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 RUclipsr. Very, very good videos -- excellent modelling of how to solve a physics problem.
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.
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.
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 :) :)
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
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 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
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.
+Melissa Z. Use the definition of cosine and sine and also note the right triangle. cos(theta) = adjacent side / hypothenuse sin(theta) = opposite side / hypothenuse
Which right triangle are they using in the problem? Because I don't understand why they labeled it they way they did. I would of thought it was the other way around.
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.
AHAHAHHA OOMG I thought there was smthing wrong with my headphones and started hitting the table with them to work until I lost hope lmaooo xD I feel dumb
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.
@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
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.
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.
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?
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 have a stupid question : consider acceleration equal 0 . how do i derive the torque equation in the point where the formal force is applied ? because only one force has a torque which means the object rotates and that doesn't make sense
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.
Michel van Biezen Can you, or have you, post a video showing how to draw the forces on a Force Diagram? To me, force diagrams get really complicated with trucks sliding on while the wheels are locked, static traffic lights, objects hanging from two springs at angle theta, etc. I am always messing up when an angle is supposed to be Tcos(theta) or Tsin(theta).
Jerry Louis-Jean The thumb nail shows all the forces acting on the object on the incline There are quite a few videos which show you how to draw the forces.
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.
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.
The friction force is caused by the component of the force which is normal to the incline. (The mg sin(theta) component cannot cause any friction since it is parallel to the surface)
If friction opposes parallel motion down the incline why is it mu times cos theta? I understand it opposes the normal force but how then can it prevent parallel motion at the same time?
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.
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.
Yep. Static friction is the friction force that is needed to be overcome in order to move an object. Once the object is in motion, kinetic friction is needed to be overcome to keep the object in motion.
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.
I might not be eager to be able to solve this type of problems as much as I would really love to be able to instantly label what is sin and what is cos. The soh.cah.toa mnemonics don't seem to help me. I am fine as long as the triangle is not rotated, but once you start to move things around, I am lost and need a few minutes to find the reference and analogy. Would you mind recording a lesson on how to master this skill, please?
Connect the vectors labeled mg and mg cos(theta) to make a triangle. Then place the angle, theta, into the upper angle of the triangle. Now can you identify the adjacent side and the opposite side with respect to the angle?
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!
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.
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.
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
If only my teacher explained the problem like you do 35 yrs back. It could change my life altogather. Thank you.
He explains this so much better than my teacher, who just made it very complicated and difficult. SMH
Thanks Prof. Biezen!
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'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
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 have spent hours watching your lectures to help me pass my physics midterm, 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've been teaching two years and have only just discovered this RUclipsr. 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
when youtube channels deserves your tuition fee more than the school does :(
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.
Cannot express how much you have helped me throughout this quarter, keep up the good work!
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
You saved my grade sir, thank you!
Excellent explanation without a single missed step. Thank you for making this so much easier to understand!
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.
I learned it effortlessly, much better than khan academy. Thank you!
I am a student from Iraq and have benefited a lot from you sir. I thank you for this effort
Welcome to the channel!
THIS DUDE IS A BEAST! THANK YOU! PAY THIS MAN!
+Pat DeCesare any time
Calm down, Phil.
TYSM! You are ways better than our teacher who can't even explain the basics !!
https: //ruclips.net/video/paVOEi7cYrA/видео.html (Mec .English and French)👍💐
Helped my last minute due assignment, dropped a like
Glad this was helpful
this video is the reason i am going to pass my physics test today! THANKYOU!!!
I have a midterm tomorrow and I just now finally understand this! Thank you!
thank you, in 5 minutes you have done what my teacher haven´t explain properly in one year.
teaching skills are top class
Love from INDIA🇮🇳
I understood the concept very well....
Thank you and welcome to the channel!
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
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.
THANK A TONS, Sir Michel!! I understood everything. Please, continue to provide such a quality lecture !
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
You make Physics easy!!! Thanks bro.
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.
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). 🙂
Love from INDIA 🇮🇳🇮🇳
Hi and welcome to the channel
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
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
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.
how could you find the difference between sin theta and cos theta when you split the x and y components up? ( at about 2:00 in the video)
+Melissa Z.
Use the definition of cosine and sine and also note the right triangle.
cos(theta) = adjacent side / hypothenuse
sin(theta) = opposite side / hypothenuse
Which right triangle are they using in the problem? Because I don't understand why they labeled it they way they did. I would of thought it was the other way around.
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: //ruclips.net/video/paVOEi7cYrA/видео.html (Mec .English and French)👍💐
From #India.... thank u for explaining clearly.. I subscribed
https: //ruclips.net/video/paVOEi7cYrA/видео.html (Mec .English and French)👍💐
anyone else's sound only come out of one headphone?
We didn't have stereo on the older videos. That is now fixed for the new ones.
ok still a fantastic video really helpful thanks!
AHAHAHHA OOMG I thought there was smthing wrong with my headphones and started hitting the table with them to work until I lost hope lmaooo xD I feel dumb
yeah man me to aahhaha
lol, I only had one head phone on
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.
Love from India sir ❤ Your teaching helped me a lot
Thank you. Welcome to the channel!
This dude is absolutely awesome.
Thanks!
Amazing! Helping me prepare for my midterm.
Wow this video cleared up so much of what confused me last year. Super helpful!!!
https: //ruclips.net/video/paVOEi7cYrA/видео.html (Mec .English and French)👍💐
Hey, I'm new to physics so I have a question. Why is the theta in the perpendicular vectors of mg at 2:21 equal to 30 degrees? I appreciate any help.
Since mg is perpendicular to the horizontal, and mg cos(theta) is perpendicular to the incline. Therefore the 2 angles must be the same.
@@MichelvanBiezen ohh right that makes sense. Thanks so much :)
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.
your videos are............life changing
Thank you for your comment. We are glad they are helping.
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
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!
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.
Thank you, Michel, for sharing your knowledge. You are making my Physics 201 much more clearer! Thank you!
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.
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: //ruclips.net/video/paVOEi7cYrA/видео.html (Mec .English and French)👍💐
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!
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?
Hope bro knows he's still helping people even after 11 years
Thank you. Glad our videos are still helpful. 🙂
i have a stupid question : consider acceleration equal 0 . how do i derive the torque equation in the point where the formal force is applied ? because only one force has a torque which means the object rotates and that doesn't make sense
https: //ruclips.net/video/paVOEi7cYrA/видео.html (Mec .English and French)👍💐
Hello I wanted to ask what would happen if the friction force is > than mgsin(theta)? Thank you 🙏
Then the block would slow down if it was moving and will remain motionless if it was not moving.
You just made my life easier, thank you
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.
Michel van Biezen Can you, or have you, post a video showing how to draw the forces on a Force Diagram? To me, force diagrams get really complicated with trucks sliding on while the wheels are locked, static traffic lights, objects hanging from two springs at angle theta, etc. I am always messing up when an angle is supposed to be Tcos(theta) or Tsin(theta).
Jerry Louis-Jean
The thumb nail shows all the forces acting on the object on the incline
There are quite a few videos which show you how to draw the forces.
I got 3m/s2 at 6:30 . is there a problem ?
What was the first decimal when you calculated it?
NO DECIMAL
My new question is why does mgcosheta cancel out he mormal force
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
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.
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! 🙂
You're God sent sir! Very much appreciated.
Glad to help 🙂
What if the object is a sort of liquid? Will the principles remain the same?
+Donald Henrietta
Liquids behave very differently. Look at fluid dynamics.
Michel van Biezen Ok thanks
Please give a equation for pulling and pushing. If F making an angle ¤(theta) with the vertical
Take a look at this playlist: PHYSICS 4.7 FRICTION & FORCES AT ANGLES
Thank you for making such a clear and easy to understand video! This helped me so much :)
wouldnt the friction be mgsin0 because it is in the x direction like the Fg force?
The friction force is caused by the component of the force which is normal to the incline. (The mg sin(theta) component cannot cause any friction since it is parallel to the surface)
@@MichelvanBiezen gotcha so whenever I use mgsin0 as the Fg I'll just remember to switch is around for Friction
If friction opposes parallel motion down the incline why is it mu times cos theta? I understand it opposes the normal force but how then can it prevent parallel motion at the same time?
Friction force is a vector. The magnitude is equal to the normal force x coefficient of friction and the direction is parallel to the incline.
U explain fabulously
This Professor is awesome! God bless you! and Thank You Sir!
The coefficient would be the kinetic friction right?
Yes, since the block would is moving.
Simply Excellent explanation
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.
Best teaching
Love from india
Thank you and welcome to the channel!
thanks sir... Please upload some more advanced problems.
+kaustav das Your welcome
the best teaching
Thank you. We appreciate the comment.
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.
Sorry sir. What's the gravitational force in this case? Is it mg cos 0 since you've resolved mg into mg cos 0 and mg sin 0? Thank you very much!
https: //ruclips.net/video/paVOEi7cYrA/видео.html (Mec .English and French)👍💐
Excellent explanation! The video series really clarified my understanding.
is there a difference between kinetic friction and static friction
Yep. Static friction is the friction force that is needed to be overcome in order to move an object. Once the object is in motion, kinetic friction is needed to be overcome to keep the object in motion.
that is static friction, right ? because when the box moves, the friction will be changing to kinetic friction
God, thank you so much! I missed the day we learned this in class so this is very much appreciated.
Thanks sir u cleared my concept...to much extent..respect frm india
Welcome to the channel!
my teacher just skipped over half of this so it made no sense, thank you so much
wow just amazing.
i like the way you work out so neat and tidy
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.
I might not be eager to be able to solve this type of problems as much as I would really love to be able to instantly label what is sin and what is cos. The soh.cah.toa mnemonics don't seem to help me. I am fine as long as the triangle is not rotated, but once you start to move things around, I am lost and need a few minutes to find the reference and analogy. Would you mind recording a lesson on how to master this skill, please?
Connect the vectors labeled mg and mg cos(theta) to make a triangle. Then place the angle, theta, into the upper angle of the triangle. Now can you identify the adjacent side and the opposite side with respect to the angle?
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!
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.