In carefully considering what is THE EARTH/ground, what is THE SUN, AND the fact that the stars AND PLANETS are POINTS in the night sky, we know that E=MC2 is CLEARLY and necessarily proven to be F=ma IN BALANCE; as ELECTROMAGNETISM/energy is gravity !!! (Gravity is ELECTROMAGNETISM/energy.) Consider what is the speed of light (c) ON BALANCE. Great. Now, very importantly, outer "space" involves full inertia; AND it is fully invisible AND black. Think. BALANCE and completeness go hand in hand !!! E=MC2 is CLEARLY AND necessarily proven to be F=ma ON BALANCE, as ELECTROMAGNETISM/energy is gravity !!! Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, as ELECTROMAGNETISM/energy is gravity; as E=MC2 is CLEARLY and necessarily F=ma IN BALANCE. By Frank DiMeglio
I actually have one doubt , won't T1= minus mgsintheta + ma since it is in the opposite direction to the motion . Otherwise this was a great video , thanks so much
No, that is not how you should look at it. Think of it in terms of what T does. It holds m1 against gravity AND T also accelerates it. Both terms are positive. You can also draw a free body diagram and use F = ma to get the same correct result.
+Michel van Biezen really you wish you would draw a force diagram because i dont understand half the stuff youre talking about when you dont include it. im trying to draw one right now and im having trouble. :/
E=mc2 is F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity IS ELECTROMAGNETISM/energy, AS E=MC2 IS F=MA. The Earth (A PLANET) is a MIDDLE DISTANCE form that is in BALANCED relation to the Sun AND the speed of light (c), AS the stars AND PLANETS are POINTS in the night sky; AS E=MC2 IS F=MA; AS TIME DILATION ultimately proves ON BALANCE that ELECTROMAGNETISM/energy is gravity. Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=MA; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=mc2 is F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. MOREOVER, a given PLANET (INCLUDING WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS F=ma, E=mc2, AND what is perpetual motion; AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. By Frank DiMeglio
@@Markism07 E=mc2 is F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity IS ELECTROMAGNETISM/energy, AS E=MC2 IS F=MA. The Earth (A PLANET) is a MIDDLE DISTANCE form that is in BALANCED relation to the Sun AND the speed of light (c), AS the stars AND PLANETS are POINTS in the night sky; AS E=MC2 IS F=MA; AS TIME DILATION ultimately proves ON BALANCE that ELECTROMAGNETISM/energy is gravity. Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=MA; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=mc2 is F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. MOREOVER, a given PLANET (INCLUDING WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS F=ma, E=mc2, AND what is perpetual motion; AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. By Frank DiMeglio
Well some people are using TH-cam the right way and some only to review stuffs, what is going on their life etc... This is what TH-cam supposed to do, education is the strongest weapon, who knows there will be next great scientist in the making, who will solve the world greatest mystery and that is "time"
I really appricate the videos about mechanics you upload. They are first class, it gets into my head straight away! I wish you was my maths teacher... I was wondering if you could do a couple tutorials on how to do Moments (without vectors) such as calculating the magnitude of the turning effect if a force applied to a rigid body; problems about bodies in equilibrium and solving problems about non-uniform bodies, statics of a particle - such as a force acting on a particle on an inclined plane at and angle (not parallel to the plane which you have uploaded already). I hope you consider doing some more A-level maths tutorials. Many thanks.
Toby, If you could give me an example or two of what types of problems you are looking for it will give me a better idea of what you are looking for. I will be doing more physics videos in the future, so it will help figure out what material to cover. Thanks.
Toby, Many of the types of problems you are requesting are already there. Look in the 2 physics - mechanics play lists. I will continue to add to the list, but that will take a little while.
The best way to think about it is to look and see what T1 is doing. If nothing was moving or accelerating, (static situation) then T1 would just be keeping m1 from sliding and T1 would be equal to mg sin(theta). But T1 is also accelerating m1 up the incline and Newton's second law tells us that F = ma thus T1 must be increased by m1a. Thus T1 = mg sin(theta) + m1a
sir that way you were using to find acceleration of the system is not so effective because its differing from when you find the acceleration by resolving the forces separately
In real life of course there is always friction. But we start problems like this in a more simple manner, by leaving off the friction forces (and yes there is no friction anywhere in the system). Later we add the additional complication of friction.
Newton's Second Law of Motion needs a slight correction since they did not know about squaring a number in Newton's era we can assume that this may have been overlooked. As such (F=ma^2) is a natural correction that can be easily proofed through calculation, formulation, and experimentation. This correction is 33% more accurate if you do not accept it proof it yourself.
There are different approaches. In one approach you can draw a free body diagram around each component and determine the vectors, and then add them (solve them simultaneously), or you can take the whole system together as shown in the video.
Hi! Am I right in thinking that you can essentially ignore the force NORMAL to the structure, m1*g*cos(30), ONLY if the coefficient of friction is nil when looking for F-net? Whereas if I had some friction, I'd need to include m1*g*cos(30)*friction in the relevant direction in my F-net calculation?
The component of the weight perpendicular to the slanted surface is cancelled out by the normal force pushing back with an equal amount of force in the opposite direction (Newton's third law). However that said, it is used to calculate the friction force if there is a coefficient of friction between the 2 surfaces.
@@MichelvanBiezen mechanics and lab touches on this then statics expands on it? There are two courses at my school one is mechanics the other is statics
In the example in this video, the tension is the same throughout the wire. But if the pulley had mass or some resistance to its motion, then the tension on both side of the pulley would be different.
Ohh. Sir actually, I am preparing for JEE advanced. Could you suggest me on how can I approach difficult questions as I have tried to solve them, I have succeed 4 out of every 10 questions. And those questions are just too out of the box. For reference, you can google 2020 iit jee advance paper.
When trying to find the tension, why don’t we consider the total mass since the m in ma is m1+m2? Can I just use one single mass? I thought they accelerate as a system so we can’t separate them...
+irwin israel Tomas I am not sure what you mean by "what if they exchange their given mass?" If you put m2 on the incline and m1 is hanging from the rope, you have a very different problem. You would have to do more than just interchanging m1g and m2g. I always recommend that you don't make those types of "quick" guesses, but work out the problem as stated in a systematic fashion as shown in the example.
Alex, If you slide the parallel component down so that it connects the tips of the vector mg and the vector mg cos(theta) then you have a triangle where mg is the hypotenuse and mg sin(theta) is the opposite side to the angle theta, which is the same angle as the angle of the inclined plane.
There is Fnet in the vertical direction: N = m2g cos theta and then there is Fnet in the direction of motion parallel to the incline Fnet = m1g - m2g sin theta
The tension in the string is always equal to the force required to hold the body against the force of gravity PLUS the force required to accelerate it against gravity. (Only when the object accelerates downward (like with m2) will you subtract that term).
Hello Michael. Looking at sin 30 degree which is located in quadrant three, I thought we should have had negative 0.5 instead of positive 0.5 which will make the acceleration 7.62 instead of 3.27. PLEASE PROVIDE CLARITY. Thanks..
The signs of the forces are referenced to the assumed direction of the acceleration. + for the forces aiding the acceleration and - for the forces opposing the acceleration.
lot of thanks for replying.Sir,but T1 and T2 holding the block together if it would have not been it just fall with acceleration m1g sine theta and m2g.
Correct, the strings holds the block together making into a single system such that the blocks move at the same speed and have the same acceleration. That is why we can solve it as a single system.
@@MichelvanBiezen the box on an inclined plane is being pulled at a constant speed up a frictionless inclined plane by the box hanging on the chord, sir.
@@MichelvanBiezen A 30N block is pulled at a constant speed up a frictionless inclined plane by a weight of 10N hanging from a chord attached to a block and passing over a frictionless pulley at the top of the plane. Find a) the slope angle of the plane, b) the tension in the chord, c) the normal force on the block by the plane.
If they are moving at a constant speed, there is no acceleration and therefore there is no net force. The weight of the hanging block (M2g) is equal to the component of the weight of the block on the incline along the incline. M1g sin(theta) M2g = M1g sin(theta) sin(theta) = M2/M1 and theta = sin^-1 (M2/M1)
everything I see videos like this, old and fun, I ask myself, isn't this like time traveling to the past? I mean it is insane how we get to see this 8 years after it is made when back then I didn't even know what physics was, today looking at this and questioning what this guy looks like today and what happened, why did he stop posting updates it is just insane how technology evolved thanks to all the ones who've seeked knowledge or is seeking it and thanks to the ones teaching and spreading it, just a wholesome world we live in you just gotta look around
@@MichelvanBiezen oh appreciated sir actually I'm new here, been watching a few videos but looking at the way you teach and explain giving all the possible shortcuts you are to remain undefeated with the high quality education you are offering, I wish you the best of life obviously one of the best teachers I've got taught by, hopefully I'll be here everyday throughout the journey
The mass is not reduced by the incline. Only the effect of the weight is reduced due to the incline. The weight still acts 8 kg mass in its full amount, but part of the weight is supported by the normal force from the incline. The inertia never ceased to exist, and the whole 8 kg mass still has to be accelerated.
this guy is a god. my physics teacher makes this problem 10x more cancer and complicated using two separate systems instead of what this guy did, which was one.
My professor asked a similar problem, but after solving for tension and acceleration he wants us to find the position of block one when it stops sliding up the ramp after block two hits the ground. I'm looking through the 5 kinematic equations, but I'm getting stuck on all of them. :(
+Olivia Wentworth You can solve that in two ways: 1) use conservation of energy 2) use equations of kinematics. After the hanging block hits the ground it no longer pulls on the string. Then find the net force on the block on the incline, use that to find the acceleration and then use the equation: V^2 = Vo^2 + 2*a*x
Interesting question would be if the wedge was massless and sitting on a frictionless flat surface. The 10 kg mass touched the vertical face of wedge. There was a coefficient of friction of mu for each of the masses with the wedge. How would system move? Compute accelerations.
m1 a_rope sin(theta) = (T - mu N1) sin(theta) - m1 g + N1 cos(theta) ----------- FBD m1 in vertical direction, a_rope = acceleration in direction of rope pull (up and right) m1 (a_rope cos(theta) + a_wedge) = (T - mu N1) cos(theta) - N1 sin(theta) ----------- FBD m1 in horizontal direction, a_wedge = acceleration of wedge horizontally (right) N1 sin(theta) + mu N1 cos(theta) = N2 + T cos(theta) ----------- FBD massless wedge in horizontal direction, "T cos(theta)" term comes from pulley N2 = m2 a_wedge ----------- FBD m2 in horizontal direction m2 g - T - mu N2 = m2 a_rope ----------- FBD m2 in vertical direction, a_rope = acceleration in direction pulling rope (downward) There are 5 variables (T, N1, N2, a_rope, a_wedge) and 5 equations, so you can then solve for a_rope and a_wedge to know how system will move. If you want to compute the effective weight of the system on the flat surface, which equals Nw (normal reaction of wedge on flat surface): Nw = N1 cos(theta) - mu N1 sin(theta) + mu N2 + T sin(theta) + T ----------- FBD massless wedge in vertical direction, "T sin(theta) + T" term comes from pulley
Tension is a force INTERNAL to the system and thus does not add to the acceleration. Only EXTERNAL forces acting on the WHOLE system affect the acceleration.
Please don't ever delete your videos! I will need them to review for my exam! :)
i wanna like but u got 69 likes
lol how did ur exam go
@@karalgh643 he probably got his bachelors already
In carefully considering what is THE EARTH/ground, what is THE SUN, AND the fact that the stars AND PLANETS are POINTS in the night sky, we know that E=MC2 is CLEARLY and necessarily proven to be F=ma IN BALANCE; as ELECTROMAGNETISM/energy is gravity !!! (Gravity is ELECTROMAGNETISM/energy.) Consider what is the speed of light (c) ON BALANCE. Great. Now, very importantly, outer "space" involves full inertia; AND it is fully invisible AND black. Think. BALANCE and completeness go hand in hand !!!
E=MC2 is CLEARLY AND necessarily proven to be F=ma ON BALANCE, as ELECTROMAGNETISM/energy is gravity !!! Gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites, as ELECTROMAGNETISM/energy is gravity; as E=MC2 is CLEARLY and necessarily F=ma IN BALANCE.
By Frank DiMeglio
@@frankdimeglio8216 Hell y e a h. That's cool. Bad ass
you can preferably take g as 10 as it will simplify all your calculations and simplify the whole solving analogy too.
Yes, but most books do not simplify like that, they use 9.8.
I actually have one doubt , won't T1= minus mgsintheta + ma since it is in the opposite direction to the motion . Otherwise this was a great video , thanks so much
No, that is not how you should look at it. Think of it in terms of what T does. It holds m1 against gravity AND T also accelerates it. Both terms are positive. You can also draw a free body diagram and use F = ma to get the same correct result.
Udin,
There are other videos in the same playlist that cover inclines with friction
+Michel van Biezen really you wish you would draw a force diagram because i dont understand half the stuff youre talking about when you dont include it. im trying to draw one right now and im having trouble. :/
E=mc2 is F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity IS ELECTROMAGNETISM/energy, AS E=MC2 IS F=MA. The Earth (A PLANET) is a MIDDLE DISTANCE form that is in BALANCED relation to the Sun AND the speed of light (c), AS the stars AND PLANETS are POINTS in the night sky; AS E=MC2 IS F=MA; AS TIME DILATION ultimately proves ON BALANCE that ELECTROMAGNETISM/energy is gravity. Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=MA; AS ELECTROMAGNETISM/ENERGY IS GRAVITY.
GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=mc2 is F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. MOREOVER, a given PLANET (INCLUDING WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS F=ma, E=mc2, AND what is perpetual motion; AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY.
By Frank DiMeglio
@@Markism07 E=mc2 is F=ma. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity IS ELECTROMAGNETISM/energy, AS E=MC2 IS F=MA. The Earth (A PLANET) is a MIDDLE DISTANCE form that is in BALANCED relation to the Sun AND the speed of light (c), AS the stars AND PLANETS are POINTS in the night sky; AS E=MC2 IS F=MA; AS TIME DILATION ultimately proves ON BALANCE that ELECTROMAGNETISM/energy is gravity. Indeed, TIME is NECESSARILY possible/potential AND actual IN BALANCE; AS E=MC2 IS F=MA; AS ELECTROMAGNETISM/ENERGY IS GRAVITY.
GRAVITATIONAL force/ENERGY IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE, AS E=mc2 is F=ma; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE, AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution. MOREOVER, a given PLANET (INCLUDING WHAT IS THE EARTH) sweeps out EQUAL AREAS in equal times consistent WITH/AS F=ma, E=mc2, AND what is perpetual motion; AS E=MC2 IS F=MA ON BALANCE; AS ELECTROMAGNETISM/ENERGY IS GRAVITY.
By Frank DiMeglio
your a savior to people like me who sleep during the university classes
( As a student I used to fall asleep in many of my classes as well in college) 🙂
LOL!
His explanations are so intuitive, ahm almost scared to believe it could be that simple
Great Job! You break it down so much easier than my physics teacher.
Well some people are using TH-cam the right way and some only to review stuffs, what is going on their life etc... This is what TH-cam supposed to do, education is the strongest weapon, who knows there will be next great scientist in the making, who will solve the world greatest mystery and that is "time"
Thank you so much, you save my life. My professor is really sucked in explaining things
You're the man!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! The only reason why I'm succeeding in Missouri S&T is because of you!
I really appricate the videos about mechanics you upload. They are first class, it gets into my head straight away! I wish you was my maths teacher...
I was wondering if you could do a couple tutorials on how to do Moments (without vectors) such as calculating the magnitude of the turning effect if a force applied to a rigid body; problems about bodies in equilibrium and solving problems about non-uniform bodies, statics of a particle - such as a force acting on a particle on an inclined plane at and angle (not parallel to the plane which you have uploaded already).
I hope you consider doing some more A-level maths tutorials.
Many thanks.
Toby,
If you could give me an example or two of what types of problems you are looking for it will give me a better idea of what you are looking for. I will be doing more physics videos in the future, so it will help figure out what material to cover. Thanks.
Toby,
Many of the types of problems you are requesting are already there. Look in the 2 physics - mechanics play lists. I will continue to add to the list, but that will take a little while.
what happens if the coefficient of friction isn't 0 but let's say like, 0.15 or something like that
We have examples in this playlist that show that with and without friction.
Thanks!! Realy helped me understant these kinds of problems
Glad it helped!
This is more helpful thank you..requesting for.more modified examples
We have dozens of examples like that on Newton's second law applications.
I can't understand when you are solving for T1. Isn't T1 and m1gsinQ oppusite direction,so why isnt T1=m1gsinQ-m1a?
The best way to think about it is to look and see what T1 is doing. If nothing was moving or accelerating, (static situation) then T1 would just be keeping m1 from sliding and T1 would be equal to mg sin(theta). But T1 is also accelerating m1 up the incline and Newton's second law tells us that F = ma thus T1 must be increased by m1a. Thus T1 = mg sin(theta) + m1a
In Turkey, these questions were easy to us :(, our education system forcing to us to solve harder questions :')
Yes, Turkey has high standards in their education system.
sir that way you were using to find acceleration of the system is not so effective because its differing from when you find the acceleration by resolving the forces separately
We have examples of both methods. I find this method to be the most effective and easy to learn for the students. .
Why would you put friction as zero?
Does it mean there’s no friction in the system?
In real life of course there is always friction. But we start problems like this in a more simple manner, by leaving off the friction forces (and yes there is no friction anywhere in the system). Later we add the additional complication of friction.
i like the way you explain because it is more simple compared to what i have learnt in my university
Keep on doing, great meticulous explanations, i appreciate effort and your passion on teaching.❤
Thank you! 😃
Newton's Second Law of Motion needs a slight correction since they did not know about squaring a number in Newton's era we can assume that this may have been overlooked. As such (F=ma^2) is a natural correction that can be easily proofed through calculation, formulation, and experimentation. This correction is 33% more accurate if you do not accept it proof it yourself.
I wish i had you for my physics teacher!
Appreciate this, I've understood these problems easily
You are welcome. 🙂
I don't know how can I put it but its nice thing and we thanking you because its when helping revising .
Hello sir, I was wondering what would happen if you cut the 10kg block?
YOU ARE THE BEST !!
THANKS A LOT ()"
just use the formula T= [m1m2/(m1+m2)]*(1+sin30)*g
Why we don’t need to plus it likes vector
There are different approaches. In one approach you can draw a free body diagram around each component and determine the vectors, and then add them (solve them simultaneously), or you can take the whole system together as shown in the video.
Timer, danko is thank you 🙏🏿
You are welcome.
But What About Is There1, Friction
2. The Third Ground MasS How We Calculete Shuch Like Question At Last But Not Least Thanks
There are dozens of examples on this channel with and without friction in all kinds of different situations. They can all be found from the home page.
@@MichelvanBiezen ok can u put the link
i am in 8th grade yet why am i studying this XD
If you understand the basics of trigonometry you can learn this. (It is never too early to learn).
This channel is the reason why i pass my mechanics
Hi! Am I right in thinking that you can essentially ignore the force NORMAL to the structure, m1*g*cos(30), ONLY if the coefficient of friction is nil when looking for F-net?
Whereas if I had some friction, I'd need to include m1*g*cos(30)*friction in the relevant direction in my F-net calculation?
The component of the weight perpendicular to the slanted surface is cancelled out by the normal force pushing back with an equal amount of force in the opposite direction (Newton's third law). However that said, it is used to calculate the friction force if there is a coefficient of friction between the 2 surfaces.
You're such an awesome teacher
Thank you. Glad you found our videos! 🙂
You're a fucking god dude. Thank you so much
How do you know if it's a sinθ or cosθ?
By using the definitions: sin = opposite side / hypotenuse cos = adjacent side / hypotenuse
Thank you so much for this video - incredibly helpful and clear
Guys quick question will I learn this in statics or mechanics and lab???
You will learn this in a physics mechanics course.
@@MichelvanBiezen mechanics and lab touches on this then statics expands on it? There are two courses at my school one is mechanics the other is statics
That sounds like two mechanical engineering classes. (typically statics comes first and then dynamics).
my physics teacher suuuucks. glad I found you videos! such a big help
Why is everyone so obsessed with the second law of thermodynamics...
thanks but i understand it now because of your other vid
Sir one doubt how can there be two diff tension in the same wire
In the example in this video, the tension is the same throughout the wire. But if the pulley had mass or some resistance to its motion, then the tension on both side of the pulley would be different.
@@MichelvanBiezen really appreciate that and i will looking forward for more of your videos
Which video has this situation with friction?
uyasebenza nsinzwa yakwethu
It does take a lot of work putting thousands of these videos on line. 🙂
hello respected sir its excellent lec. but can u make little more clear screen tq
Man, I wanna ask one thing, are these university level questions in America??
Both high school and university. The difference is that the problems are more difficult at the university level.
Ohh. Sir actually, I am preparing for JEE advanced. Could you suggest me on how can I approach difficult questions as I have tried to solve them, I have succeed 4 out of every 10 questions. And those questions are just too out of the box. For reference, you can google 2020 iit jee advance paper.
Did you find our JEE advanced test question videos? You can find them from the home page.
@@MichelvanBiezen Thanks a lot sir. Will help me a lot for my exam.
When trying to find the tension, why don’t we consider the total mass since the m in ma is m1+m2? Can I just use one single mass? I thought they accelerate as a system so we can’t separate them...
This playlist will shed some more light on that: PHYSICS 4.8 FREE BODY DIAGRAMS
why is the sin and cos there ( i know trig but I just don't understand-isn't the velocity going down the slope the adjacent?)
we don't get u so please explain what u want to be told clearly
what if they exchange their given mass? can i say m1g - m2g? because my 'a' positive is going to the left and my 'a' negative is going to the right?
+irwin israel Tomas
I am not sure what you mean by "what if they exchange their given mass?"
If you put m2 on the incline and m1 is hanging from the rope, you have a very different problem. You would have to do more than just interchanging m1g and m2g.
I always recommend that you don't make those types of "quick" guesses, but work out the problem as stated in a systematic fashion as shown in the example.
shouldn't the parallel component @2:08 be m_1g/sin(30) ?isnt that component technically the hypotenuse of the triangle?
Alex,
If you slide the parallel component down so that it connects the tips of the vector mg and the vector mg cos(theta) then you have a triangle where mg is the hypotenuse and mg sin(theta) is the opposite side to the angle theta, which is the same angle as the angle of the inclined plane.
You are the best teacher!
You are an amazing teacher, beats my college professor for which I pay money
Thank you. Glad you found our videos.
This is amazing sir, keep it up!
Glad you found our videos. 🙂
sir you are an angel i cant thank you enough
i just wondered that u wrote a=Fnet/Mt. my question is why did u use mg - mg theita. actually, Fnet is mgcos theita... I'm just confused...
There is Fnet in the vertical direction: N = m2g cos theta
and then there is Fnet in the direction of motion parallel to the incline
Fnet = m1g - m2g sin theta
Why we can assembly it in one equation cause force that act on the system is not the same axis
Because we are only looking for the net force acting on the syatem along the line of motion of the system
But force is in the different axis
The pulley changes the direction of the string. So we consider the forces ALONG THE DIRECTION OF MOTION FOR EACH PART OF THE SYSTEM
At 10:00 Why are you adding the two forces? m1*g*sin theta is in the opposite direction of the motion. Thanks!
The tension in the string is always equal to the force required to hold the body against the force of gravity PLUS the force required to accelerate it against gravity. (Only when the object accelerates downward (like with m2) will you subtract that term).
Thank you so much
You're most welcome 🙂
What would you do if you were solving for acceleration but the coefficient of friction was unknown?
That would depend on what else is known / given.
life saver...
Hello Michael. Looking at sin 30 degree which is located in quadrant three, I thought we should have had negative 0.5 instead of positive 0.5 which will make the acceleration 7.62 instead of 3.27. PLEASE PROVIDE CLARITY. Thanks..
The signs of the forces are referenced to the assumed direction of the acceleration. + for the forces aiding the acceleration and - for the forces opposing the acceleration.
THANKS!
Good job! Thank you Michel!
These videos are life savers!
Very nice video! Exam helper!
You have set another ember for me in physics. My teacher destroyed it since the start of the semester. THANK YOU!
thank you so much!
Glad it helped!
Awesome!!! I really enjoyed your lecture.
Thank you so much for such explanations ......u made it really easy for me man.....
Love your tactics sir
Thank you so much
You're most welcome
I like your way of teaching with my respect and thanks
thankyou very much for associatig us...........may god live u long ameen
THANK YOU
I gotta Say i love your video. Very helpful for my physics course. hopefully i would apply it in my engineering career.
Sir,why you not included T1 and T2 as the part of the net force.though you have taken m1g sine theta and m2g why not T1 and T2
Those are internal forces to the system and do NOT contribute any force to the acceleration of the SYSTEM
lot of thanks for replying.Sir,but T1 and T2 holding the block together if it would have not been it just fall with acceleration m1g sine theta and m2g.
Correct, the strings holds the block together making into a single system such that the blocks move at the same speed and have the same acceleration. That is why we can solve it as a single system.
Think you
You are welcome.
Lovely❣️
Thank you
How to solve the overall questions if coefficient of friction is not equal to zero but equal to 0.2?
Sir can you plize add for finding individual acceleration for each masses
We have lots of examples like that in the playlist. (See the home page).
Michel van Biezen thank you very much sir for your time these videos are very helpful to clear basic concepts thank you very much sir
Sir, I hope you can help me with my problem. How can I find the slope angle if the only given in the problem are the weight of the two boxes?
They should have also told you what the boxes are doing (moving, staying still, accelerating) and if there is friction or not.
@@MichelvanBiezen the box on an inclined plane is being pulled at a constant speed up a frictionless inclined plane by the box hanging on the chord, sir.
@@MichelvanBiezen A 30N block is pulled at a constant speed up a frictionless inclined plane by a weight of 10N hanging from a chord attached to a block and passing over a frictionless pulley at the top of the plane. Find a) the slope angle of the plane, b) the tension in the chord, c) the normal force on the block by the plane.
If they are moving at a constant speed, there is no acceleration and therefore there is no net force. The weight of the hanging block (M2g) is equal to the component of the weight of the block on the incline along the incline. M1g sin(theta) M2g = M1g sin(theta) sin(theta) = M2/M1 and theta = sin^-1 (M2/M1)
theta = sin^-1 (10/30)
听不懂（i cant unstained）
everything I see videos like this, old and fun, I ask myself, isn't this like time traveling to the past? I mean it is insane how we get to see this 8 years after it is made when back then I didn't even know what physics was, today looking at this and questioning what this guy looks like today and what happened, why did he stop posting updates it is just insane how technology evolved thanks to all the ones who've seeked knowledge or is seeking it and thanks to the ones teaching and spreading it, just a wholesome world we live in you just gotta look around
Thank you for your comment. We are actually still posting a video every day and have done so now for almost 10 years.
@@MichelvanBiezen oh appreciated sir actually I'm new here, been watching a few videos but looking at the way you teach and explain giving all the possible shortcuts you are to remain undefeated with the high quality education you are offering, I wish you the best of life obviously one of the best teachers I've got taught by, hopefully I'll be here everyday throughout the journey
Thank you and welcome to the channel!
what about if there is a spring at where m2 is?
It depends on the question. It can be a simple harmonic motion problem or a static problem, again depending on how it is worded.
Why is it Fnet/8kg + 10kg and not Fnet/(8kg(sin30) + 10 kg)? It seems that the mass of 8kg would be reduced by the angle of incline.
The mass is not reduced by the incline. Only the effect of the weight is reduced due to the incline. The weight still acts 8 kg mass in its full amount, but part of the weight is supported by the normal force from the incline.
The inertia never ceased to exist, and the whole 8 kg mass still has to be accelerated.
How to solve the overall questions if coefficient of friction is not equal to zero but equal to 0.2?
What should I do if there is friction?
There are lots of examples in the playlists where there is friction as well.
thank u
Thank you so much for your videos.🙏🏽
Ily thank u
this guy is a god. my physics teacher makes this problem 10x more cancer and complicated using two separate systems instead of what this guy did, which was one.
very very helpful
My professor asked a similar problem, but after solving for tension and acceleration he wants us to find the position of block one when it stops sliding up the ramp after block two hits the ground. I'm looking through the 5 kinematic equations, but I'm getting stuck on all of them. :(
+Olivia Wentworth You can solve that in two ways: 1) use conservation of energy 2) use equations of kinematics. After the hanging block hits the ground it no longer pulls on the string. Then find the net force on the block on the incline, use that to find the acceleration and then use the equation: V^2 = Vo^2 + 2*a*x
Interesting question would be if the wedge was massless and sitting on a frictionless flat surface. The 10 kg mass touched the vertical face of wedge. There was a coefficient of friction of mu for each of the masses with the wedge. How would system move? Compute accelerations.
m1 a_rope sin(theta) = (T - mu N1) sin(theta) - m1 g + N1 cos(theta) ----------- FBD m1 in vertical direction, a_rope = acceleration in direction of rope pull (up and right)
m1 (a_rope cos(theta) + a_wedge) = (T - mu N1) cos(theta) - N1 sin(theta) ----------- FBD m1 in horizontal direction, a_wedge = acceleration of wedge horizontally (right)
N1 sin(theta) + mu N1 cos(theta) = N2 + T cos(theta) ----------- FBD massless wedge in horizontal direction, "T cos(theta)" term comes from pulley
N2 = m2 a_wedge ----------- FBD m2 in horizontal direction
m2 g - T - mu N2 = m2 a_rope ----------- FBD m2 in vertical direction, a_rope = acceleration in direction pulling rope (downward)
There are 5 variables (T, N1, N2, a_rope, a_wedge) and 5 equations, so you can then solve for a_rope and a_wedge to know how system will move.
If you want to compute the effective weight of the system on the flat surface, which equals Nw (normal reaction of wedge on flat surface):
Nw = N1 cos(theta) - mu N1 sin(theta) + mu N2 + T sin(theta) + T ----------- FBD massless wedge in vertical direction, "T sin(theta) + T" term comes from pulley
a= 7.6 m/s²
thanks sir
Which grade physics is this in the US?
Either 12th grade in high school or first year in college.
Why you are not considering tension?
Tension is a force INTERNAL to the system and thus does not add to the acceleration. Only EXTERNAL forces acting on the WHOLE system affect the acceleration.
Okay Sir Thank you.
You are The best.
Perfect thank you so much
You're welcome 😊 Glad you found it helpful.
thank you!!!!
This question was easy. Do tougher ones
Excellent. Thanks.
Glad you liked it!