Thank you for the testimonial. Yes, the key to understanding any problem (or anything in science for that matter) is to pick out the fundamentals of what you know to be true and correct and then piece them together.
Gareth, This is an algebra question. The key to working out algebra problems (as well as any mathematical problem) correctly they should always be in equation format. That way you can see what changes from line to line. In this case, the only change in the equation is that the a2 changed to (1/2)a1, which would imply that a2 = (1/2)a1. Thus I did not multiply by 2, but simply replaced one term by an equivalent term.
I have a mechanics exam in two weeks time. I'd like to thank you for all the videos you have shared to the public to help us students out. You sir, are a genius :)
my teacher went over this same exact situation on monday. didnt explain half of the things you did. made everything seem 100x more complicated than it actually is. i didnt even take notes on it because i hoped that you had a video on it which is 10000000x better than taking his notes. please adopt me.
For those who wonder why "m1g" is positive while "m1a" is negative thought the gravitational acceleration and "a" are both in the same direction, Fnet = ma. The forces which are acting on mass 1 are tension and gravity. If we take the acceleration direction as our positive direction, then the equation should be: +m1g - T = ma >> subtract "m1g" from both sides, multiply by -, and you should get the same equation as the video.
I also see why the relationship between the two acceleration is like that. From h = 1/2 at^2. The one moving down will always cover twice the distance of the one on the table and that's because of how the rope is connected in the pulley. Gracias 👍
This is really helpful! How would I factor in the Inertia of the pulley? For example if the moment of inertia, I, and the angular acceleration, alpha, need to be included in the question. How would this affect the tension in the rope?
Yes the technique is very different. There are a number of examples in the CHAPTER 13.1 Playlist: Physics - Application of the Moment of Inertia (8 of 11) Acceleration=? When Pulley Has Mass (mu=0) ruclips.net/video/p6ZlBp_vQ6c/видео.html
kylie, Good question. It is often difficult to comprehend or visualize what is actually going on and our intuition sometimes leads us astray. In this problem we are using and "ideal" pulley which means that it does not have mass. If the pulley does not have mass then there isn't an object to accelerate and when drawing the free body diagram around the pulley the forces on both sides must be equal. If the pulley had mass, then you would be correct.
i didnt understand the part where you say that the acceleration of m2 will be half of that of m1....why would it be like that ? i didnt get it through the pulley reference also...😞
Since m2 is connected to the midpoint of the pulley, you have what we call a 2:1 mechanical advantage. You need 1/2 the force, but will require twice the pulling distance to move m2. It is the way pulleys work. If you have one, try it and see that it works that way.
Simply put, if you draw the free body diagram for pulley 2, then you could clearly see, 2 tensions of T1 on the right-hand side and thus it must be T2 = 2 x T1 on the left-hand side.
Just a point of clarification Michael. If you're multiplying m2(1/2a1) by 2 in order to get rid of the 1/2a1 doesn't that give 2m2 a1? Because don't you multiply the whole expression by 2 i.e multiplying m2 x(1/2a1) by 2 so you get 2(m2 x 1/2a1) = 2m2 x a1
Alexkey Ov, Vectors have both magnitude and direction. In the equation you reference, we just use the magnitude part and determine the sign by figuring out which way the block is moving. The best way to figure out the equation is this way: 1) if the block is not moving the T1 = m1g 2) if the block is accelerating upward you will need additional force to cause the acceleration T1 = m1g + m1a 3) if the block is accelerating downward you don't need all of the force to hold it against gravity T1 = m1g - m1a
Could you pleas do similar problem,but on an inclined plane, what I mean is m2 hanging and m1 is one the plane inclined at an angel of 30. Thank you for all your help sir.
To find the direction of the tension in the string you must express it in terms of the object relative to. Note that every string has tension in both directions simultaneously.
The direction of the tension only makes sense if you have a reference point. (tension relative to what?) Thus the tension T1 just above the 5 kg block is directed upward RELATIVE to the 5 kg block. But T1 on the top string is directed to the right relative to the left pulley and to the left relative to the right pulley.
so for this problem, why couldn't you do m1g1 -m1a1 = m1a1. This is the F= ma but instead of T2 i just used T1. Using this formula you move m1a1 to the other side hence giving m1g1 = m1a1 + m1a1. Here you factor a out and move both m1 to the other side. This will give you (m1g1/m1+m1). Doing it this way also gave me 4.9. So is this way also acceptable for future problems?
If you draw a free body diagrams, then identify all the forces in and out of the diagram, and if they match your equation, you have the correct equation. (Better not use the "can this always be used")
I have the same question as another commenter ... how does T2 = 2 * T1 because if the masses are being accelerated, arent the pulleys then also being accelerated, meaning forces aren't balanced?
+Catherine Flynn Whenever you draw a free body diagram around an object (like the dashed line box around the pulley), the sum of the forces (added as vectors) must satisfy F = m * a Since the pulley is assumed not to have mass in this example, the sum of the forces must equal zero. Therefore 2 T1 = T2
Please sir I cannot understand the idea of a2=0.5 a1 , though mass 1 is actually 25% of mass 2 , is not this a factor affects both the disatnce and acceleration ? Waiting for your reply 🤔🤔
That is the result of the mechanical advantage of a pulley. The tension in the string connected to m2 is double the tension in the string connected to m1, and the velocity of m2 is half the velocity of m1. Hence the acceleration of m2 is also half of the acceleration of m1. See the sketch on the right side of the board.
@@MichelvanBiezen so this means that the mass connected with rope doesnot affect the acceleration as I understand , it is about the mechanical advantage . I truely appreciate this , gratful for your help❤🌷
If T2=2T1 and T2=m2g=196N. Then can T1=T2/2= 98N. ? Then use T1=m1g-m1a1 to find a1? Which I found to be 9.8=[a1] ? then use a2=1/2 a1 to find a2. Which in this case would be 4.9 m/s^2? Am I wrong if I do this? Please respond?
i have question. i understand the diagram on the right where if we pull it upward for 1m, the distance will only be half at the end of the pulley. but i don't understand how we can use that length reference in an acceleration problem. is it possible for the pulley to "take" half of the acceleration as well?
You can prove mathematically that it has to work out this way, or you can do it by inspection as the professor did, simply reasoning through the proportions of changes in length. The changes in length are always proportional, which means time derivatives of any order of these changes in length will also be proportional. Below, I've written up how we know that, with a calculus proof. Define the position of block m1 to be distance y below the pulley. Define the position of block m2 to be distance -x from the pulley, such that x is a negative number when the block is to the left of the pulley. We assign a negative sign to this one, so that its acceleration can be a positive number, which you'll see later why this helps. The length of the string among the pulleys, the fixed point, and mass m1 is as follows (neglecting the diameter of the pulley, and the length of the string with tension T2): L = | 2*x | + y Since x is negative, replace the absolute value bars with a negation: L = -2*x + y Solve for y in terms of x: y = L + 2*x The acceleration is defined as the second derivative of each block's position, relative to time. Therefore: a2 = d^2/dt^2 x a1 = d^2/dt^2 y Substitute y = L + 2*x: a1 = d^2/dt^2 (L + 2*x) Since L is a constant, it's derivative is zero, and likewise, its second derivative is zero. The leading constant of 2, can be pulled out in front on the 2*x term: a1 = 2 * d^2/dt^2 x We recognize that this is the expression for a2 in this equation, which therefore means that: a1 = 2*a2 And that concludes how we can prove that acceleration a1 has to equal 2*a2. It is based on the constraint that the length of the rope is a constant.
There are two choices. It either accelerated downward, or the friction between the table and the other block are sufficient to keep it from accelerating. Therefore assume the block accelerates and see what the acceleration is. If the answer is negative (compared to the assumed direction of the acceleration) then the block would not move.
If you draw a free body diagram around the pulley on the left, you'll see that the one force to the left must be equal to the 2 forces on the right. Therefore both tensions on the left mus be equal and they must be half of the tension on the left side of the pulley.
+Dwight D Since there is no friction between the large mass and the table, there is no friction force, so how would the large mass keep the small mass from accelerating downward? Always go back to the F= ma equation and find all the forces aiding the acceleration and all the forces opposing the acceleration. If the forces causing the acceleration are greater, there will be an acceleration.
+Roshan Varghese I am not sure about the question but I'll try to answer it this way: Tension acts in both directions. (The string pulls on each object that it is connected to.)
T2 is the tension in the rope that connects the 20 kg mass to the pulley on the left. It is T2 that accelerates the 20 kg mass to the right at acceleration = a2. If you draw a free body diagram around that pulley you have 3 forces contained within the free body diagram. T2 pulling to the left and 2 T1 forces pulling to the right (relative to the pulley). These forces must equal one another, thus T2 = 2 T1
If you look at the free body diagram around the pulley then you can determine that the forces pulling to the left must equal to the forces pulling to the right. Thus T2 = T1 + T1
Tension means stretching force. A flexible rope, chain, cable, cord, or string has to carry a stretching force if it carries any force at all. If you attempt to push a rope, it will become slack, and be incapable of carrying any force. Therefore, the tension force has to point away from the object to which it is applied.
Because the pulley by mass M2 is treated as an idealized pulley, which means it has negligible mass and negligible friction. Forces acting on an object of negligible mass must add up to zero, otherwise it would have infinite acceleration. Two segments of the rope carrying tension T1, act on this pulley to the right. Both segments carry equal tension, because the pulley is idealized as massless and frictionless. The only remaining force acting on this pulley, is tension T2, which acts to the left. This means 2*T1 - T2 = 0, and we can solve for T2 = 2*T1.
Due to the mechanical advantage of a pulley. M2 is attached to the midpoint of the pulley. Therefore as the string is pulled on one end, the point where M2 is attached only moves half as far. (try it on a pulley and you will see). Mechanical efficiency requires that F1 * d1 = F2 * d2 (otherwise you create work and energy out of nothing).
Since the force required to lift up the object attached to the pulley is only half the weight of the object, you must pull twice the distance in order to accomplish the same amount of work. W = F * d = (1/2)F * 2d
Thank you for the testimonial.
Yes, the key to understanding any problem (or anything in science for that matter) is to pick out the fundamentals of what you know to be true and correct and then piece them together.
Gareth,
This is an algebra question. The key to working out algebra problems (as well as any mathematical problem) correctly they should always be in equation format. That way you can see what changes from line to line.
In this case, the only change in the equation is that the a2 changed to (1/2)a1, which would imply that a2 = (1/2)a1.
Thus I did not multiply by 2, but simply replaced one term by an equivalent term.
I have a mechanics exam in two weeks time. I'd like to thank you for all the videos you have shared to the public to help us students out. You sir, are a genius :)
my teacher went over this same exact situation on monday. didnt explain half of the things you did. made everything seem 100x more complicated than it actually is. i didnt even take notes on it because i hoped that you had a video on it which is 10000000x better than taking his notes. please adopt me.
I don't know how, but you make what I initially expected to be a difficult problem easy-peasy...you're smthg special.
For those who wonder why "m1g" is positive while "m1a" is negative thought the gravitational acceleration and "a" are both in the same direction, Fnet = ma. The forces which are acting on mass 1 are tension and gravity. If we take the acceleration direction as our positive direction, then the equation should be: +m1g - T = ma >> subtract "m1g" from both sides, multiply by -, and you should get the same equation as the video.
This problem is hard, hopefully I'll see it on the test. I'm sure lots of ppl will be dead.
You make physics exciting to learn, Sir. Thank you.
I also see why the relationship between the two acceleration is like that. From h = 1/2 at^2. The one moving down will always cover twice the distance of the one on the table and that's because of how the rope is connected in the pulley.
Gracias 👍
That is correct. You understand the problem.
I've never understood Newton this well... thank you so much!!!
at 6:39 you "factor out an a1" and it seems like you delete an "a1" out of the equation and just stop counting it... how could you do that ?
I factured out the a1 and then divided both sides of the equation by m2 + 4 m1
You are the definition of Big Brain Time
This is why I love mechanical physics.....so Practical.
simply superb professor.ur teaching is awsome
This is really helpful! How would I factor in the Inertia of the pulley? For example if the moment of inertia, I, and the angular acceleration, alpha, need to be included in the question. How would this affect the tension in the rope?
Yes the technique is very different. There are a number of examples in the CHAPTER 13.1 Playlist: Physics - Application of the Moment of Inertia (8 of 11) Acceleration=? When Pulley Has Mass (mu=0) ruclips.net/video/p6ZlBp_vQ6c/видео.html
Michel van Biezen Thanks Michel! I will watch that video and many more of yours.
how does T2 = 2 * T1 because if the masses are being accelerated, arent the pulleys then also being accelerated, meaning forces arent balanced ..?
kylie,
Good question.
It is often difficult to comprehend or visualize what is actually going on and our intuition sometimes leads us astray.
In this problem we are using and "ideal" pulley which means that it does not have mass.
If the pulley does not have mass then there isn't an object to accelerate and when drawing the free body diagram around the pulley the forces on both sides must be equal.
If the pulley had mass, then you would be correct.
+Kylie Rose The pulleys here were massless and frictionless.
i didnt understand the part where you say that the acceleration of m2 will be half of that of m1....why would it be like that ? i didnt get it through the pulley reference also...😞
Since m2 is connected to the midpoint of the pulley, you have what we call a 2:1 mechanical advantage. You need 1/2 the force, but will require twice the pulling distance to move m2. It is the way pulleys work. If you have one, try it and see that it works that way.
Simply put, if you draw the free body diagram for pulley 2, then you could clearly see, 2 tensions of T1 on the right-hand side and thus it must be T2 = 2 x T1 on the left-hand side.
Nandha Kumar usne dusra question pucha tha..
Hi Michel. Great video, but there is something I don't quite understand and it's why is T2=m2a2 not equal to T2-m2g=m2a2 at 4:22 in your video?
Since m2 is sitting on the table, there is no tension component due to gravity, only to accelerate the block.
Thanks i appreciate am now doing well in physics and math thanks so much for the great expectations
You're Welcome!
Just a point of clarification Michael. If you're multiplying m2(1/2a1) by 2 in order to get rid of the 1/2a1 doesn't that give 2m2 a1? Because don't you multiply the whole expression by 2 i.e multiplying m2 x(1/2a1) by 2 so you get 2(m2 x 1/2a1) = 2m2 x a1
Hi. Can You explain way T1=m1g-m1a (g and a same direction)
Alexkey Ov,
Vectors have both magnitude and direction. In the equation you reference, we just use the magnitude part and determine the sign by figuring out which way the block is moving.
The best way to figure out the equation is this way:
1) if the block is not moving the T1 = m1g
2) if the block is accelerating upward you will need additional force to cause the acceleration T1 = m1g + m1a
3) if the block is accelerating downward you don't need all of the force to hold it against gravity T1 = m1g - m1a
Thanks for the answer. You are a very good teacher.
Sir, You are simply Amazing!...
Could you pleas do similar problem,but on an inclined plane, what I mean is m2 hanging and m1 is one the plane inclined at an angel of 30. Thank you for all your help sir.
We have dozens of examples in all kinds of physical situations in the playlists on this topic.
@@MichelvanBiezen How would the equations change if the mass one the left was pulled to the left.
Sir, Why is the direction of both T1 and T2 to the right? I think that the tension of the string below is to the left
To find the direction of the tension in the string you must express it in terms of the object relative to. Note that every string has tension in both directions simultaneously.
Wouldn't the tension of the t1 above the other t1 act towards the left not the right ? Thanks
The direction of the tension only makes sense if you have a reference point. (tension relative to what?) Thus the tension T1 just above the 5 kg block is directed upward RELATIVE to the 5 kg block. But T1 on the top string is directed to the right relative to the left pulley and to the left relative to the right pulley.
Michel van Biezen thank you for the explanation
nice problem - insightful solution
Thank you. Glad you liked it. 🙂
so for this problem, why couldn't you do m1g1 -m1a1 = m1a1. This is the F= ma but instead of T2 i just used T1. Using this formula you move m1a1 to the other side hence giving m1g1 = m1a1 + m1a1. Here you factor a out and move both m1 to the other side. This will give you (m1g1/m1+m1). Doing it this way also gave me 4.9. So is this way also acceptable for future problems?
If you draw a free body diagrams, then identify all the forces in and out of the diagram, and if they match your equation, you have the correct equation. (Better not use the "can this always be used")
I have the same question as another commenter ... how does T2 = 2 * T1 because if the masses are being accelerated, arent the pulleys then also being accelerated, meaning forces aren't balanced?
+Catherine Flynn
Whenever you draw a free body diagram around an object (like the dashed line box around the pulley), the sum of the forces (added as vectors) must satisfy F = m * a Since the pulley is assumed not to have mass in this example, the sum of the forces must equal zero. Therefore 2 T1 = T2
Ah I see, thanks so much! P.S. your videos are an amazing help!
Please sir
I cannot understand the idea of a2=0.5 a1 , though mass 1 is actually 25% of mass 2 , is not this a factor affects both the disatnce and acceleration ?
Waiting for your reply 🤔🤔
That is the result of the mechanical advantage of a pulley. The tension in the string connected to m2 is double the tension in the string connected to m1, and the velocity of m2 is half the velocity of m1. Hence the acceleration of m2 is also half of the acceleration of m1. See the sketch on the right side of the board.
@@MichelvanBiezen so this means that the mass connected with rope doesnot affect the acceleration as I understand , it is about the mechanical advantage .
I truely appreciate this , gratful for your help❤🌷
@@MichelvanBiezen my dr solved the same example but she considered a2=2a1 so this confused me can you please explain
@@hibazreik2228 maybe professors M¹ and M² are her M² and M¹ respectively
Really nice and it will never go ebb of my mind
Amazing explanation
Glad you think so!
@@MichelvanBiezenafter 10 year I am watching a masterpiece
Sir, can I use T2=2T1 ro prove why the acceleration a2= ½a1?
what the book that you use???
We use many books for reference and for my classes
Ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooh.....
Is that correct??
If T2=2T1 and T2=m2g=196N. Then can T1=T2/2= 98N. ? Then use T1=m1g-m1a1 to find a1? Which I found to be 9.8=[a1] ? then use a2=1/2 a1 to find a2. Which in this case would be 4.9 m/s^2? Am I wrong if I do this? Please respond?
Never mind just found the mistake
@@neworld2.073 which mistake please?
i have question. i understand the diagram on the right where if we pull it upward for 1m, the distance will only be half at the end of the pulley. but i don't understand how we can use that length reference in an acceleration problem. is it possible for the pulley to "take" half of the acceleration as well?
You can prove mathematically that it has to work out this way, or you can do it by inspection as the professor did, simply reasoning through the proportions of changes in length. The changes in length are always proportional, which means time derivatives of any order of these changes in length will also be proportional. Below, I've written up how we know that, with a calculus proof.
Define the position of block m1 to be distance y below the pulley.
Define the position of block m2 to be distance -x from the pulley, such that x is a negative number when the block is to the left of the pulley. We assign a negative sign to this one, so that its acceleration can be a positive number, which you'll see later why this helps.
The length of the string among the pulleys, the fixed point, and mass m1 is as follows (neglecting the diameter of the pulley, and the length of the string with tension T2):
L = | 2*x | + y
Since x is negative, replace the absolute value bars with a negation:
L = -2*x + y
Solve for y in terms of x:
y = L + 2*x
The acceleration is defined as the second derivative of each block's position, relative to time. Therefore:
a2 = d^2/dt^2 x
a1 = d^2/dt^2 y
Substitute y = L + 2*x:
a1 = d^2/dt^2 (L + 2*x)
Since L is a constant, it's derivative is zero, and likewise, its second derivative is zero. The leading constant of 2, can be pulled out in front on the 2*x term:
a1 = 2 * d^2/dt^2 x
We recognize that this is the expression for a2 in this equation, which therefore means that:
a1 = 2*a2
And that concludes how we can prove that acceleration a1 has to equal 2*a2. It is based on the constraint that the length of the rope is a constant.
@@carultch wow,
Mr Carl ,I just started a physics course , can I get you on WhatsApp maybe , I need your sort of help in my physics life 😔
How do you know that blick (5kg) accelerate downward
There are two choices. It either accelerated downward, or the friction between the table and the other block are sufficient to keep it from accelerating. Therefore assume the block accelerates and see what the acceleration is. If the answer is negative (compared to the assumed direction of the acceleration) then the block would not move.
If T2 is equals to 2T1 then can we use (m1g-m1a)=2(m1g-m1a) to find a, instead of equating it to m2a2?
That equation cannot be correct. It is like writing x = 2x
Right. Thanks alot.
Sir can you please tell me,
-What is the condition for the tension to be equal?
If you draw a free body diagram around the pulley on the left, you'll see that the one force to the left must be equal to the 2 forces on the right. Therefore both tensions on the left mus be equal and they must be half of the tension on the left side of the pulley.
thank you sir...
why is g not considered -9.8m/ss since it is a downward acceleration?
Since we are only finding the magnitude of the acceleration.
oh my god man, thank you very much for explaining about the acceleration being halved.
Yes, it is tricky.
THANK YOU SO MUCH!
Why would it be accelerating downward if the mass is 4 times heavier? It would be stationary
+Dwight D
Since there is no friction between the large mass and the table, there is no friction force, so how would the large mass keep the small mass from accelerating downward? Always go back to the F= ma equation and find all the forces aiding the acceleration and all the forces opposing the acceleration. If the forces causing the acceleration are greater, there will be an acceleration.
@@MichelvanBiezen nicely noted
What will be a2 if m2 is at inclined plane??
We have a number of examples with inclined planes and will make some more in the future.
Sir please make videos on complex problem including inclined plane
Oh sir you've Blown up my mind XD >>> But it's Enjoyable ! Thank you :)
Sir how do you know the direction of both the T1?
+Roshan Varghese
I am not sure about the question but I'll try to answer it this way: Tension acts in both directions. (The string pulls on each object that it is connected to.)
Shouldnt T1= m1g+m1a ?
No, it is T1 = m1g - m1a
@@MichelvanBiezen Thanks.
But the directions of the two magnitudes are the same. Only tension has the direction.
I meant only tension has an opposite direction.
The direction of the tension depends on the reference point.
If you draw a free body diagram around m1 and use F = ma you get the following equation mg - T = ma Solve for T.
Why does T2 = 2T1?
I am also confused what T2 even does
T2 is the tension in the rope that connects the 20 kg mass to the pulley on the left. It is T2 that accelerates the 20 kg mass to the right at acceleration = a2. If you draw a free body diagram around that pulley you have 3 forces contained within the free body diagram. T2 pulling to the left and 2 T1 forces pulling to the right (relative to the pulley). These forces must equal one another, thus T2 = 2 T1
sir could you elaborate how T2=2*T1
If you look at the free body diagram around the pulley then you can determine that the forces pulling to the left must equal to the forces pulling to the right. Thus T2 = T1 + T1
sir can u please explain me the directions of T1
Tension means stretching force. A flexible rope, chain, cable, cord, or string has to carry a stretching force if it carries any force at all. If you attempt to push a rope, it will become slack, and be incapable of carrying any force.
Therefore, the tension force has to point away from the object to which it is applied.
Thank you
Glad these videos are helping.
how is a2 is half of a1?
Pulleys give you a 2:1 mechanical advantage which causes the distance you pull on one side to be twice the distance the pulley moves.
Michel van Biezen thank you sir
Hello cher.
Why T2=2T1
Because the pulley by mass M2 is treated as an idealized pulley, which means it has negligible mass and negligible friction. Forces acting on an object of negligible mass must add up to zero, otherwise it would have infinite acceleration.
Two segments of the rope carrying tension T1, act on this pulley to the right. Both segments carry equal tension, because the pulley is idealized as massless and frictionless. The only remaining force acting on this pulley, is tension T2, which acts to the left. This means 2*T1 - T2 = 0, and we can solve for T2 = 2*T1.
Thanks a lot now iget the point
This is a good example for that.
It's impossible to read the text..,
can you explain why T2=2*T1??
See the other comments below.
thanks man
WHY IS ACCELERATION OF M2 HALF
Due to the mechanical advantage of a pulley. M2 is attached to the midpoint of the pulley. Therefore as the string is pulled on one end, the point where M2 is attached only moves half as far. (try it on a pulley and you will see). Mechanical efficiency requires that F1 * d1 = F2 * d2 (otherwise you create work and energy out of nothing).
Obakeng Moribe differentiate displacement twice with respect to time
why 1/2ΔL?
takkis Nassiopoulos
Think about it. How far does the other object move, if the first object moves 1 m?
I just said 5×9.8 is F and 25 is m and I found the a as 1.96... I dont understand why it is wrong
Did you follow the method in the video?
I bet
That's awesome ;-)
why∆l/2 sir
Since the force required to lift up the object attached to the pulley is only half the weight of the object, you must pull twice the distance in order to accomplish the same amount of work. W = F * d = (1/2)F * 2d
You can also try it yourself and you'll see that it is a two to one ratio
thanx sir
your videos are very helpful
Sir T1=m1g+m1a, u wrote wrong sir.
Are you sure about that?
Yes
It is T1=m1g-m2a
I bet what u wrote is wrong
gotdamn shawty
This was actually easy. I understood how to solve it right before sir started to solve it.
Wow
Could I get through to you on WhatsApp maybe , I just started a physics course ,and it's not that easy for me
l like
I bet that what you wrote is wrong
I like ur lectures but the answer is wrong
It is not correct