I appreciate that they left in the part where he gets frustrated with his calculations not adding up. It's encouraging to know that even the best can struggle with Physics.
AliGay Nice guys... I don't get too heated at a fellow student(or even teacher) for making a mistake or there, especially if for that one mistake there's a million times they've solved,plugged, memorized, derived and experimented flawlessly. Also, why are you watching a simple tutoring video about algebraic physics if you weren't 'struggling' a bit? I mean cm'on. This is easy stuff(This paragraph is satire boys. Think Samuel Clemens on this one.).
wow sir, your explaination is simply the best of all, even better than khan academy. your pace of teaching isn't too fast nor too slow and the way you explain makes everything so much understandable in the name of physic. Please continue to upload videos. Thank you so much!.
I just wanted to say thank you for making these videos. You may have single-handedly saved my Physics grade. You make it so much easier to understand than my professor and the way you break it down and explain it makes sense to me. You've earned a follow from me for sure.
Professor van Biezen, your videos flow so effortlessly, at such a smooth and steady pace, and I have been watching one after the other, so it was not until the moment of frustration we see in this video that I finally wake up to all the work that you and your helpers must put into making one great video after the other. There are so many videos, and they are so good!! Yours is the best science/math channel on UA-cam! I want to thank you for taking the time (and these must have taken you SO much time) that you have put into sharing your excellent teaching style and knowledge with us all, and the effort that you have made to do so. I don't know why you do this other than to help those of us who are struggling to learn. Thank you so much!!! Kudos to you, Sir!
PolyPixel,Thank you for this very heartwarming comment. Yes, between my wife and I we have invested 7500 hours of our time into this venture and it has grown far beyond our expectation. Every one of us has been given a skill and we believe that a big part of our life and existence is to use that skill to give back to society in response to the blessing of life that we have received. It is great to know that these videos have helped many thousands of students around the world.
You absolutely blow my professor out of the water. You are so much more clear with all the steps you take and it makes understanding the topics much easier. Thanks so much.
If the pulley has mass you must solve the problem using the moment of inertia of the pulley. Instead of F - m x a you must use Torque = moment of inertia x angular acceleration. Look at the videos on moment of inertia. There are some examples where the pulley has mass.
Laurelindo It looks like the "editor" didn't take out that part of the video. (oops) This is one of the early videos that we made. Notice how they didn't all go that smooth.
David Learmonth I actually kinda like bloopers like that, I am not bothered by them at all and gladly watch the events unfold! =D They also show that even professors can make mistakes.
Murathan, g = 9.8 m/sec^2 g is the acceleration due to gravity, experienced by all objects on the surface of the earth. Weight = mass x g or w = mg weight is a force and therefore, mg is the force pulling an object with mass m to the Earth.
His videos are so amazing, and it is very refreshing to know the reasons behind equations! My physics teacher barely explains anything, and jumps all over the place without telling us why.
I would like to thank you because I was in a crying panick attack because I did not understand anything and your video was the light of my night. I wish you the best of blessings in your life and in your families life. Thank you.
First of all, thank you for the great explanation. Second of all, the bit where you have a calculator issue was hilarious and I applaud you for leaving it in. Bravo good sir. I learned something today!
I was able to perfectly understand both parts. These videos are excellent, and you are an excellent teacher. Please continue to enrich the science education of others the way you did here. Physics and Math and their related fields are often seen as too daunting, and it's heroes like you that make them accessible to those who would otherwise believe they don't have the brains to do them.
you sir, are awesome! I almost cried watching your videos, I really appreciate people like you who explains things so simple and such organized manner! This is what makes learning fun! kuddos to professors like you! 👏👏👏👍👍👍
Sjaak, I wouldn't elevate it to the level of "dispair", but you can see that I have my moments that my brain gets stuck and I can't see where to go next. That makes understanding it afterwards so much better. Yes, we are all human.......
Dear professor Van Biezen, thank you and your, beautiful supporting, wife. I wish I could express my gratitude, for helping hundreds of thousands if not millions of students like myself, even more. Your efforts in making these videos and the encouragement your wife gave you is a symbol of your good intentions. I wish you both nothing but happiness and success.
+Michel van Biezen hey but there is a slight confusion. you gave a coefficient of friction right? but you need to make it clear whether it is a static or kinetic. for the question we need a static coefficient.
+Michel van Biezen , okay if you take it as "kinetic " , try to use the static friction if it makes the object move . Unfortunately, it won't make it move .
Mass m = 90.8 kg sits on an inclined plane that makes an angle α = 25.3o with the horizontal. A massless string is tied to m, passes over a frictionless, massless pulley, and is tied to mass M = 56.8 kg hanging on the other end. Assume: - the coefficient of kinetic friction between m and the surface of the plane is μs = 0.269. - the string is parallel to the surface of the incline. If m is initially moving up the incline find the acceleration (magnitude in m/s2 and direction) of the system. (To give direction: give the answer as positive if the acceleration of m is up the incline, and negative if the acceleration is down the incline.) it gives 0.31 but it says its wrong, do you know why?
A block rests on a plane that is inclined at angle of 30degrees to the horizontal. Coefficient of kinetic friction between the block and the plane is 0.2 . if the block had a mass of 10kg and a man wishes to push it up the plane : (a) what is the direction of the frictional force? (b)what force parallel to the incline is necessary to keep it sliding up the plane ? (c) what horizontal force would be required to push it up the plane ?
how do we judge the direction of motion before starting the problem? We get a different equation if the motion is assumed to be in the opposite direction - i.e. if the mass on the incline is going down. in that case a = (m1+m2 sin theta - m1cos theta * mu) * g / (m1 + m2).
+Thun Guns Pro The definition of the friction force is: F fr = N * mu where N is the normal force and mu is the coefficient of friction. The normal force on an inclined plane is equal to the perpendicular component of the weight of the object, the reactionary force to the component. Thus F fr = N * mu = mg cos (theta) * mu
Hello professor, I just started taking dynamics and since we have to break up forces into their x and y components we also need to account for the x and y component of the Friction force, which never really happened with kinematics
On an inclined plane, the components of the forces are written in terms of the parallel and perpendicular components (relative to the surface of the plane).
I am working in an experiment where I placed a car on an inclined, the car is attached to a string and the string passes through a pulley, the string is holding a bigger mass that is pulling the car to the highest point of the inclined. But we measured the tension of the string with a sensor as we held the car to avoid moving. Now I have to calculate the Tension applying Newton's First Law and compare the theoretical Tension with the measured Tension. Where I am struggling with it is with my diagram, I don't know if I have to include the force I am applying onto the car. Any tip would be appreciated.
+Glecie Lopez consider a vector, find the angle (theta) with respect to the x axis, then take resultant of the vector, the one with angle theta is costheta and the (90-theta) one is sintheta
You are too great sir I love the fact that u portrait a good quality teaching sir and that moment of despair change my way of thinking it really great u didn't edit it sir 🥰😋 A question on the tension I can't picture a situation that the pulley that will have a mass and how it will affect the tension on the system
We have several playlists with examples like that. See this video: ua-cam.com/video/qu46xL4KPvU/v-deo.html in the playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS
That depends on the direction of motion, or what the direction of the motion would be if there was no friction. The direction of friction typically is in the opposite direction to the motion if there was no friction.
If the object is at rest, we only use the static coefficient of friction. Once the object begins to slide, then the static coefficient of friction is no longer used and we only use the static coefficient of friction.
Michel van Biezen Do you mean we don't use the static coefficient of friction and use the kinetic coefficient of friction once the object begins to slide?
It depends on how much bigger. If it is big enough to cause an acceleration in the other direction ,you would simply reverse the order Aiding force = m1gsin(theta) Opposing force m2g and the friction force.
We probably should make some videos where we combine forces pushing at an angle while the system is connected over a pulley. We have them as separate examples now.
If you are given the acceleration and you want to find the frictional force, you would solve it in the EXACT SAME WAY. Set up the equation as shown and then you'll have the coefficient of friction as the unknown that you must solve for instead of the acceleration.
These are not Ax and Ay vectors. The vector components of the weight (mg) are parallel and perpendicular to the incline. Look at an example when the angle is 80 degrees.
Since one block goes up the incline and the other block moves downward, it makes more sense to express the acceleration in terms of the magnitude and show the direction with the curved arrow.
@@MichelvanBiezen Alright, I understand! Thank you. Also, I just took my physics midterm and I got the highest grade and I really want to thank you because it's all possible due to your videos! Cheers and keep doing what you're doing :D
In the numerator, we have 3 forces. One that aids the acceleration, the other 2 oppose the acceleration. One of those 2 is the friction force. The definition of the friction force is that it is equal to the normal force (mg cos(theta)) x coefficient of friction (u).
Neither. The friction force is parallel to the slanted surface. And by definition the friction force = normal force to the surface x coefficient of friction. We have videos that go into more detail on the friction force.
if we take direction of accelaration or assume that it is in opp to what taken in vedio we would see that friction is opp to mgsintheta .. so we subtract it from it . to get froce in that direction . then subtract it from the block hanging{froce of it} . then we would get resultant froce . then we can find accelaration. i tried like this i did not get answer. but we need to know direction of accelaration . if we take opp to what in vedio we wont get ans. as friction is opp to motion . it may differ in casses.
Don't forget to change the sign of m2g and m1gsin(theta). If you change the direction of the assumed acceleration, m2g now opposes the acceleration and m1gsin(theta) now aides the acceleration. You will get the same answer.
if we take froce of m1 right side of the plane then friction would be opp or to the left..... if we subtract froces we would some froce to the left of the m1 . that froce can be subtracted from m2[force] . to get resulting force. is the method correct?
To know the direction of acceleration, i need to calculate mg for hanging mass then calculate the mg sin theta for another mass, the biggest weight will be the direction of acceleration? Is that true, Could you please explain it to me?
Yes, you need to compare the weight of the hanging mass to the mg sin(theta) of the mass on the slope. Off course you also need to take into account the friction force. Or you can just assume a direction, work out the problem and then see if you get a realistic answer.
Always start from the basic equations. In this case we use Newton's second law (F = ma) which means that a = F/m which means we divide the net force applied by the total mass (not the weight).
I appreciate that they left in the part where he gets frustrated with his calculations not adding up. It's encouraging to know that even the best can struggle with Physics.
+Stefan Bekker I did cringe quite hard tho. Physics cringe compilation may 2016(ap test times of course)???
ha gay Oh, so you've never cringed after making a mistake?
nah only dumb people struggle at physics
AliGay Nice guys... I don't get too heated at a fellow student(or even teacher) for making a mistake or there, especially if for that one mistake there's a million times they've solved,plugged, memorized, derived and experimented flawlessly.
Also, why are you watching a simple tutoring video about algebraic physics if you weren't 'struggling' a bit? I mean cm'on. This is easy stuff(This paragraph is satire boys. Think Samuel Clemens on this one.).
Stefan Bekker you are damn right man
I love that you didn't edit the frustration out. Shows that it's normal to be drained by a problem and you just gotta keep plugging away at it.
Yes, I too get confused at times, until I can get my brain unstuck.
wow sir, your explaination is simply the best of all, even better than khan academy. your pace of teaching isn't too fast nor too slow and the way you explain makes everything so much understandable in the name of physic. Please continue to upload videos. Thank you so much!.
+Thy Nguyen Nah fam. Khan academy is a new level. You can't compare
That is my opinion
+Thy Nguyen well your opinion is wrong. don't give wrong opinions
+Danish Ashar wrong about?
+Danish Ashar kid, if you dnt like what i say you can fk off :)
"my mind is not on this"
same
I just wanted to say thank you for making these videos. You may have single-handedly saved my Physics grade. You make it so much easier to understand than my professor and the way you break it down and explain it makes sense to me. You've earned a follow from me for sure.
Thank you. We are glad you found our videos. :)
well at 7:00 I learnt not only Newton's Law but also "Perfection is Imperfection"
the dopest Professor! Thanks ü
Professor van Biezen, your videos flow so effortlessly, at such a smooth and steady pace, and I have been watching one after the other, so it was not until the moment of frustration we see in this video that I finally wake up to all the work that you and your helpers must put into making one great video after the other. There are so many videos, and they are so good!! Yours is the best science/math channel on UA-cam! I want to thank you for taking the time (and these must have taken you SO much time) that you have put into sharing your excellent teaching style and knowledge with us all, and the effort that you have made to do so. I don't know why you do this other than to help those of us who are struggling to learn. Thank you so much!!! Kudos to you, Sir!
PolyPixel,Thank you for this very heartwarming comment. Yes, between my wife and I we have invested 7500 hours of our time into this venture and it has grown far beyond our expectation. Every one of us has been given a skill and we believe that a big part of our life and existence is to use that skill to give back to society in response to the blessing of life that we have received. It is great to know that these videos have helped many thousands of students around the world.
G-d bless you and your wife and family. I owe you a debt of gratitude. All the best for continued success in your teaching endeavors and everything!
You absolutely blow my professor out of the water. You are so much more clear with all the steps you take and it makes understanding the topics much easier. Thanks so much.
If the pulley has mass you must solve the problem using the moment of inertia of the pulley. Instead of F - m x a you must use Torque = moment of inertia x angular acceleration. Look at the videos on moment of inertia. There are some examples where the pulley has mass.
What happened on 7:00?
Laurelindo
It looks like the "editor" didn't take out that part of the video. (oops) This is one of the early videos that we made. Notice how they didn't all go that smooth.
Laurelindo LOL he forgot to multiply the numerator by gravity. He didn't mess up on the board, he just plugged it in to his calculator wrong.
Quick editing tip. Even in UA-cam, you can fairly easily trim out the bit in the middle where you stumbled on the calculator. Thanks for the videos!
David Learmonth
I actually kinda like bloopers like that, I am not bothered by them at all and gladly watch the events unfold! =D
They also show that even professors can make mistakes.
He wanted to start again respect for a perfectionist
Murathan,
g = 9.8 m/sec^2
g is the acceleration due to gravity, experienced by all objects on the surface of the earth.
Weight = mass x g
or w = mg
weight is a force and therefore, mg is the force pulling an object with mass m to the Earth.
His videos are so amazing, and it is very refreshing to know the reasons behind equations! My physics teacher barely explains anything, and jumps all over the place without telling us why.
6:53 my reaction everytime i press my calculator.
nevertheless, the best account for teaching physics so far!
I would like to thank you because I was in a crying panick attack because I did not understand anything and your video was the light of my night. I wish you the best of blessings in your life and in your families life. Thank you.
Thank you and we are glad we could help.
at 7 minutes, that is me with every physics problem. :P
That was before we learned how to edit videos. Now we'd just edit it out. But it is kind of funny. XD - Editing "Crew"
First of all, thank you for the great explanation. Second of all, the bit where you have a calculator issue was hilarious and I applaud you for leaving it in. Bravo good sir. I learned something today!
i love the way you explain. really my mechanics became a lot easier.
I was able to perfectly understand both parts. These videos are excellent, and you are an excellent teacher. Please continue to enrich the science education of others the way you did here. Physics and Math and their related fields are often seen as too daunting, and it's heroes like you that make them accessible to those who would otherwise believe they don't have the brains to do them.
Thank you! 😃 We are glad you find these videos helpful!
you sir, are awesome! I almost cried watching your videos, I really appreciate people like you who explains things so simple and such organized manner! This is what makes learning fun! kuddos to professors like you! 👏👏👏👍👍👍
I love this moment of despair! We are all human, aren't we? ;-)
Sjaak,
I wouldn't elevate it to the level of "dispair", but you can see that I have my moments that my brain gets stuck and I can't see where to go next. That makes understanding it afterwards so much better. Yes, we are all human.......
:) I agree with you Dr, we must keep going and trying until we success
Dear professor Van Biezen, thank you and your, beautiful supporting, wife. I wish I could express my gratitude, for helping hundreds of thousands if not millions of students like myself, even more. Your efforts in making these videos and the encouragement your wife gave you is a symbol of your good intentions. I wish you both nothing but happiness and success.
Thank you Mahammud, we really appreciate your comment.
I like how you explain this. You're going at the perfect pace and it's really helpful. Thank you!
You are much better than my professor
Thank you Sir.It helped me a lottt.You made the difficult things very easy to understand.Am grateful to you.
great video, my left ear is very thankful for your help
hahahahahahah
hahahhahahahaha
Thank you for explaining it so simple.
You make physics a lot easier ! THANKS SO MUCH !
good job sticking with the problem. I saw it...I'm glad you took a second to see it for yourself.
just wanted to say thank you for explanations. You have helped me understand physics better than any other book or teacher. Keep up the great work!
By far the best video on physics that expalins everyhting......m ready for this part of the test (TEEHEE)
Thanks Michel, your way of explaining has improved my understanding.
forgot to edit out the short period of frustration at 7:02
I'm glad that's in there. Real people.
Agreed.
+Michel van Biezen hey but there is a slight confusion. you gave a coefficient of friction right? but you need to make it clear whether it is a static or kinetic. for the question we need a static coefficient.
+Robel Fitiwi Since the objects are in motion, this is kinetic coefficient of friction.
+Michel van Biezen , okay if you take it as "kinetic " , try to use the static friction if it makes the object move . Unfortunately, it won't make it move .
+Michel van Biezen which makes using kinetic friction zero percent necessary.
I mean you literally can't get acceleration because it won't move .
Well just give it a little push with your hand and the system will move, and the rest is just as calculated.
Thanks bro. U make it so easy when u break it down to just each force opposing the other without adding extra stuff. bless
Glad to help
"My mind is not on this.." Me: Same energy. BUt thank you so much for this videoooo
Mass m = 90.8 kg sits on an inclined plane that makes an angle α = 25.3o with the horizontal. A massless string is tied to m, passes over a frictionless, massless pulley, and is tied to mass M = 56.8 kg hanging on the other end. Assume:
- the coefficient of kinetic friction between m and the surface of the plane is μs = 0.269.
- the string is parallel to the surface of the incline.
If m is initially moving up the incline find the acceleration (magnitude in m/s2 and direction) of the system. (To give direction: give the answer as positive if the acceleration of m is up the incline, and negative if the acceleration is down the incline.)
it gives 0.31 but it says its wrong, do you know why?
Really useful lessons, please keep on going sir you are aiding the education of a lot of students!
A block rests on a plane that is inclined at angle of 30degrees to the horizontal. Coefficient of kinetic friction between the block and the plane is 0.2 . if the block had a mass of 10kg and a man wishes to push it up the plane :
(a) what is the direction of the frictional force?
(b)what force parallel to the incline is necessary to keep it sliding up the plane ?
(c) what horizontal force would be required to push it up the plane ?
Love you videos! Just came across them with a physics exam tomorrow... Big help!
Great video, easy to understand
Thanks very much sir.you have help me so much for am having exams of the same things.... watching from the university of Zambia
Wow, it is great to welcome you to the channel!
how do we judge the direction of motion before starting the problem? We get a different equation if the motion is assumed to be in the opposite direction - i.e. if the mass on the incline is going down. in that case a = (m1+m2 sin theta - m1cos theta * mu) * g / (m1 + m2).
He became so relatable at 6:55
Why is it umgcostheta, not umgsintheta? I thought the force of friction is in the x direction.
+Thun Guns Pro
The definition of the friction force is: F fr = N * mu where N is the normal force and mu is the coefficient of friction.
The normal force on an inclined plane is equal to the perpendicular component of the weight of the object, the reactionary force to the component. Thus F fr = N * mu = mg cos (theta) * mu
Very clear ! Congratulations and many thanks from Santiago, Chile
This is an excellent explanation
Thank you. Glad you liked it. 🙂
I love this man.
Hello professor, I just started taking dynamics and since we have to break up forces into their x and y components we also need to account for the x and y component of the Friction force, which never really happened with kinematics
On an inclined plane, the components of the forces are written in terms of the parallel and perpendicular components (relative to the surface of the plane).
This is comparatively simple. Mechanics is not that hard. Just have to be careful. Thank you sir, and please keep it up.
That is a good way of putting it. "It is not that hard, but you just have to be careful". Great comment.
You make everything very easy..Thank you so much.
what is we were asked to find the tension of the string?
It's great. I have no other words to say !
it is very very very very very very very clear thanks very much
You are most very very very very bery welcome!
I am working in an experiment where I placed a car on an inclined, the car is attached to a string and the string passes through a pulley, the string is holding a bigger mass that is pulling the car to the highest point of the inclined. But we measured the tension of the string with a sensor as we held the car to avoid moving. Now I have to calculate the Tension applying Newton's First Law and compare the theoretical Tension with the measured Tension. Where I am struggling with it is with my diagram, I don't know if I have to include the force I am applying onto the car. Any tip would be appreciated.
That seems like a dangerous experiment. You should be able to find the information you need from these videos.
clear , easy . thank you for your explanation .
Thank you very much ,
You make everything easy,
you are the best.
I am confused of placing sine and cosine.
Glecie Lopez
There are playlists on the channel that can help you understand:
Take a look at:
TRIGONOMETRY BASICS
TRIGONOMETRY - THE RIGHT TRIANGLE
+Glecie Lopez consider a vector, find the angle (theta) with respect to the x axis, then take resultant of the vector, the one with angle theta is costheta and the (90-theta) one is sintheta
This video helped me so much! Thank you
Im a fan professor, great teacher
Thank you
your explainations are so clear sir . i became more powerful on solving problems . thank you so much sir
Good for you. Keep up the hard work.
sir u doing a fab work... u got struck fo a while... doesn't matters at all! i love the way u spreading knowledge!!! keep moving ahead... love u sir
You are too great sir I love the fact that u portrait a good quality teaching sir
and that moment of despair change my way of thinking it really great u didn't edit it sir 🥰😋
A question on the tension
I can't picture a situation that the pulley that will have a mass and how it will affect the tension on the system
We have several playlists with examples like that. See this video: ua-cam.com/video/qu46xL4KPvU/v-deo.html in the playlist: PHYSICS 13.1 MOMENT OF INERTIA APPLICATIONS
Enlightening lecture!
However I would like to recommend using the value of 'g' as 10 m/s^2 just for easier calculations and explanation.
Thanks.
What direction is friction if the box on the incline has a greater mass than the hanging object?
That depends on the direction of motion, or what the direction of the motion would be if there was no friction. The direction of friction typically is in the opposite direction to the motion if there was no friction.
Thank you!
Love to see professionals like yourself get frustrated at physics... thanks for leaving this in the video
HaHa, yes, sometime the brain just locks up and the obvious is not seen.
sir it is very clear for me now I can solve any pully problem thank u sir
Thank you very much for this wonderful video! help me a lot on figuring out this type of problem..
Great! Glad you found our videos! 🙂
+Michel van Biezen; And what if the object was at rest and now moving. Ther will be static friction too, right? In that situation, what should I do?
If the object is at rest, we only use the static coefficient of friction. Once the object begins to slide, then the static coefficient of friction is no longer used and we only use the static coefficient of friction.
Michel van Biezen Do you mean we don't use the static coefficient of friction and use the kinetic coefficient of friction once the object begins to slide?
What would the equation look like if the mass of m1 was bigger than m2
It depends on how much bigger. If it is big enough to cause an acceleration in the other direction ,you would simply reverse the order Aiding force = m1gsin(theta) Opposing force m2g and the friction force.
@@MichelvanBiezen okay thank you that really helps
Now, i know angels do walk on earth. Thank you sir for guiding us, this is a tremendous help for us struggling students.
How would you go about doing the same problem but there is an angled force applied directly to m1 going up the ramp?
We have similar examples like that in the playlists.
@@MichelvanBiezen Thank you for the quick response. The closest video I can find isn’t a pulley system ua-cam.com/video/Hjz3e1yummQ/v-deo.html
We probably should make some videos where we combine forces pushing at an angle while the system is connected over a pulley. We have them as separate examples now.
What if we find the acceleration, and the Frictional Force? How would we find that?
If you are given the acceleration and you want to find the frictional force, you would solve it in the EXACT SAME WAY. Set up the equation as shown and then you'll have the coefficient of friction as the unknown that you must solve for instead of the acceleration.
love you video .thank you somuch
Thanks for watching!
Hats off sir u r. The bestest
Thank you, I am glad you think so
I love the 7:00 part, cuz I just realized that I'm not the only one always in panic when doing Physics haha
Really amazing jobs explaining it!
Glad you think so!
l now understand better. THANK YOU Sir
also i had a question, in vectors Ax = ACos(theta) and Ay = ASin(theta), why is x and y switched with Sin and Cos now?
i found the answer myself, it's an upside down triangle to match side with side, ha!
These are not Ax and Ay vectors. The vector components of the weight (mg) are parallel and perpendicular to the incline. Look at an example when the angle is 80 degrees.
Thanks u so much sir.... God always with u... Stay blessed.
Thank you so much sir crystal clear explanation!
Really great explanation to help with reviewing!
It's nice to know that even the best screw up.
Won't the friction affect the tension of the inclined object?
The concept of the tension is only applied to the strings (not the object). And yes, the friction force will affect the tension.
Im confused wouldn't there be another force which is the Tension force, also would the tension force be equal to m2g?
Tension is internal to the system and does not affect the acceleration of the system. (It is not an outside force acting on the system)
Can you please explain why the acceleration is not negative, it's going in the downward direction!!
Since one block goes up the incline and the other block moves downward, it makes more sense to express the acceleration in terms of the magnitude and show the direction with the curved arrow.
@@MichelvanBiezen Alright, I understand! Thank you. Also, I just took my physics midterm and I got the highest grade and I really want to thank you because it's all possible due to your videos! Cheers and keep doing what you're doing :D
hi, why isnt tension included in the equation?
Tension is internal to the system. The system only accelerates due to a net force acting on the system.
Wow Michael!
Well explained.
But why did you multiply the costheta with the friction only and not the whole sum of the forces?
In the numerator, we have 3 forces. One that aids the acceleration, the other 2 oppose the acceleration. One of those 2 is the friction force. The definition of the friction force is that it is equal to the normal force (mg cos(theta)) x coefficient of friction (u).
Why dont masses cancel out in numerator and denominator
You cannot cancel per the algebraic rules. (M - m) / (M + m) cannot be reduced
2022 am still this is great thank you so, but if the camera will go close to the board so we can see clearly.
Yes, we made some improvements in recent years, moving the camera closer to the board, unlike this older video.
yeah. i am feeling the same but i have much of the semester left. I have to do better.
Good ..nice waybto explain
Thank you.
Why is friction force = m1gCOS theta and not m1gSIN theta? Is the friction force on the y axis? Why not on x axis?
Neither. The friction force is parallel to the slanted surface. And by definition the friction force = normal force to the surface x coefficient of friction. We have videos that go into more detail on the friction force.
This is very helpful. Thank You.
if we take the direction of accelaration the other way he took we wont get answer as friction will be add instead of subtraction .
Friction can never "add" to the system. It is a retarding force and will slow the object(s) down.
if we take direction of accelaration or assume that it is in opp to what taken in vedio we would see that friction is opp to mgsintheta .. so we subtract it from it . to get froce in that direction . then subtract it from the block hanging{froce of it} . then we would get resultant froce . then we can find accelaration. i tried like this i did not get answer. but we need to know direction of accelaration . if we take opp to what in vedio we wont get ans. as friction is opp to motion . it may differ in casses.
Don't forget to change the sign of m2g and m1gsin(theta). If you change the direction of the assumed acceleration, m2g now opposes the acceleration and m1gsin(theta) now aides the acceleration. You will get the same answer.
if we take froce of m1 right side of the plane then friction would be opp or to the left.....
if we subtract froces we would some froce to the left of the m1 . that froce can be subtracted from m2[force] . to get resulting force. is the method correct?
Thanks sir, really appreciated your help and contribution to my studies
You are welcome. Glad our videos are helping. 🙂
Very helpful Sir. Thanks you very much.
To know the direction of acceleration, i need to calculate mg for hanging mass then calculate the mg sin theta for another mass, the biggest weight will be the direction of acceleration?
Is that true, Could you please explain it to me?
Yes, you need to compare the weight of the hanging mass to the mg sin(theta) of the mass on the slope. Off course you also need to take into account the friction force. Or you can just assume a direction, work out the problem and then see if you get a realistic answer.
@@MichelvanBiezen still replying after 9 years!? you're such a G.O.A.T. my guy!
Sir thank you so much
I finally understand the different relations
why?are we not multiplying the total mass x9.81
Always start from the basic equations. In this case we use Newton's second law (F = ma) which means that a = F/m which means we divide the net force applied by the total mass (not the weight).
I seriously think all he needs is a real calculator...