Hat's off you Sir. U r doing a great job❤️ keep Posting!! A small suggestion: It will be more useful if you explain a concept/ theory and do a problem related to that concept which makes us even more understanding.👍🏻
I watched many different lectures, all of them said the same thing that to take opposite directions as positive in pulley cases like you explained in this video. But I had difficulty in understanding what happens if we dont. Is it wrong to do that or just inconvenient, meaning if i take upwards as positive for both the masses, how to then solve the problem
The reason you can’t treat up as positive for both of them is that then the acceleration for the two objects would not be the same. Since they are moving together, it’s important to set it up so that either both are moving in the positive direction or both are moving in the negative direction so that they have the same acceleration.
teacher I always write vector with negatif or positive such as I write fg(vector sign on top ) -fg(vector sign on top) is it correct? I prefer to write with - or + sign all of them
It's ok to have vector signs on the vectors when drawing the free body diagram. However, once we are plugging into Fnetx=max and Fnety=may there should not be vector signs since we are dealing with the components of the vectors.
They incorrectly think of the two masses, pulling down on the rope as opposed to realizing that the rope is simply transmitting the force from one mass to the other. That’s why you always have to draw the free body diagram for a particular object and really pay attention to the forces acting just on that object.
hey I find it difficult to comprehend these topics in physics 😢. Can you please share a technique I can use to enhance my understanding in these physics concepts?
Watch through the video the first time to get the basic ideas. Make sure that you’ve watched the videos on Newton laws before the current one to get the background you need. Then go through the video a second time pausing it each time I’m about to solve a problem. Make yourself try to work out the solution. If you get stuck, then watch the video just up to the point where you got stuck, then pause the video again and keep going. Learning physics is like learning a sport. You have to actually practice the process of doing it. You can imagine if I told you all about how to play a sport like golf but you never actually got out there and swung the club that you wouldn’t be able to do it. So that’s the key.
Have a test on this in two days. It feels like my life depends on it. This is the last test that will be going into the grade book before the end of the semester of my physics class. I need to get my B to an A.
@@smithjomiddlesexedu Got a 76 😞… However I am ready to get my test back, identify my mistakes and weak points, and then do better on the retake. Managed to get my grade from a B to a B+ at least.
hey man ur explantion is great and all but at 33:27 u didnt mention that the car exerted a force on the brick wall which caused the wall to give it a force if u didnt mention this then that meant the car would just accelerate after it crashed in the direction that the wall hit it too
You are correct that if the brick wall exerts a force on the car, the car exerts an equal and opposite force on the wall (Newton's 3rd Law). However, when we are looking at the motion of the car, we only care about the forces acting ON the car. For example, the car is also pulling up on the earth through gravity (reaction to Earth's gravity pulling down) and pushing down on the surface of the Earth (reaction to Normal Force). If we were looking at the brick wall, then we would care about the force the car is exerting on it.
The box is not moving at all in the y direction, so it’s definitely not accelerating in the y direction. That’s how we know that may =0. In other words, all of the forces in the y direction have to add to exactly 0. You’ll notice that’s how we then are able to solve for the normal force. Since the applied force is partly in the upward direction, notice that the normal force is less than mg. The key point here is that if an object is moving only in the x direction then you know the acceleration in the y direction has to be equal to zero
Terminal velocity is responsible for weightlessness. Since everything is falling inside the elevator including the air, air resistance is not a factor, so there wouldn't be a terminal velocity. Let me give a quick explanation of terminal velocity. When an object falls with air resistance there are now two forces acting - gravity down and air resistance up. Air resistance increases as the speed of the object increases until it is equal in magnitude to the force of gravity. Once this occurs, the net force on the object is zero, so it will no longer accelerate - it will now fall with a constant velocity (the "terminal" velocity which is the highest velocity it reaches). In general objects that have low density (think - feather) will reach terminal velocity very quickly. Dense objects (think bowling ball) will take much longer to reach terminal velocity. If we are dropping a dense object just a few meters, we can ignore air resistance and assume the object will fall with a constant acceleration of -9.8 m/s^2
@ oh ok I get that but please why would it fall with a constant acceleration of -9.8 since gravity isn’t acting opposite to how it normally does (sorry for the continuous questions )
@@OtamaRecaps The acceleration due to gravity is always a=-g where g=9.8m/s^2. g is the magnitude of the acceleration. We say a=-g=-9.8 m/s^2 since the velocity is increasing in the downward (or negative) direction.
The blocks will move together, so they will have the same acceleration. Another way to think about it is to picture the two blocks together as one big mass of 30 kg. So the acceleration of both blocks together is equal to the applied force divided by the combined mass. This is what my solution shows.
I understand that it's given visually right to the question. But, what if there's no visuual representation? (I'm not hating, I'm just genuinely asking. I'm cramming this, I have a test in few hours HAHAHA)
If the angle weren't given, then there would not be enough information to solve the problem. However, I could have given you the acceleration and then asked you to find the angle. The analysis would be the same, but you would plug in acceleration to solve for theta. That's really the key - no matter what you are given, the analysis works the same way, then just solve for the variable you need based on the variables given.
18:46 i though tension will be zero as from both sides 9.8 on the same rope but if we image the tension going upwards they will cancel each other out, can you clear my this doubt please
Start by looking at just one mass. Since the mass is at rest (not accelerating) the net force on it must be zero from Newton's 2nd Law. Therefore the magnitude of the tension must be equal to the weight of the mass (mg). Think of the tension in the rope as the way the two masses communicate their presence with each other. So each mass is pulling up on the other mass with its weight.
uhm i have a concern about the problem on the car. you forgot to include the 0.85m in your calculations hehe. hope that will address my concern. thank you sir. God bless.
I used the 0.85m to calculate the acceleration. Go to: ruclips.net/video/PgrohbgwxxU/видео.html to see where the 0.85m gets plugged in to the kinematic equation to find the acceleration. Thanks
sorry sir but if yiu say symbols represents the magnitude then when then represents the direction because magnitude is basically size and from vector we know that signs represents the direction
Yes, the sign represents the direction. So, for example, the magnitude of the force of Friction is Fk. If it is directed to the left, when we put it in Newton’s second law we would write -Fk.
Think of the tension as the way each mass exerts a force on the other. So these have to be equal (think Newton's 3rd law). What will be different for each mass is the net force
Heroes don't wear capes, they explain physics in a simple and quick way, big thanks from Belgium! ;)
Wow, thanks!
Wow! I am beyond thankful to you and your descriptive explanations. I struggle so much with these problems. You are the best!
Thanks!
bro i have been studying for 4 hours now, did not understand shit till now Thanks a lot
Thanks!
Spent 4 months studying dynamic , couldn't get an equation right, Your one lecture and I finally got the concepts right
Glad it helped!
You are great professor, thank you
I am abandoning my physics teacher. From now on I will learn physics from this guy 💁♀️
Thanks!
Thank you so much, you're out here helping so many people
Happy to help!
Hat's off you Sir. U r doing a great job❤️ keep Posting!!
A small suggestion: It will be more useful if you explain a concept/ theory and do a problem related to that concept which makes us even more understanding.👍🏻
Thanks for the suggestion
I watched many different lectures, all of them said the same thing that to take opposite directions as positive in pulley cases like you explained in this video. But I had difficulty in understanding what happens if we dont. Is it wrong to do that or just inconvenient, meaning if i take upwards as positive for both the masses, how to then solve the problem
The reason you can’t treat up as positive for both of them is that then the acceleration for the two objects would not be the same. Since they are moving together, it’s important to set it up so that either both are moving in the positive direction or both are moving in the negative direction so that they have the same acceleration.
Thank you so much for this easy to understand video, now i'm confident that I'll ace my physics quiz
Glad to help!
thanks man, i have a pilot assesment with an airline on 25th needed this to brush up
Your welcome!
teacher I always write vector with negatif or positive such as I write fg(vector sign on top ) -fg(vector sign on top) is it correct? I prefer to write with - or + sign all of them
It's ok to have vector signs on the vectors when drawing the free body diagram. However, once we are plugging into Fnetx=max and Fnety=may there should not be vector signs since we are dealing with the components of the vectors.
Thanks MANNNN ur a lifesaver
You’re welcome!
AAAAA THIS HELPED ME SO MUCH IN MY FINALS
Great!
I have a quiz tomorrow hopefully this will help me
Good luck!
@@smithjomiddlesexedu I did very well ! 19/20 Thank you professor
Your welcome!
19:43 how do people even get b? The net force is 0, so mg=T, same mass and mass=1. I'm curious how some got that wrong.
They incorrectly think of the two masses, pulling down on the rope as opposed to realizing that the rope is simply transmitting the force from one mass to the other. That’s why you always have to draw the free body diagram for a particular object and really pay attention to the forces acting just on that object.
this video was super helpful, than you so much professor.
You're very welcome!
hey I find it difficult to comprehend these topics in physics 😢. Can you please share a technique I can use to enhance my understanding in these physics concepts?
Watch through the video the first time to get the basic ideas. Make sure that you’ve watched the videos on Newton laws before the current one to get the background you need. Then go through the video a second time pausing it each time I’m about to solve a problem. Make yourself try to work out the solution. If you get stuck, then watch the video just up to the point where you got stuck, then pause the video again and keep going. Learning physics is like learning a sport. You have to actually practice the process of doing it. You can imagine if I told you all about how to play a sport like golf but you never actually got out there and swung the club that you wouldn’t be able to do it. So that’s the key.
So helpful!! You are amazing!
Glad it was helpful!
Have a test on this in two days. It feels like my life depends on it. This is the last test that will be going into the grade book before the end of the semester of my physics class. I need to get my B to an A.
Good luck! I hope my videos help you out!
@@smithjomiddlesexedu Got a 76 😞… However I am ready to get my test back, identify my mistakes and weak points, and then do better on the retake. Managed to get my grade from a B to a B+ at least.
I love you so much professor! ❤
Thanks!
hey man ur explantion is great and all but at 33:27 u didnt mention that the car exerted a force on the brick wall which caused the wall to give it a force if u didnt mention this then that meant the car would just accelerate after it crashed in the direction that the wall hit it too
You are correct that if the brick wall exerts a force on the car, the car exerts an equal and opposite force on the wall (Newton's 3rd Law). However, when we are looking at the motion of the car, we only care about the forces acting ON the car. For example, the car is also pulling up on the earth through gravity (reaction to Earth's gravity pulling down) and pushing down on the surface of the Earth (reaction to Normal Force). If we were looking at the brick wall, then we would care about the force the car is exerting on it.
Would this work for AP physics c?
Absolutely!
College level calculus based course for science and engineering majors
For example of the box being pulled it literally shows the box being pulled in a diagonal direction, how is may 0????
The box is not moving at all in the y direction, so it’s definitely not accelerating in the y direction. That’s how we know that may =0. In other words, all of the forces in the y direction have to add to exactly 0. You’ll notice that’s how we then are able to solve for the normal force. Since the applied force is partly in the upward direction, notice that the normal force is less than mg.
The key point here is that if an object is moving only in the x direction then you know the acceleration in the y direction has to be equal to zero
if thats free fall wont your weghtlessness be caused by you reaching terminal velocity
Terminal velocity is responsible for weightlessness. Since everything is falling inside the elevator including the air, air resistance is not a factor, so there wouldn't be a terminal velocity. Let me give a quick explanation of terminal velocity. When an object falls with air resistance there are now two forces acting - gravity down and air resistance up. Air resistance increases as the speed of the object increases until it is equal in magnitude to the force of gravity. Once this occurs, the net force on the object is zero, so it will no longer accelerate - it will now fall with a constant velocity (the "terminal" velocity which is the highest velocity it reaches). In general objects that have low density (think - feather) will reach terminal velocity very quickly. Dense objects (think bowling ball) will take much longer to reach terminal velocity. If we are dropping a dense object just a few meters, we can ignore air resistance and assume the object will fall with a constant acceleration of -9.8 m/s^2
@ oh ok I get that but please why would it fall with a constant acceleration of -9.8 since gravity isn’t acting opposite to how it normally does (sorry for the continuous questions )
@@OtamaRecaps The acceleration due to gravity is always a=-g where g=9.8m/s^2. g is the magnitude of the acceleration. We say a=-g=-9.8 m/s^2 since the velocity is increasing in the downward (or negative) direction.
Am not sure about acceleration on the second example
The blocks will move together, so they will have the same acceleration. Another way to think about it is to picture the two blocks together as one big mass of 30 kg. So the acceleration of both blocks together is equal to the applied force divided by the combined mass. This is what my solution shows.
I don't understand where did you get 32° at 6:32
I understand that it's given visually right to the question. But, what if there's no visuual representation? (I'm not hating, I'm just genuinely asking. I'm cramming this, I have a test in few hours HAHAHA)
If the angle weren't given, then there would not be enough information to solve the problem. However, I could have given you the acceleration and then asked you to find the angle. The analysis would be the same, but you would plug in acceleration to solve for theta. That's really the key - no matter what you are given, the analysis works the same way, then just solve for the variable you need based on the variables given.
thank you brother, god bless you
Thank you
At first quesiton, why we didn't use Fn for calculating acceleration.(NOT FNETX)
cause at the y direction the acceleration is zero there is no acceleration there, that is why we did not use the Fn because Fn is in y direction
18:46 i though tension will be zero as from both sides 9.8 on the same rope but if we image the tension going upwards they will cancel each other out, can you clear my this doubt please
Start by looking at just one mass. Since the mass is at rest (not accelerating) the net force on it must be zero from Newton's 2nd Law. Therefore the magnitude of the tension must be equal to the weight of the mass (mg). Think of the tension in the rope as the way the two masses communicate their presence with each other. So each mass is pulling up on the other mass with its weight.
uhm i have a concern about the problem on the car. you forgot to include the 0.85m in your calculations hehe. hope that will address my concern. thank you sir. God bless.
I used the 0.85m to calculate the acceleration. Go to: ruclips.net/video/PgrohbgwxxU/видео.html to see where the 0.85m gets plugged in to the kinematic equation to find the acceleration. Thanks
super helpful video thank you
you're welcome!
in the 2nd example how does F12 = F21 while both have different masses
From Newton’s Third Law. The force that object one exerts on object two must be equal in magnitude to the force that object two exerts on object one
tnks professor
😭😭what if you're required to calculate the pull force itself?
You would have to be given the acceleration. Then you could calculate the pull force from that and the mass using Newton's 2nd Law.
Thank you sir
sorry sir but if yiu say symbols represents the magnitude then when then represents the direction because magnitude is basically size and from vector we know that signs represents the direction
Yes, the sign represents the direction. So, for example, the magnitude of the force of Friction is Fk. If it is directed to the left, when we put it in Newton’s second law we would write -Fk.
@smithjomiddlesexedu ohhhhh
Sergey hocama selamlar
Thanks!
23:56 why the tensions are equal the masses are different>???
Think of the tension as the way each mass exerts a force on the other. So these have to be equal (think Newton's 3rd law). What will be different for each mass is the net force
I still don’t get it 😢
Where are you getting stuck?