchangeable?
If you want to find the net force on this system, you need to establish a list of external forces acting on the system from objects outside it. The external forces are as follows: the weight of m1, the weight of m2, and the bearing reaction force (B) that supports the pulley. The tension force is not an external force, because it exclusively acts among objects within the system (mass 1, mass 2, the massless pulley, and the massless string). Add these forces up, defining upward as positive: Fnet = B - m1*g - m2*g Since the pulley is massless, the forces acting on it have to add up to zero. This means that B = 2*T, since it has two tension forces pulling it downward (two equal tension forces), and bearing reaction force B will be as large as necessary to hold the pulley in place. This means: Fnet = 2*T - g*(m1 + m2) Recall our previous expression for tension in the string: T = 2*g*m1*m2/(m1 + m2) Substitute to find mass: Fnet = 4*g*m1*m2/(m1+m2) + g*(m1+m2) Simplify: Fnet = g*(4*m1*m2 + (m1+m2)^2)/(m1+m2) You will find that this net force is consistent with the total mass multiplied by the acceleration of the center of mass between m1 and m2.
If the pulley has mass, you need to use a different technique. See this video: Physics - Application of the Moment of Inertia (10 of 11) Acceleration=? When Pulley Has Mass th-cam.com/video/TiZh9DYZ744/w-d-xo.html
There is nowhere on this planet that g equals exactly 10 m/s^2, or rather 10 Newtons/kilogram. 9.8 N/kg is close to the global average value, and is what most teachers expect you to use in Physics problems if it is not otherwise specified. The value ranges from 9.77 N/kg to 9.83 N/kg across the planet, as the apparent gravitational field in a reference frame for an observer who is stationary on the Earth's surface.
Since the original information was given with 1 significant figure, the answer should only have one significant figure. We were just pointing that out.
@@MichelvanBiezen i see. thank you for your answer! i guess i was just wondering whether a teacher would consider the answer incorrect if the student were to keep the 0.26.
What would be an example of an "internal torque"? In the case that something in the system needs to rotate, (and it has mass), then you have to take it into account. See the moment of inertia videos on this playlist.
@@MichelvanBiezen I meant torque from internal forces. We can consider the system pulley, rope, masses m1 and m2. So only the weights of the objects would contribute to the change of angular momentum. Wouldn't this help us to predict the direction of rotation of the pulley without taking into account the tension?
We have a lot of good examples in the 3 or 4 playlists that cover the topic of moment of inertia and angular momentum and angular acceleration. If you look at all those examples you'll get a good understanding of how that works. Much better than me trying to explain in a few sentences that which is already explained in systematic fashion in those videos. Do you know where to find them? th-cam.com/users/ilectureonlineplaylists?view=50&sort=dd&shelf_id=4
when do the two masses come to rest ? like once they accelerate is their a point where they balance and dont move or does one pulley (the heavier) one fall down and the lighter one stick on the wall
@@hritkandel8080 It's a simple matter of setting up an expression for the length of the string in terms of the vertical positions of each of the masses. The length of the string is a constant, which sets the kinematics of each mass (distance traveled from starting point, velocity, and acceleration) to be equal and opposite. The masses will therefore have to cross at exactly half the original height of the mass that started high. If we had a more complicated pulley system with a mechanical advantage it would get a lot more interesting, since the kinematic constraints would be more complicated than one being equal and opposite of the other.
Since the pulley does not have mass and does not have friction, no force is needed to make it rotate. Therefore the tension on both sides is equal. If the pulley has mass, the tension will NOT be the same.
Because one mass is accelerated UPWARD and the other is accelerated DOWNWARD Take a look at this playlist (it will help you understand) : PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS th-cam.com/play/PLX2gX-ftPVXV0Hc1CGnYKaPq19P3RX0Lx.html
You need an external force to accelerate the system (which is being held together by the string, thus the tension in the string simply keeps the two masses together)
Extremely helpful video, Mr. Biezen. Very clear and I appreciate how you solved algebraically before plugging in values; I usually define my formulas and jump in plugging numbers.
mhhh ..., since to get the expression of acceleration a_x, we need to assume that T1 = T2, thus a_x = (m1-m2)g / m1+m2. So why T1 isn't equal to T2, since we first assumed that T1 = T2 ?
+Ayush Poudel That is an interesting question. When a object is pulled from both directions (a rope, or a beam) the object is placed under "tension". The dictionary defines tension as: either of two balancing forces causing or tending to cause extension By that definition I would consider a reaction force.
+Ayush Poudel If the net force is zero, the acceleration will be zero. Which means that whatever motion existed before, that same motion will continue.
If the pulley is massless, then we don't care if it's frictionless. If the pulley is frictionless, we don't care if it's massless. In other words, if it's either frictionless or massless, things will work out the same. Also, if it's either of those things, it doesn't make it obvious that the tension is the same throughout the string. It requires that the string be massless and is a not-so-trivial thing to explain. Also: if you consider two blocks as the "system", one should carefully explain what the net force on the system is. If you add m1g and m2g, which are, after all, in the same direction, you get m1g + m2g, not (m1-m2)g. Of course, the answer IS m1-m2)g - but why? Should be explained more carefully, I think.
If the pulley has mass you have to take its moment of inertia into account. If the pulley has friction you have to take the friction force into account. They are very different.
If the pulley has mass BUT is frictionless, its mass (and rotational inertia) don't matter since the string will be sliding along, without exerting a torque. If the pulley is massless, then not being frictionless doesn't matter: there would still be no friction and no torque. When I say "frictionless", I mean, of course, the rim where the string is sliding, not the axle (which better be frictionless since, hey, intro physics). For the tension to be equal on both sides, there should be no friction between the pulley and the string AND the string must be massless. I am sure we agree on this.
Teacher, sorry but to me it does not sound Okay except if the diff probably is between 6032 and 6030N. Practically, tensions should be the same only if the masses are the same?.
+Sil Bombardi If the pulley has no mass and there is no friction, then the tension in the string will be the same on both sides even when the mass is not the same on both sides.
If the initial constants are given with only 1 significant figure of accuracy, then the final answer should only have one significant figure of accuracy. (This was just to illustrate that principle).
It is good to see that these videos are peaking your interest in physics and science in general. It does appear that you are re-discovering the principles and understanding them. Good for you!
I wish that you teached me when I was in high school, I always loved physics but my grade are not that good, but now thanks to god that I know your channel...
Michel van Biezen thank u
Thanks, sir!!!
Your teaching is excellent but I do have one question, why did you make the gravity positive? I would have made the gravity negative because its going downwards, but it would give me the wrong answer.
@@abdulnaqvi8284 Because gravity itself, fundamentally, is a DOWNWARD force. (As it *PULLS* objects toward earth)
So the positives value will always be downwards, and negative value will be upwards because it is against gravity.
Hope u get it, correct me if I'm wrong.
i've learned more from you in 11 minutes than i have from my professor in 9 hours.
Thank you!
You are welcome. Thanks for the comment.
Suvrath Hegde
Sure. It makes everything a lot easier. If you are just trying to understand the concepts or show that you know how to solve a problem 10m/sec^2 makes it much simpler. But when the "exact" answer is asked for you must use 9.8 m/sec^2
Faisal,
At first only using variables (and no numbers) may look strange and confusing.
But it is the best way to learn how to solve problems in physics.
I recommend that you learn this technique, because in the long run, it will greatly improve your ability to solve physics problems.
Just take it one step at a time and make sure you understand each step before going on to the next step. After giving it some time it will make sense
it is pretty nice to see that you are still replying to comments considering the fact that the video was uploaded 9 years ago! I am generally solving these questions by turning these two masses into one body and finding the acceleration from there.
Yes, that is the technique we use in the videos as well; look at the whole system as one entity.
My left ear learned a lot from this video.
lol i thought it was just me !
I would have been puzzled the whole video if I didn't see your comment xD I only had my right one in
i thought my earphones have broken!
I also thought my earphones were broken xD
My right 👂 is angry
There is a formula for this type of question which is actually a derivation.
Here the mass of object 1 should be greatest than mass of object 2 (m1 > m2) and should be in Vertical motion
For acceleration
a = [(m1 - m2)÷(m1 + m2)]×g
For tension
T= [(2m1m2) ÷(m1 + m2) ]×g
(I mentioned in magnitude)
If m1 = m2
Then
a=0
If i am wrong correct me
Nice work. We try to avoid assigning equations to everything, instead teaching the techniques to derive the equation in any situation.
@MichelvanBiezen Sir, it's actually very helpful. Your way of teaching is great and beginner friendly too and you explain the concept crystal clear.I really appreciate it.All I wanted to convey is that if direct formula and tricks are given it will be helpful for students appearing in competitive exams that's it. And Sir, I have been watching your videos for a long time and it really helped me to ace my exams😊. Thank you so much. Wow it's been 11years and you still reply. Hats off
And I am going to attend an entrance exam. Wish me luck please🙏
Best wishes and good luck!
First day discovering the channel. 995,000 subscribers!!! I bet when I come back in a couple days you’re over the mill mark 👏
Welcome to the channel. There are 9900 videos, so that will keep you busy for a while. 🙂
YOU JUST SAVED MY PHYSICS GRADE BLESS YOU
Thank! You had to do some heavy lifting too.
Thankyou my school teacher sucks so i study from you 🙂👍
Glad we are able to help with our videos. 🙂
Why this tutorial did't come when I was young. I'm now learning together with my daughter from your tutorial. It's very useful. Thank you.
why did he write m1g minus m2g and not plus? if you look at the diagram both forces act in the same direction so the resultant force on the system is g(m1 + m2). what is the explanation for putting a minus?
To find the net force you must add all the forces that aid the acceleration (they act in the same direction as the acceleration) and SUBTRACT all the forces OPPOSING the acceleration (they act in the opposite direction of the acceleration).
+Michel van Biezen yes because even though m2g is acting down, acceleration for M2 is upwards so it opposes. thank you very much, i understand now :)
Damn this is high quality video for a physics tutorial. Subbed
this saved my life from a high school teacher that doesnt know how to teach properly, basicaly a self tought class. This helped me very much thank you!
Thank you your explanations are very in depth with a simple and easy to follow delivery. Keep up the good work.
Really like how you explain simple pulleys. Make my life alot easier
i know that the m2g is slow down the systyem acceleration but the definition when we sum the force is force in the same vector need to add not subtract
Because you want to differentiate between which forces are aiding the acceleration and which forces are opposing the acceleration.
Yeah but how it can controversial the definition
Thank you, Sir! This was great. I had two physics teachers that tried to explain this to me and they made it so complicated by subbing equations into each other. You made it so easy and I have a test tomrrow.
Glad it helped!
Sir when will tension be different and how will we find acceleration when tension is different
We have examples of that when the pulley has mass. Physics - Mechanics: Application of Moment of Inertia and Angular Acceleration (2 of 2) th-cam.com/video/wrhT5xGS-f8/w-d-xo.html
The tension in the string is 60.3, the resultant force is 60.3*2 = 120.6(N), which is less than the sum of the two weights (5+8)*9.8 = 127.4 (N). Please explain why!
The tension in a string can only equal the weight of the object havning from it, IF the object is not moving or moving at a constant speed. If the object is accelerating, the tension will not be the same. (See the other videos in the Newton's application videos).
Thank you very much!
Can i know why the tension is same on both the strings even though two different masses are acting on either side
For the tension to change from one part of the string to another part of the string there must be a force (from outside the system) somewhere that pulls on the string other than at each end by the two masses. If there isn't (like in this example), then the tension must be the same on both ends. Newton's third law states that for every action, there must be an equal and opposite reaction. The pull by the one mass is therefore equal and in the opposite direction to the pull of the other mass.
@@MichelvanBiezen sir if there is a string where one end of the string is attached toa rigid support and other end is passed through a pulley and is attached some load now in between pulley and one end of the string that is attached to the rigid support there attached another load directly to the string, now will the tension be same over the whole string sir
@@aryasomayajularajeswarisra9794 Tension is a "stretching force" that opposes a stretched structural member from increasing its length. In this example with an idealized string that is massless and inextensible, the string length remains constant. Tension will be as large as necessary to prevent the distance along the string between the two masses from increasing. If one tries to move away from the other along the path between them, the string's tension will oppose this motion. The string length remains constant, and the tension is the same on both sides of the string.
The tension is the same throughout the string because the string is idealized to have no mass of its own, and the pulley is idealized to be massless and frictionless. All the pulley does in this example, is redirect the tension so that it pulls both masses up. There is a bearing that supports the pulley, which applies a force of 2*T to allow the pulley to support the tension from below, and be supported from above.
@@carultch thank you sir
@@aryasomayajularajeswarisra9794 No problem. Let me know if you have any further questions.
The mg force is in the Same direction so we can’t calculate in one system because F =ma can use only F make the object has Same acceleration but we need to calculate in each object and add 2 equation so we can get your equation does it correct
There are 2 ways to solve a problem like this. You can take the whole system at once and calculate the net force on the whole system, or you can draw a free body diagram around each mass and calculate the total force on each block, and then solve the the two equations simultaneously. The first method is easier and faster, but they are both correct methods to solve the problem. We show the second method in a different video.
@@MichelvanBiezen but the whole system solution has the source from the each mass equation rigg
What do you mean by the "source". We are only dealing with mass and force, not sources. Also, what do you mean by "solution"? We are either looking for acceleration of the system, or the tension in the string.
Question: Why would the tensions of both of the ropes be equal the same?
For the tension to change from one part of the string to another part of the string there must be a force (from outside the system) somewhere that pulls on the string other than at each end by the two masses. If there isn't (like in this example), then the tension must be the same on both ends. Newton's third law states that for every action, there must be an equal and opposite reaction. The pull by the one mass is therefore equal and in the opposite direction to the pull of the other mass.
@@MichelvanBiezen But aren't the masses of the two objects different? So it would have different tensions per rope?
The size of the mass doesn't matter. The rope is pulling on each mass with the same force and hence the tension is the same everywhere in the rope. It is physically not possible for the tension to be different, unless there is an external force (not part of the system) exerting a force on the string along the direction of the string.
@@MichelvanBiezen Does that mean the acceleration due to the ropes on the objects are also the same?
And what if I wanted to include friction and mass of the pulley? How would the equation look?
We usually ignore friction (since it tends to be insignificant), but if the pulley has mass you must take into account its moment of inertia: see these two chapters: PHYSICS 13 MOMENT OF INERTIA APPLICATIONS and PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 starting at: th-cam.com/video/dEAtTpIG8AM/w-d-xo.html and th-cam.com/video/XHkDJwQ4Xng/w-d-xo.html
Let us say I'm making a 2:1 ratio pulley system. 1st mapping is: (A) 10kg weight on ground rope tied to it, (B) Pulley anchored on ceiling, and (D) I am hand pulling from free end on ground over top pulley to move (A) to a balanced point and progress capture. Here I get 10kg to move, 20kg on pulley anchor, while I tugged at 10kg. Correct? OK... Map 2: What if I add another identical size pulley hanging same height @ 60cm away... then run atop both pulleys still in a 2:1 (no direction change). Even if angle of the two pulleys is more or less out of the equation... Am I reducing friction? Does this have potential for preventing potential vectoring or twist of my tugging position? Did I make any difference like changing acceleration or ease I might feel by possibly sort of creating a larger bullwheel so there is less bite over just atop the single pulley? I know that is a lot to ask. Thank you for considering my inquiries.
MY first questions get transformed into another practical or not reason to know where I am going with this. Map X: I am standing on the shore trying to free a pinned raft. I am fortunate enough in this scenario to have the pulley attached to the Raft. I make a 2:1 Point (A) = Shoreline anchor Point (B) = Pulley secured to Raft Point (C) = Me tugging on the free end of rope. Here I should get mechanical advantage of shoreline anchor receiving same load as me tugging, while the pulley to Raft is doubled. Correct? SO... what if I employ a second identical pulley inline with first, same 60cm apart (maybe/likely even more) still 2:1 as in no direction change for second pulley. Same inquiry: am I achieving anything other than a backup pulley? Thank you again.
We call it Atwood's machine because Mr Atwood came up with this machine for the purpose of slowing down the acceleration of gravity, so that it would be practical to measure it accurately when subject to human reaction time. He also used it as a way to experimentally verify the constant acceleration kinematic formulas.
Nice comment. Like the insight.
Shadd,
I believe I made a problem like that.
3:10
6:30
What happens at these 2 times?
A very interesting lecture. My question:
How would you explain this by the old system? Since you only examine a static condition, the forces in the rope’ s sides while the pulley is not rotating.
I spent 60 years in static engineering and I am glad that I got out of it in the right time. One of my daughter is a physicist from Cambridge.
In this example, the pulley is rotating and the masses are accelerating.
Michael thanks for your reply. my phone is this year's ZTE . iPhones may allow changing resolution but as far as I know I can not change the resolution .
We just checked it out on one of my kid's phone and it looks fine on that one as well on the regular setting. And she can increase the resolution as needed. (that is part of the youtube settings and it should not depend on your phone).
Video was very helpful, helps an AP high school student more than the teacher does.
How can the acceleration be the same if the mass is bigger on one end meaning the acceleration is bigger would have to be bigger for m2 to have the same force?
The two masses are connected. Therefore they must have the same speed and same acceleation.
Can you do this problem including the friction and mass of the pulley ?
Look in this playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS 2 th-cam.com/video/TiZh9DYZ744/w-d-xo.html
Anyone else cramming some last minute studying before an exam?
Lol me
Lol me
Hahaha
Akuu
How is it that the tensions are the same if the masses are different weights?
Unless the pulley has mass or friction, there is nothing between the left end of the string and the right end of the string to make the tension different. Draw a free body diagram around the pulley (and not around the 2 masses) and you'll see that the tension must be the same unless there is another force to make them different. Watch the other videos in the playlists to get a better understanding.
Michel van Biezen so is it a misconception to say that the masses have the effect on tension?
It is not a misconception, because they do cause the tension in the string. Take a look at the other videos to get a better understanding.
Thank you sir.Your lecture is the savior.I really hate my physics teacher and like a rubbish.Thank you so much for my physics test!
Glad you found our videos. Good luck on your test. 🙂
Wow this video was God sent! Someone recommended this video in an MCAT practice question thread and this just explained everything so well. Thank you :)
Glad it helped and glad you found our videos. They should be a big help for getting ready for the MCAT. 🙂
Thank you so much.If I were not so daft I should have realised that.However, as far as I can see, we always seem to deal with a light string.Well, supposing that under the same conditions, as in the the simple example of 1;2 we deal with a heavy rope of mass m.Could we answer the question by the proportionality within the rope? I have tried this but still fail to find the right answer.
You are so welcome
Sir, I also did a problem like this on the test and I have rounding errors as well. I'm kind of curious about how those rounding errors are created because the tension from the two sides (same rope) are really really close but it's not exact.
Just keep more decimal places, or just use the calculator results without rounding.
@@MichelvanBiezen Okay thank you! I kept more decimal places the last time I calculated it, and it perfected the result although they're still missing by 0.02.
i have one doubt . Here its said that tension in the string is same in both side of the pulley .but a pulley will be in static equilibrium if the tension is equal on both sides as turning effect clockwise and anticlockwise cancels out the rotation chance . here in this question the pulley will rotate anyway . so how we can say tension in the string is same on both sides ?plz help
The tension in the string will be the same on both sides, even when the object are accelerating.
@@MichelvanBiezen but how a pulley can rotate if the tension in the string on both sides are equal ?turning effect will cancel out and pulley will be at stuck na ? my email id is arunkumar8c@gmail.com . if u can elaborate with any sketches u can mail me to it . will be helpful if ur helping me .
If the pulley does not apply a force to the string (as in this example), then the tension on both side is the same.
How do we know when to include tension or not?
Tension is an internal force to the system. Only external forces can accelerate the system.
This man deserves a damn award, saving my grade one video at a time.
I have unsuccessfully tried to calculate the length of the string in such problems.Can anyone suggest a method?
The string can be any length (assuming we can ignore the mass of the string) and the acceleration would be the same
But the whole system
Has the source from each of mass equation right?
I am not sure what you are asking. The mass of the whole system = m1 + m2
I wish my professor would teach these concepts straightforward like Biezan, he does it in like 10min too. My professor just explains how the equations are derived and has us do practice questions as HW, without actually doing at least one problem to show us how its done. Going though step by step of how to solve physics questions in these videos helps me understand more and gives me hope to pass my physics class
Thank you so much! I understand you much more than my AP Physics teacher... hopefully i'll do fine on my test tomrrow!
Jonathan che my words exactly
How’d it go??
It's been four years. Did you pass?
Very Nice Video
But , I have a doubt sir.
Isn't the Tension in the string same because it is the same string used ?
Yes, as shown in the video.
At 6.45 you assume the tension on side to be T1 and the other side to be T2 .Why? Can we solve it without using this method ?
The tension on both sides is the same due to two conditions, both of which are necessary: (1) the pulley is massless/frictionless so that there is no friction between the string and the pulley and (2) the string is massless. ONLY under those conditions is the force of tension constant throughout the string. It can be shown by writing N2L for any piece of string. Because the quantity (ma) for any piece of string is zero (due to its mass being zero), the tensions pulling it in the opposite directions must be equal. There are problems in AP Physics C where the tension in a string is NOT constant throughout.
THe video does mention, of course, that the tension forces are of equal magnitude - but I don't think it ever explains why. In fairness, most online videos and even many textbooks gloss over this as well - which is fine unless you encounter something less trivial.
Also, check this out: en.wikipedia.org/wiki/Atwood_machine
This video is exponentially better than the SAT study books I just bought.
my you tube settings don't have resolution settings it may be available on you tube red not going that route
While the video is playing there is a spoked wheel symbol at the bottom of the picture. That is the "settings" button.
Does the first mass have to be the heavier on or is it nonimportant?
One mass must be greater than the other mass or there will not be an acceleration.
@@MichelvanBiezen oh no. I mean if mass 1 or m1 is required to be the heavier mass in the equation. For example, on one end of the rope has 20kg of weight and the other has 25kg.
Would the 25kg object take the place of mass 1? Or is it interchangeable?
So if we want to find the nett force, the bigger mass should minus the smaller mass?
If you want to find the net force on this system, you need to establish a list of external forces acting on the system from objects outside it. The external forces are as follows: the weight of m1, the weight of m2, and the bearing reaction force (B) that supports the pulley. The tension force is not an external force, because it exclusively acts among objects within the system (mass 1, mass 2, the massless pulley, and the massless string).
Add these forces up, defining upward as positive:
Fnet = B - m1*g - m2*g
Since the pulley is massless, the forces acting on it have to add up to zero. This means that B = 2*T, since it has two tension forces pulling it downward (two equal tension forces), and bearing reaction force B will be as large as necessary to hold the pulley in place.
This means:
Fnet = 2*T - g*(m1 + m2)
Recall our previous expression for tension in the string:
T = 2*g*m1*m2/(m1 + m2)
Substitute to find mass:
Fnet = 4*g*m1*m2/(m1+m2) + g*(m1+m2)
Simplify:
Fnet = g*(4*m1*m2 + (m1+m2)^2)/(m1+m2)
You will find that this net force is consistent with the total mass multiplied by the acceleration of the center of mass between m1 and m2.
i've learned more from you in 11 minutes than i have from my professor in 9 hours.
What a perfect video for the night before my exam. Thank you so much for this!
All the best on your exam.
Thank you, You have helped my understanding and with that i have improved my grade significantly.Thanks
Would it be solved the same way if the pulley had mass? Thank you so much, this was very helpful!
If the pulley has mass, you need to use a different technique. See this video: Physics - Application of the Moment of Inertia (10 of 11) Acceleration=? When Pulley Has Mass th-cam.com/video/TiZh9DYZ744/w-d-xo.html
For simplified calculation, can't we assume the value of g as 10. Bcoz in Indian entrance exams (for universities) we are asked to assume g= 10m/s^2
There is nowhere on this planet that g equals exactly 10 m/s^2, or rather 10 Newtons/kilogram. 9.8 N/kg is close to the global average value, and is what most teachers expect you to use in Physics problems if it is not otherwise specified. The value ranges from 9.77 N/kg to 9.83 N/kg across the planet, as the apparent gravitational field in a reference frame for an observer who is stationary on the Earth's surface.
its always the old videos, thank you so much
Glad you like them!
what? you have to round down to a=2m/s2 because they gave you whole numbers?? what is that all about...
Since the original information was given with 1 significant figure, the answer should only have one significant figure. We were just pointing that out.
@@MichelvanBiezen i see. thank you for your answer! i guess i was just wondering whether a teacher would consider the answer incorrect if the student were to keep the 0.26.
pleas uplod video how to solve eqution of tnsion
We have: PHYSICS 17 TENSION AND WEIGHT th-cam.com/play/PLX2gX-ftPVXXc4Evk5apxnDaiNREn14m9.html
"times g which is 9.8 meters per second squared"
where did 9.8 meters per second squared come from? Thanks
g is the acceleration due to gravity
It has been obtained experimentally and is a known value
thanku sir...love and respect from india...
Welcome to the channel.
Great teaching understood it in 13 min. Wish my school had a good Physic teacher.
Glad it was helpful!
Merry Christmas 🌲 thanks for all the wonderful videos and explanations.
Merry Christmas to you as wll. 🙂
@@MichelvanBiezen 🙂
Dear lord, I was trying to figure out how to answer a problem, surfed for an hour and half, and you helped me. THANK YOU !
thanks ,but have you eqution of tnsion vedio
Take a look at these videos: PHYSICS 17 TENSION AND WEIGHT th-cam.com/play/PLX2gX-ftPVXXc4Evk5apxnDaiNREn14m9.html
Can we say that internal torques don't change the angular momentum in the same way that internal forces don't change linear momentum?
What would be an example of an "internal torque"? In the case that something in the system needs to rotate, (and it has mass), then you have to take it into account. See the moment of inertia videos on this playlist.
@@MichelvanBiezen I meant torque from internal forces. We can consider the system pulley, rope, masses m1 and m2. So only the weights of the objects would contribute to the change of angular momentum. Wouldn't this help us to predict the direction of rotation of the pulley without taking into account the tension?
We have a lot of good examples in the 3 or 4 playlists that cover the topic of moment of inertia and angular momentum and angular acceleration. If you look at all those examples you'll get a good understanding of how that works. Much better than me trying to explain in a few sentences that which is already explained in systematic fashion in those videos. Do you know where to find them? th-cam.com/users/ilectureonlineplaylists?view=50&sort=dd&shelf_id=4
@@MichelvanBiezen Thanks Sir, I will check it out.
when do the two masses come to rest ? like once they accelerate is their a point where they balance and dont move or does one pulley (the heavier) one fall down and the lighter one stick on the wall
+Devious Minion The will continue to accelerate until one of them crashes on the floor or at the pulley.
So is there a way we can know the distance at which they will cross each other after accelerating ? Is it half the original distance between them ?
@@hritkandel8080 It's a simple matter of setting up an expression for the length of the string in terms of the vertical positions of each of the masses. The length of the string is a constant, which sets the kinematics of each mass (distance traveled from starting point, velocity, and acceleration) to be equal and opposite. The masses will therefore have to cross at exactly half the original height of the mass that started high.
If we had a more complicated pulley system with a mechanical advantage it would get a lot more interesting, since the kinematic constraints would be more complicated than one being equal and opposite of the other.
Thank you so much! Had a problem like this on Khan academy, the hints and solutions really weren’t helping glad I found this!!!
Glad it helped!
why the tension of each sides of the pully is same
?
Since the pulley does not have mass and does not have friction, no force is needed to make it rotate. Therefore the tension on both sides is equal. If the pulley has mass, the tension will NOT be the same.
sir,but the mass and forces are not equal....how the same tension is working both side...?
Because one mass is accelerated UPWARD and the other is accelerated DOWNWARD Take a look at this playlist (it will help you understand) :
PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS th-cam.com/play/PLX2gX-ftPVXV0Hc1CGnYKaPq19P3RX0Lx.html
thank u....
Thank you so much for this! I completely did not understand the lesson from my teacher, but this video explained it wonderfully.
Glad you found our videos! 🙂
I was asked bandwidth for same problem but only with single mass. Any idea on how to do it?
What did you mean by tension is within the system?
You need an external force to accelerate the system (which is being held together by the string, thus the tension in the string simply keeps the two masses together)
Extremely helpful video, Mr. Biezen. Very clear and I appreciate how you solved algebraically before plugging in values; I usually define my formulas and jump in plugging numbers.
mhhh ...,
since to get the expression of acceleration a_x, we need to assume that T1 = T2, thus a_x = (m1-m2)g / m1+m2.
So why T1 isn't equal to T2, since we first assumed that T1 = T2 ?
Is the pulley diameter is dependent or independent in the pulley related problems???
Unless we include the mass (and the moment of inertia) of the pulley, we don't have to worry about the diameter of the pulley.
Beautifully explained
what is they question asked to find the acceleration of each block ?
+MrDoYouWannaBeOnTop
Each block will have the same acceleration (since they are connected)
+Michel van Biezen Same magnitude, directions will be opposite!
Thank you so much. Repeating the video over and over has helped me a lot!
sir, what is the magnitude of the tension of the uppermost string(connecting the ceiling and the pulley)?
Is the tension due to reaction force sir
+Ayush Poudel That is an interesting question. When a object is pulled from both directions (a rope, or a beam) the object is placed under "tension". The dictionary defines tension as: either of two balancing forces causing or tending to cause extension By that definition I would consider a reaction force.
+Michel van Biezen Sir then the net force on system is zero but there is still downward motion ??
+Ayush Poudel If the net force is zero, the acceleration will be zero. Which means that whatever motion existed before, that same motion will continue.
Very clear explanation from Papua new Guinea
Glad you think so! Welcome to the channel!
I love to learn physics specially when mathematical equations are involved. You do such a great job at that. Thanks you for sharing the knowledge!
You are very welcome. Glad you found our videos! 🙂
Thank you sir. You are better than my real Physics teacher.
If the pulley is massless, then we don't care if it's frictionless. If the pulley is frictionless, we don't care if it's massless. In other words, if it's either frictionless or massless, things will work out the same.
Also, if it's either of those things, it doesn't make it obvious that the tension is the same throughout the string. It requires that the string be massless and is a not-so-trivial thing to explain.
Also: if you consider two blocks as the "system", one should carefully explain what the net force on the system is. If you add m1g and m2g, which are, after all, in the same direction, you get m1g + m2g, not (m1-m2)g. Of course, the answer IS m1-m2)g - but why? Should be explained more carefully, I think.
If the pulley has mass you have to take its moment of inertia into account. If the pulley has friction you have to take the friction force into account. They are very different.
If the pulley has mass BUT is frictionless, its mass (and rotational inertia) don't matter since the string will be sliding along, without exerting a torque. If the pulley is massless, then not being frictionless doesn't matter: there would still be no friction and no torque. When I say "frictionless", I mean, of course, the rim where the string is sliding, not the axle (which better be frictionless since, hey, intro physics).
For the tension to be equal on both sides, there should be no friction between the pulley and the string AND the string must be massless. I am sure we agree on this.
damn entered my left ear and left
+Dhruv Pandey
Sorry but our early video are for left-ear only. Then we discovered we could make our videos for left- AND right-ears!
+Michel van Biezen lool
Entered your ear and left 🤣🤣😂😂
what video is before this?
Intuitively I would think it’s masses subtracted as well on the bottom
No, the force must act on the total mass of the system. F = ma
Thanks 🙏
You’re welcome 😊
Could you do another example where this system is being supported by another pulley son the actual system is moving verticallt
Mind expanded thank you 🙏
Glad you found our videos. 🙂
I have an exam in 10 minutes this saved my skin
I have clearly understood whatever he explained me...now I can do my exam well.
Thank you so much, you saved my test grade
Good luck on your test.
Where do you get that 9.8 m/cm2 from?
MrCharlesdale Not cm but sec. I use the abbreviation "sec" to represent seconds.
Teacher, sorry but to me it does not sound Okay except if the diff probably is between 6032 and 6030N. Practically, tensions should be the same only if the masses are the same?.
+Sil Bombardi If the pulley has no mass and there is no friction, then the tension in the string will be the same on both sides even when the mass is not the same on both sides.
how?
If I could like this video twice I would. Thank you so much Mr Van Biezen for these
Just watched an entire 2 min studypug ad AND clicked the link, cuz you deserve it. Saved my ass
Thank you
I didn't get why he told us to write the acceleration as 2 and not 2.26.
If the initial constants are given with only 1 significant figure of accuracy, then the final answer should only have one significant figure of accuracy. (This was just to illustrate that principle).
Michel van Biezen thank you so much for the quick response sir!
In college physics, this is amazing. thank you.
You are welcome. Glad you found our videos. 🙂