the whole semester i didn't learn anything in the class but from your 7 min video i learned everything you are the best teacher i have ever seen, the way you teach i swear thats what teaching is called.
I've been teaching two years and have only just discovered this UA-camr. Very, very good videos -- excellent modelling of how to solve a physics problem.
This made so much more sense than my teacher, one question though, why is the perpendicular component of gravity cos and the parallel sin? What angle are you using to determine that?
We are using the angle between the verical (mg) and the perpendicular component to the incline of mg (mg cos (theta)) which is the same angle as the incline.
Thank you very much for reminding me that the friction force is Normal force by coefficient cos I needed this to submit for assignment and its due today
Hi professor Van Biezen, Today I don't have a physics question. But I wanted to thank you for all your videos and your explanation. I was studying physics for my MCAT examen of medicine. And for the part physics my results were 8/10. That is a very high score. And this result is thanks to your videos! So thanks a lot. And now I can do medicine and become a doctor because I passed my exam :) :)
Despite the fact that you mention at moment 4:00 about the magnitude of the tangent force and the friction force and even "show it" by visually measuring the length of the vectors, the friction vector length's is NOT, graphically, shorter than the tangent vector's length.
Im not sure how much to thanks you sir, i wish my teacher was 1/2 of you. My teacher was not even half of doing his work in the teaching. Im sorry for my english. Thankss$ssss
what a wonderful Teacher you are... you are amazing. you really have saved many people from failure. may GOD bless you. I am so happy for knowing such a wonderful teacher like you.
Sir: Thank you for drilling in both theory and applying it in practice with your wonderful and very well thought out example. I say to myself at what angle will the box stop sliding? Well from one view when both Mg (sin theta) and Mg (cos theta)(mew) equalize. Perhaps taking the limit of theta as it approaches another value less than 30 degrees might reveal an answer. Please provide a nudge on how to go about doing this.
As you described it, when the net force goes to zero, the acceleration goes to zero and if the object starts at rest, it will not move when the two forces you described are balanced.
If the coefficient of friction is greater than 1 or so that it will result in a negative number. (Ally on Ally dry and clean is 1.4) Does that mean that the friction is so high, the object won't slide? Thank you for your amazing way of teaching ❤
A coefficient of friction greater than 1 means that a force greater than the object's weight would be needed to move the object across the flat surface.
Dear Michel - Isn't friction an eccentric force? Therefore shouldn't there also be a couple acting about the blocks COM? Your example is assuming that the friction force is going through the COM , so are you making the assumption that the block can be considered a particle with point mass?
I had Fg|| + Fs = ma|| ( || = parallel) ( let down be negative direction, andup be possitive direction). Then divided by m to get acceleration. Same thing but less number to type in calculator which would give chances to make errors. Either way, Great lector.
Hello Sir,could you solve this problem and add it into one of your videos please ? A block of mass m = 2.00 kg is released from rest at h = 0.500 m above the surface of a table, at the top of a 30.0° incline The frictionless incline is fixed on a table of height H = 2.00 m. (a) Determine the acceleration of the block as it slides down the incline. (b) What is the velocity of the block as it leaves the incline? (c) How far from the table will the block hit the floor? (d) What time interval elapses between when the block is released and when it hits the floor? (e) Does the mass of
One thing that always puzzles me is why is the angle theta between the slope and the surface (30deg in the above video) the same between mg and mgcostheta. Geometry wasn't my forte!! Thanks.
Notice that the direction of mg is perpendicular to the flat part of the wedge and the component mg cos(theta) is perpendicular to the slanted portion of the wedge. Therefore the angle between the flat and slanted portion of the wedge must be the same as the angle between mg and mg cos(theta).
@baldy hardnut haha, not really. When I got to the unie one of first things I have learned were significant figures. Its important to know what you can and cant live without.
Sir u have explained it very well,can u please upload an example in which the inclined plane also moves with some acceleration its very tough to understand that concept can u please please upload a video on the same
That would be an interesting problem indeed. This is covered with the example: Physics - Mechanics: Newton's Laws of Motion (12 of 20) Second Law: Example 5
Michael if rather than multiply (.866x.2) you saw(.866/5) on a student paper would you mark it procedure error even though you obtained the same results 5 being the reciprocal of.2? Thanks for your time and effort
I give my students complete freedom to work out the problems on the test. I will suggest the best method but will never take off any points for doing it by any other method, including the example you gave me. There is off course one exception if the question specifically states to work it out in a particular manner.
I am an 8th grader and I have a question for Prof. Biezen. What is the problem if we solve this by taking a vector with a magnitude of 50X9.8 Newtons and a direction of 240 degrees. Calculate the two component 490cos240 and 490sin240 and go from there... I tried it and it gave me the exact same answer as you got. Plz comment.
Make your separate video in Playlist including all chapters of Physics with each other starting from beginning chapter(Part in India) to end chapter (Part in India) As in 90s Decade Syllabus in Science College,B.N. College,St.Xavier's College and Gossner College in India.
What if the friction force is larger than the parallel Force? Say the object is 4kg, coefficient of friction is 0.64 and the angle is 28 degrees. Ff= 22.17 N and the Parallel Force = mgsin(theta0 = 18.42 N. Net force = -3.75 N, acceleration is -0.9377 m/s^2. Is not sliding down, but it has a negative acceleration, that means when is pushed, it would slow down with an acceleration of -0.9377m/s^2 right!? Thanks!
Then the block would slow down and eventually stop. At that point the block would be at rest and the problem would be exactly the same as shown in the video.
Thanks so much for this video! Helped a ton! there is also an app on the Google Play Store and I believe the App Store as well that does these problems for you which has helped me a ton. Its call "phizX Calculator". Physics in spelled phizX. Just wanted to let you guys know, it could probably help you out!
Can you please help me? I've got a question. An object of given mass at an angle of 30 is in equilibrium under limiting friction. How to find the force required to keep the object in equilibrium if the angle is increased upto 60?
Force of friction = mg x cos(theta) x (coefficient of friction). Calculate that for an angle of 30 degrees and an angle of 60 degrees and the force required will be the difference.
Hello sir, I have a question, so no matter how heavy the object is, the acceleration gonna be the same? Like the masses 10kg, 20kg, 100kg, are they all gonna be the same acceleration because in the formula the mass is reduced...
@@MichelvanBiezen Thank you so much for your prompt reply, but Prof. Bienzen I have a question is the resultant force (mgsin0 - mgcos0u)? And if the resultant force gets bigger does the acceleration goes up as well? Or is the acceleration always the same?
If the angle is 10 degrees, the distance (x) is 3.5m, the block of wood weighs 0.5kg, and the time it took to reach the bottom was 6s; how do I find frictional force then?
For this question, if would the work done by gravity increase as we increase the angle of the ramp, since mgsin(theta) would increase? Also, in contrast, the work done by the frictional force would decrease as we increase the angle because normal force (= mgcos(theta)) would decrease? Thanks a lot. I appreciate your videos. I find them helpful as I have been studying for my MCATs which I will be taking this Friday.
Mizelleful The amount of work done against gravity is equal to the potential energy gained W = PE = mgh And you are correct about the work done against friction.
sir you mentioned that if frictional force is greater than mgsin(thita) then block wont move downward.sir i want to know if friction force is more then Fnet will be in upward direction so block should move in upward direction or not?
When we say: "the friction force is greater than mg sin(theta) we mean the calculated friction force. The actual friction force can never be greater than mg sin(theta) it can only be equal to or smaller than. (The friction force is a reaction force (Newton's third law) and can only match the mg sin(theta) up to its maximum theoretical force.
Your handwriting. It's beautiful. I've never had a science or math teacher with legible handwriting.
I know that feel bro
PR dan PT oopoppoppp PO 98gxxcvvjitjkhcv
Large Marge,
Because I am very appreciative for what the US has done for me and my family.
I'll take this explanation over Khan Academy's any day.
I'll take any free education resource from either gentleman any day! hehe :)
Also shoutout to math bff too!
What's so bad about khan academy.. I think it's pretty awesome..
yes exactlyyyyyy
If only my teacher explained the problem like you do 35 yrs back. It could change my life altogather. Thank you.
This guy is to physics what Tyler Dewitt is to Chemistry. Thank you so much.
Bozeman - Bio
math - physics - chemistry - mechanics :
the organic chemistry tutor
@Letter L nah math Examsolutions that guy is awesome
Math = PatrickJMT
@@yousifnabil9091 Sooooooo Truuuuueeee. That guys the GOAT
WHY CANT THIS MAN BE MY TEACHER!
For. Real.
Facts
He explains this so much better than my teacher, who just made it very complicated and difficult. SMH
Thanks Prof. Biezen!
lizbeth diaz
No incline means that mg sin(theta) = 0 and thus there is no acceleration
Thanks this really helped
True but this is an inclined plane problem.
the whole semester i didn't learn anything in the class but from your 7 min video i learned everything you are the best teacher i have ever seen, the way you teach i swear thats what teaching is called.
This man condensed 2 hours of lecture and text book nonsense into a 7 minute video that's easy to digest and understand. Thank you
I've been teaching two years and have only just discovered this UA-camr. Very, very good videos -- excellent modelling of how to solve a physics problem.
We wish you all the best with your teaching.
I'm abt to cry bc of how happy I am now that I finally understand
This made so much more sense than my teacher, one question though, why is the perpendicular component of gravity cos and the parallel sin? What angle are you using to determine that?
We are using the angle between the verical (mg) and the perpendicular component to the incline of mg (mg cos (theta)) which is the same angle as the incline.
oh okay, thank you so much
In the beginning I had wanted to cry manly tears.
In the end I was crying manly tears - of joy.
lol
I have spent hours watching your lectures to help me pass my physics midterm, thank you so much.
I learned it effortlessly, much better than khan academy. Thank you!
Excellent explanation without a single missed step. Thank you for making this so much easier to understand!
teaching skills are top class
Thank you very much for reminding me that the friction force is Normal force by coefficient cos I needed this to submit for assignment and its due today
10 years later still helping people, what an absolute legend. My physics final is in 24 hours. Thank you so much.
Thank you and all the best on your final!
Mines in an hour and a half 😅
Haha, mines in 2 hours.
You make Physics easy!!! Thanks bro.
Love from INDIA🇮🇳
I understood the concept very well....
Thank you and welcome to the channel!
Had a problem with to case on each other., on tth incline. Used this video and got the answer. Thank you very much.
Glad it helped! 🙂
Hi professor Van Biezen, Today I don't have a physics question. But I wanted to thank you for all your videos and your explanation. I was studying physics for my MCAT examen of medicine. And for the part physics my results were 8/10. That is a very high score. And this result is thanks to your videos! So thanks a lot. And now I can do medicine and become a doctor because I passed my exam :) :)
That is great news. Congratulations! Keep up the great effort.
Despite the fact that you mention at moment 4:00 about the magnitude of the tangent force and the friction force and even "show it" by visually measuring the length of the vectors, the friction vector length's is NOT, graphically, shorter than the tangent vector's length.
tf u talking bout dawg dont confuse me more i have a hard test tmrw
Congratz on getting 100k subs, you truly deserve it :)
Thanks. It has been quite a journey these last 3 years.
he deserves atleast a million
THANK A TONS, Sir Michel!! I understood everything. Please, continue to provide such a quality lecture !
You saved my grade sir, thank you!
I don't know if you'll read this, but you have genuinely saved my life
Glad we were able to help. All the best.
Helped my last minute due assignment, dropped a like
Glad this was helpful
Cannot express how much you have helped me throughout this quarter, keep up the good work!
I am a student from Iraq and have benefited a lot from you sir. I thank you for this effort
Welcome to the channel!
I have a midterm tomorrow and I just now finally understand this! Thank you!
THIS DUDE IS A BEAST! THANK YOU! PAY THIS MAN!
+Pat DeCesare any time
Calm down, Phil.
Im not sure how much to thanks you sir, i wish my teacher was 1/2 of you. My teacher was not even half of doing his work in the teaching. Im sorry for my english.
Thankss$ssss
this video is the reason i am going to pass my physics test today! THANKYOU!!!
what a wonderful Teacher you are... you are amazing. you really have saved many people from failure. may GOD bless you. I am so happy for knowing such a wonderful teacher like you.
https: //ua-cam.com/video/paVOEi7cYrA/v-deo.html (Mec .English and French)👍💐
From #India.... thank u for explaining clearly.. I subscribed
https: //ua-cam.com/video/paVOEi7cYrA/v-deo.html (Mec .English and French)👍💐
TYSM! You are ways better than our teacher who can't even explain the basics !!
https: //ua-cam.com/video/paVOEi7cYrA/v-deo.html (Mec .English and French)👍💐
Sir: Thank you for drilling in both theory and applying it in practice with your wonderful and very well thought out example. I say to myself at what angle will the box stop sliding? Well from one view when both Mg (sin theta) and Mg (cos theta)(mew) equalize. Perhaps taking the limit of theta as it approaches another value less than 30 degrees might reveal an answer. Please provide a nudge on how to go about doing this.
As you described it, when the net force goes to zero, the acceleration goes to zero and if the object starts at rest, it will not move when the two forces you described are balanced.
If the coefficient of friction is greater than 1 or so that it will result in a negative number. (Ally on Ally dry and clean is 1.4)
Does that mean that the friction is so high, the object won't slide?
Thank you for your amazing way of teaching ❤
A coefficient of friction greater than 1 means that a force greater than the object's weight would be needed to move the object across the flat surface.
@@MichelvanBiezen Thank you
Dear Michel - Isn't friction an eccentric force? Therefore shouldn't there also be a couple acting about the blocks COM? Your example is assuming that the friction force is going through the COM , so are you making the assumption that the block can be considered a particle with point mass?
This dude is absolutely awesome.
Thanks!
I had Fg|| + Fs = ma|| ( || = parallel) ( let down be negative direction, andup be possitive direction). Then divided by m to get acceleration. Same thing but less number to type in calculator which would give chances to make errors. Either way, Great lector.
Wow this video cleared up so much of what confused me last year. Super helpful!!!
https: //ua-cam.com/video/paVOEi7cYrA/v-deo.html (Mec .English and French)👍💐
U explain fabulously
your videos are............life changing
Thank you for your comment. We are glad they are helping.
Simply Excellent explanation
Love from India sir ❤ Your teaching helped me a lot
Thank you. Welcome to the channel!
Thank you so much!!!! I had no idea what I was doing because my prof didn’t show us how to get to those formulas.
https: //ua-cam.com/video/paVOEi7cYrA/v-deo.html (Mec .English and French)👍💐
thank you, in 5 minutes you have done what my teacher haven´t explain properly in one year.
Amazing! Helping me prepare for my midterm.
This is brilliant
Hello Sir,could you solve this problem and add it into one of your videos please ?
A block of mass m = 2.00 kg is released from
rest at h = 0.500 m above the surface of a table, at the
top of a 30.0° incline
The frictionless incline is fixed on a table of height
H = 2.00 m. (a) Determine the acceleration of the
block as it slides down the incline. (b) What is the
velocity of the block as it leaves the incline? (c) How far
from the table will the block hit the floor? (d) What
time interval elapses between when the block is
released and when it hits the floor? (e) Does the mass of
the block affect any of the above calculations?
You're God sent sir! Very much appreciated.
Glad to help 🙂
You just made my life easier, thank you
One thing that always puzzles me is why is the angle theta between the slope and the surface (30deg in the above video) the same between mg and mgcostheta. Geometry wasn't my forte!! Thanks.
Notice that the direction of mg is perpendicular to the flat part of the wedge and the component mg cos(theta) is perpendicular to the slanted portion of the wedge. Therefore the angle between the flat and slanted portion of the wedge must be the same as the angle between mg and mg cos(theta).
Thank you.
Omg!!! Thank you so much!!! In my university, it has shut down due to Covid 19 and I was struggling with this!!! New subscriber!
Hang in there. We'll make it through.
the best teaching
Thank you. We appreciate the comment.
Significant figures might have to be the most pointless rule in the world
Hahahaha
In real life, like in a laboratory, they're extremely important.
@baldy hardnut haha, not really. When I got to the unie one of first things I have
learned were significant figures. Its important to know what you can and cant live without.
Sir u have explained it very well,can u please upload an example in which the inclined plane also moves with some acceleration its very tough to understand that concept can u please please upload a video on the same
That would be an interesting problem indeed. This is covered with the example: Physics - Mechanics: Newton's Laws of Motion (12 of 20) Second Law: Example 5
Thank you, Michel, for sharing your knowledge. You are making my Physics 201 much more clearer! Thank you!
when youtube channels deserves your tuition fee more than the school does :(
Michael if rather than multiply (.866x.2) you saw(.866/5) on a student paper would you mark it procedure error even though you obtained the same results 5 being the reciprocal of.2?
Thanks for your time and effort
I give my students complete freedom to work out the problems on the test. I will suggest the best method but will never take off any points for doing it by any other method, including the example you gave me. There is off course one exception if the question specifically states to work it out in a particular manner.
If you teacher is approachable you may want to request a review of the grade if the question didn't specify how you worked out the problem.
I am an 8th grader and I have a question for Prof. Biezen. What is the problem if we solve this by taking a vector with a magnitude of 50X9.8 Newtons and a direction of 240 degrees. Calculate the two component 490cos240 and 490sin240 and go from there... I tried it and it gave me the exact same answer as you got. Plz comment.
There are many different ways in which a problem like this can be solved. Ultimately use the method you are most comfortable with.
Make your separate video in Playlist including all chapters of Physics with each other starting from beginning chapter(Part in India) to end chapter (Part in India) As in 90s Decade Syllabus in Science College,B.N. College,St.Xavier's College and Gossner College in India.
That would be a HUGE task. We placed the videos in the order as found in most college text books used in the US
@@MichelvanBiezen Why not in India when during my birth In Akhand Bihar as mentioned in the above Colleges.
What if the friction force is larger than the parallel Force?
Say the object is 4kg, coefficient of friction is 0.64 and the angle is 28 degrees.
Ff= 22.17 N and the Parallel Force = mgsin(theta0 = 18.42 N.
Net force = -3.75 N, acceleration is -0.9377 m/s^2.
Is not sliding down, but it has a negative acceleration, that means when is pushed, it would slow down with an acceleration of -0.9377m/s^2 right!?
Thanks!
the joe leonard of physics. Just needs to get ripped and tell people to get off their phones. Love all these lessons!
Thank you. (working on the "get ripped" part). 🙂
Thank you,sir,it clearly explains my problem
Great! Glad you found our videos. 🙂
wow just amazing.
i like the way you work out so neat and tidy
GREAT TEACHER 10/10
Excellent explanation! The video series really clarified my understanding.
can u please tell me what if the the block was not released from rest, but instead travelling up the slope
Then the block would slow down and eventually stop. At that point the block would be at rest and the problem would be exactly the same as shown in the video.
@@MichelvanBiezen that i understand but what about, if the question tells us to get the driving force
Thanks so much for this video! Helped a ton! there is also an app on the Google Play Store and I believe the App Store as well that does these problems for you which has helped me a ton. Its call "phizX Calculator". Physics in spelled phizX.
Just wanted to let you guys know, it could probably help you out!
Can you please help me? I've got a question. An object of given mass at an angle of 30 is in equilibrium under limiting friction. How to find the force required to keep the object in equilibrium if the angle is increased upto 60?
Force of friction = mg x cos(theta) x (coefficient of friction). Calculate that for an angle of 30 degrees and an angle of 60 degrees and the force required will be the difference.
@@MichelvanBiezen Thank you so much. Got it!
Thank you for this beautiful lesson 😊
You're welcome 😊 Glad you liked it.
omg this helped me SO MUCH IVE BEEN STUCK ON THIS STUFF FOR DAYS!
https: //ua-cam.com/video/paVOEi7cYrA/v-deo.html (Mec .English and French)👍💐
Best teaching
Love from india
Thank you and welcome to the channel!
Happy new Year , Sir and Many thanks for your videos.
Alberto from Tuscany!
Happy New Year to you as well. Tuscany is a beautiful part of the world. Welcome to the channel.
that is static friction, right ? because when the box moves, the friction will be changing to kinetic friction
I love this explanation. Thank you so much
Glad it was helpful!
Hello sir, I have a question, so no matter how heavy the object is, the acceleration gonna be the same? Like the masses 10kg, 20kg, 100kg, are they all gonna be the same acceleration because in the formula the mass is reduced...
That is correct. Even with friction, it the mass doesn't matter. The acceleration will be the same.
@@MichelvanBiezen Thank you so much for your prompt reply, but Prof. Bienzen I have a question is the resultant force (mgsin0 - mgcos0u)? And if the resultant force gets bigger does the acceleration goes up as well? Or is the acceleration always the same?
Since the equation for "a" does not contain the mass, mass does not have an effect on the acceleration.
Very clean and neat explanation.
I don't know how to be thankful from your work to be honest with you, If you ever come to Ottawa, Canada, I will take you for a special dinner :)
Your thank you was sufficient and appreciated. Welcome to the channel!
thanks soooo much this video has been my savin grace
This Professor is awesome! God bless you! and Thank You Sir!
Life Saver! Thank you so much!!
Amazing video, thanks
If the angle is 10 degrees, the distance (x) is 3.5m, the block of wood weighs 0.5kg, and the time it took to reach the bottom was 6s; how do I find frictional force then?
For this question, if would the work done by gravity increase as we increase the angle of the ramp, since mgsin(theta) would increase? Also, in contrast, the work done by the frictional force would decrease as we increase the angle because normal force (= mgcos(theta)) would decrease? Thanks a lot. I appreciate your videos. I find them helpful as I have been studying for my MCATs which I will be taking this Friday.
Mizelleful
The amount of work done against gravity is equal to the potential energy gained W = PE = mgh
And you are correct about the work done against friction.
thank you for uploading the video
I'll subscribe because I understood two of your videos
Thanks for subscribing.
Thanks sir u cleared my concept...to much extent..respect frm india
Welcome to the channel!
This video helped me so much! Thank you!
Thank you for making such a clear and easy to understand video! This helped me so much :)
Sir you are a genius
this was a great help thankyou
Thankyou sir. Great help.
Most welcome!
sir you mentioned that if frictional force is greater than mgsin(thita) then block wont move downward.sir i want to know if friction force is more then Fnet will be in upward direction so block should move in upward direction or not?
When we say: "the friction force is greater than mg sin(theta) we mean the calculated friction force. The actual friction force can never be greater than mg sin(theta) it can only be equal to or smaller than. (The friction force is a reaction force (Newton's third law) and can only match the mg sin(theta) up to its maximum theoretical force.
thank you mr professor ....