This is like a crash course but nothing about it is too clumped. Perfect, straight-to-the-point, and easy. Will subscribe and listen to you religiously throughout my physics course. Thank you.
I did a short experiment. I watched this video 2 times: yesterday night and tonight. Yesterday, I understood everything he said but couldn't get anything inside my head. So I used the whole day today to read the book and solved many maths related to frictions. Tonight as I watched this video again I was able to understand everything loud and clear😁. Thanks professor Dave
Professor Dave, your videos got me through physics 1 and 2 in my post-bacc, and as I'm studying for the MCAT I found myself needing additional explanation. The jingle came to my mind and I knew exactly where to go. So thankful for you and your videos!
Both. Impressive for Dave but sad for your professor and you guys. I'm in the same boat with my professor. His lectures are really disorganized. I was getting a little scared that I was getting confused so early in the semester, especially when I'm way ahead of the math requirements. Watching just a few of Dave's videos and I feel much more confident. I've learned more from from short UA-cam videos than from 2 hour lectures.
Unfortunately, my prof doesn't teach anything... so I am very thankful for your videos with visual aids! I hope you keep making more!!! Yours are by far the best on youtube! :)
thanks a lot bro. this new subscriber is from the Philippines, the way you deliver the topic is so smooth, wonderful and easy to understand. thanks a lot. static & kinetic friction topic. great!
7:05 aren't the x & y components mg*csc(theta) and mg*sec(theta) respectively? The trig here isn't checking out for me. ? to denote x-component of mg sin(theta) = mg/? ?*sin(theta) = mg ? = mg/sin(theta) ? = mg*csc(theta)
Thanks,dave. At this current time, i have a hard time learning about frictional force in orthodontic. And ur explanation makes me a little bit confidence. I understand the first 5:30 minutes,the rest remains a question for me. Thx
6:54 I wonder if the Fnet=mgcos(theta)-Fsf resulted in a negative number (meaning the Fsf is greater than the mgcos), then what will happen? Will the box stay at rest or will the box move upwards? Depending on the context though, i'd guess the box would remain at rest.
As said in the video, any applied force less than the maximum static frictional force will cancel out resulting in no overall movement of the object even if frictional force is much larger than applied force because it is simply meant to maintain the motion of the object (or lack thereof).
it's true, it could be addition as long as you then make the frictional force negative, we just need opposite signs on the two horizontal forces as they are in opposite directions. in the end, as long as the signs make sense you're all good!
There is no certain direction for a frictional force to be acted. It is always resistive to a motion or to an attempt to the motion. When the block moves or tends to move down, frictional force would act up the incline. When the block moves or tends to move up, the frictional force would act down the incline. Thanks.
Professor Dave, I have a question! If friction is caused by forces of attraction between atoms on a molecular, and electrostatic attraction is one of of these forces of attraction. Does this mean that friction can change based on the charge of the object?
May I have a question. At 1:27 you said the interaction between atoms in the object and atoms in the surface is an attractive interaction. But shouldn't it be repulsive instead? Because the interaction here is between electrons of the object's atoms and electrons of the surface's atoms, both are negatively charged, so they should repel each other instead of attract, right? And if they repel each other, logically this interaction should help facilitate the motion by pushing the two surfaces away from each other. But in fact, motion is hindered. Could you explain how it works?
Atoms that are stable have no net charge. The atom consists of protons as well which cancels the effect of negatively charged electrons. In an atom (which is not an ion) the number of protons and electrons are equal and both of them have the same charge.(i.e charge in a single proton= charge in a single electron) . ( This is 100% true) But the thing is that, even if two charged bodies are interacting still friction will act upon them. Considering that the bodies are big(big enough that we can see them). This is classical physics. Here we deal with objects that are not very small. (Although I am not 100% sure about this one)
Oh sir... Thanks for saving me ... Actually i love physics but the concept of cofficent "i was unable to understand 8t properly but after you video .. everythings is clear... Thanks sir! !!!!
i think there is an error in checking comprehension. if the object is at rest, we wont know whether it moves unless we look at the static friction coefficient (as opposed to kinetic friction as stated)
@@Mr3mosThe friction is minus because it's a vector quantity and remember that the friction is always opposite. If we define the mg*sin(35)'s direction as positive
Basically dave is not wrong either you can add it but since the motion is acting in different direction the other one would act as negative and the answer will be net force going to the direction of the larger one👌👌
What Dave did is correct. The frictional force is always opposite to the applied force that's why has negative sign and the sine component of the weight is positive because it has the same direction with the applied force
@@suzuki12334 The easiest way to decide what needs to be sine and cosine, is to inspect the limiting cases. When theta = 0, the sine = 0, and the cosine = 1. Vice-versa happens, when theta = 90 degrees. With your intuition, think about what will realistically happen when the angle is zero, and when the angle is 90 degrees. If the force in question should also be zero when theta =0, then it is most likely going to be sine. If the force in question should be its full amount when theta=0, then it should be cosine. You might have a tangent instead of sine, so also check the other limit. See if it should be cosine, when theta=90 degrees, to conclusively determine that it should be sine.
This model works very well for hard surfaces. I wonder how the contact area and other factors start to be important when analyzing friction in deformable surfaces like rubber or human tissue? Would be interesting to know :)
I think he has used calculator. How can someone know the values of sin 35 and cos 35 and use their decimal values in an equation? The equation becomes long and we have multiply big numbers. I think giving angles like 30,45 or 60 is good because their values are easy to work with at the time of calculation.
True and false - 1. The limiting force of static friction depends on area and contact and independent of material 2. The limiting force of kinetic friction depends upon area of contact and independent of material 3. The limiting force of kinetic friction is independent of area of contact and it depends on material 4. The limiting force of static friction is independent of area and contact and depends on material. 5. Both limiting friction and static friction depends on area of contact and nature of material
Professor Dave Explains the Induced magnetic field in a coil is a non conservative field. Still work is stored in the form of magnetic potential energy. Is it the same thing? Thanks :)
It is very clear that, any conservative force must be accompanied with a kind of potential energy. But Nonconservative forces are not. By the way, conservative force is defined by the equation of F = dU/dx. Where dU is the differential of potential energy. Work done against a conservative force will result an increase in the kind of potential energy and vise versa. Work done by a nonconservative force will definitely waste the energy which is not reversible as dU. Thank you Sir.
To be fair, mass is not required to solve the check comprehension problem. Net force equals Fnet = mgsin(α) - μmgcos(α) = mg(sin(α)-μcos(α)). Now a = Fnet/m therefore masses cancel out and we have final formula a = g(sin(α)-μcos(α)).
Prof Dave. At :58 s. This is not correct. You state that "every surface has different coefficient." No single surface has a coefficient of friction. Mu is a comparison between two surfaces, never just one. One surface, alone, does not have friction.
Hello. I'm just confuse because I read the book named "Conceptual Physics" that says that Increasing contact points doesn't add to the friction. Can someone clarify thanks
I'm not positive but I think it has more to do with the force of gravity on the contact points than the points themselves. If I have two boxes of the same size on the ground (therefore both boxes have equal contact points with the ground) and one weighs more, the normal force will be greater, and therefore the frictional force will also be greater.
It’s Newton‘s 3rd law, every action has an equal reaction. However, every surface has a limit on how much force it can offer an equal reaction to. Once that is exceeded, the object moves. Hope that clears it up :)
Thank you, Prof Dave! One quick question, As you mentioned Fnet = Ff +mg sinθ But in the last example The solution is Fnet = mg sin 35° - Ff I want to know why it is negative (or you subtract Ff in stead of add) Is it because mg sin 35° and Ff is in opposite direction so that you make it negative? Thank you very much for your effort and I am looking forward to complete the whole series of 43 videos!!:D
As you know, F=MA And the block travels downwards with acceleration "a". Frictional force always opposes the motion of a surface over another surface. So as the frictional force increases, the acceleration of the block decreases. Therefore, the friction force will be negative because it affects the acceleration. And when acceleration is affected, the force will also be affected. This means Fnet = mg sin 35° - Ff.
Anyone else confused about the equations at the end? At 6:35 it says Fnet=Ff + mg sin(theta) . . . yet at 6:56, it said Fnet= mg sin(theta) - Ff So which is it?! Do I add or subtract the frictional force?!
@@barakatalhinai3541 physics is definitely a jojo reference. think about Pucci’s stand power to change the physics of gravity 😮 and thanks for the answer!
Sir why are we subtracting friction forces from mgsin35 to have friction net force instead of adding to satisfy the static condition of forces in the x- axis
I have a question: Why is static friction used to walk forward? I know our foot pushes backwards, and the friction lets us go forward, but i thought that static friction only resisted the initiation of motion. Is there an another definition for static friction or am i just not understanding this concept correctly?
It is only locally that static friction opposes the initiation of motion. Locally at the point of contact between two objects. The static friction in this example, is what stops your foot from moving backwards as you push it backwards while trying to walk. If you were on a truly frictionless surface, your foot would slide backwards and you would run in place if you tried to walk. Think of the Earth's surface a "work mirror". You push backward on the ground, by pushing your foot backward while it is in contact with the ground. By Newton's third law, the ground also pushes forward on your foot with an equal and opposite force. The force on the ground by your foot, cannot practically do any measurable amount of work on the Earth, because of the Earth's inertia is so large that its motion due to this force is immeasurably small. Instead what happens, is that the ground reflects this force back to you, through its third law pair of this force, as the ground applies a forward force to your foot, while your body moves.
yes atomic is the root of the constact known as friction. its all part to understand as a whole like an ecosystem. so u can build with perfect natural law in harmony and then we can multiply this to sustain energy. but all this is to be equated in order by function amd presence naturally according to purpose relative to said application. its constant. like anything u cant have just peices and parts but as a whole .
mg sin 35° = 20 (9.8) sin 35° Here value of sin 35 is irrational. He has taken it till 4 to 5 decimal places I think which is 0.5735. So we have 20 (9.8)(0.573)= 112.4(approx) Actually the problem with this question is that how we can find the values of trigonometric ratios of angles like 35°.
this guy carried me through sophomore year chemistry and now he's carrying me through junior year physics
Same
so you read Tintin books?
@@floydwynn-jones4043 ii
@@floydwynn-jones4043 i
@@floydwynn-jones4043 i
This is like a crash course but nothing about it is too clumped. Perfect, straight-to-the-point, and easy. Will subscribe and listen to you religiously throughout my physics course. Thank you.
ua-cam.com/video/DE6QvdAO9Qw/v-deo.html
Reason of why static friction is more than kinetic or dynamic friction
I did a short experiment. I watched this video 2 times: yesterday night and tonight.
Yesterday, I understood everything he said but couldn't get anything inside my head. So I used the whole day today to read the book and solved many maths related to frictions.
Tonight as I watched this video again I was able to understand everything loud and clear😁. Thanks professor Dave
Professor Dave, your videos got me through physics 1 and 2 in my post-bacc, and as I'm studying for the MCAT I found myself needing additional explanation. The jingle came to my mind and I knew exactly where to go. So thankful for you and your videos!
I think of the formula for heat in terms of mass, temperature and specific heat, when I think of MCAT. Q=m*c*∆T
the jingle lol
Is it sad or impressive Dave can teach in 5 minutes what my professor can’t in 3 3 hour class periods?
Both. Impressive for Dave but sad for your professor and you guys. I'm in the same boat with my professor. His lectures are really disorganized. I was getting a little scared that I was getting confused so early in the semester, especially when I'm way ahead of the math requirements. Watching just a few of Dave's videos and I feel much more confident. I've learned more from from short UA-cam videos than from 2 hour lectures.
YESSSSSSSSSSSSSSSSSSSSS
dude thats literally what happened to me in class. 3 hour class, dont understand much. i watch this, and it makes a lot of sense.
@Learn First very true, they want you to know how to do problems and smart students will read about it more at home
Am i the only kid watching this...
Unfortunately, my prof doesn't teach anything... so I am very thankful for your videos with visual aids! I hope you keep making more!!! Yours are by far the best on youtube! :)
You can't listen, that's why
@@lemondrop8203someone’s never experienced a bad teacher
thanks a lot bro. this new subscriber is from the Philippines, the way you deliver the topic is so smooth, wonderful and easy to understand. thanks a lot. static & kinetic friction topic. great!
I swear without this video I would be super lost in my physics class
This Guy explained these concepts in just 8 minutes which I am Trying to understand from 2 weeks
7:05 aren't the x & y components mg*csc(theta) and mg*sec(theta) respectively? The trig here isn't checking out for me.
? to denote x-component of mg
sin(theta) = mg/?
?*sin(theta) = mg
? = mg/sin(theta)
? = mg*csc(theta)
thank you so much! you saved a university newbie!
Thanks,dave. At this current time, i have a hard time learning about frictional force in orthodontic. And ur explanation makes me a little bit confidence. I understand the first 5:30 minutes,the rest remains a question for me. Thx
GOOD EEVENING PROFFESOR DAVE at checking comprehension The formula Fnet=Ff+mgsin change into Fnet=mgsin-Ff 6:34 / 7:36
I had the same doubt too.
thank you so much for sharing your knowledge with the world :D! now it all makes sense :)!!
1:31 "This attractive interaction can hinder motion to some degree" This can easily be taken out of context.
💀
thank you man, i am teaching my nephew vectors and this video helps a lot to illustrate parts of the whole idea. thank you.
ua-cam.com/video/DE6QvdAO9Qw/v-deo.html
Reason of why static friction is more than kinetic or dynamic friction
6:54 I wonder if the Fnet=mgcos(theta)-Fsf resulted in a negative number (meaning the Fsf is greater than the mgcos), then what will happen? Will the box stay at rest or will the box move upwards? Depending on the context though, i'd guess the box would remain at rest.
As said in the video, any applied force less than the maximum static frictional force will cancel out resulting in no overall movement of the object even if frictional force is much larger than applied force because it is simply meant to maintain the motion of the object (or lack thereof).
Hello! In the comprehension check you use Fnet= mg sin theta - Ff, I am confused as to why you are using subtraction instead of addition.
Thanks!
it's true, it could be addition as long as you then make the frictional force negative, we just need opposite signs on the two horizontal forces as they are in opposite directions. in the end, as long as the signs make sense you're all good!
There is no certain direction for a frictional force to be acted. It is always resistive to a motion or to an attempt to the motion. When the block moves or tends to move down, frictional force would act up the incline. When the block moves or tends to move up, the frictional force would act down the incline. Thanks.
Is it safe to assume then, that in problems like these as long as the box is going downwards the frictional force will be negative ?
KB yes. It’s in the opposite direction of the motion so it’ll be negative.
Thank you, You’ve helped me understand the difference between the frictions.
Subbed not only because useful information but because Subaru 😎
Lots of new ideas thank you . It's relevant in my study
This is my first video of yours and I subscribed as soon as I saw it at a glance 😊
NOW THIS IS CLEAR THANKS PROFESSOR
Your videos are so helpful, thank you so much!
We could see how much effort you put into creating a video.
Super.👍
We A4Q team with your growth.
Full support 💪
Professor Dave, I have a question! If friction is caused by forces of attraction between atoms on a molecular, and electrostatic attraction is one of of these forces of attraction. Does this mean that friction can change based on the charge of the object?
Is it specified that the attraction is really electrostatic?
I was in tension abt tomorrow's exam but your intro made me free and calm
Professor please make a video on tension it is an important topic please make it I love your teachings
Professor Dave explains ❤️❤️❤️🥰👍🏻 love this
Wow..ur teaching is superb with practical learning
Bruh, how does this guy manage to learn something that is hard to explain in such ease and how long did it took for him to learn this stuff?
May I have a question. At 1:27 you said the interaction between atoms in the object and atoms in the surface is an attractive interaction. But shouldn't it be repulsive instead? Because the interaction here is between electrons of the object's atoms and electrons of the surface's atoms, both are negatively charged, so they should repel each other instead of attract, right?
And if they repel each other, logically this interaction should help facilitate the motion by pushing the two surfaces away from each other. But in fact, motion is hindered. Could you explain how it works?
Atoms that are stable have no net charge.
The atom consists of protons as well which cancels the effect of negatively charged electrons. In an atom (which is not an ion) the number of protons and electrons are equal and both of them have the same charge.(i.e charge in a single proton= charge in a single electron) . ( This is 100% true)
But the thing is that, even if two charged bodies are interacting still friction will act upon them. Considering that the bodies are big(big enough that we can see them). This is classical physics. Here we deal with objects that are not very small. (Although I am not 100% sure about this one)
thank you dave my physics teacher didn't explain in this way as you explained . thank you very much
exactly what i needed before my test! thanks!
Thank you man, you nailed it. My teacher can’t even explain it clearly 🙄
The best intro ever seen😄👍🏻
Thank you, Professor Dave
Thank you so much, I was stuck on a question and this video helped me a lot!
ua-cam.com/video/DE6QvdAO9Qw/v-deo.html
Reason of why static friction is more than kinetic or dynamic friction
Thanks for helping Dave
Amazing explanation, clear words, but a small request. If you could explain the numericals, it would be more helpful. Thank you.
Oh sir...
Thanks for saving me ... Actually i love physics but the concept of cofficent "i was unable to understand 8t properly but after you video .. everythings is clear...
Thanks sir! !!!!
Coefficient just means a number multiplied by another number. Like in y = A*x^2 + B*x + C, the A and B are examples of coefficients.
Thanks for explaining i was confused now i understand these topic it helps me to solve jee questions
i think there is an error in checking comprehension. if the object is at rest, we wont know whether it moves unless we look at the static friction coefficient (as opposed to kinetic friction as stated)
6:30
Fnet = Ff - mg*Sin(ceta)
sorry, i'm confused - do we add parallel vectors or minus them from each other? Dave's equation is Fnet = Ff + mg*sin(θ)???
@@Mr3mosThe friction is minus because it's a vector quantity and remember that the friction is always opposite. If we define the mg*sin(35)'s direction as positive
Basically dave is not wrong either you can add it but since the motion is acting in different direction the other one would act as negative and the answer will be net force going to the direction of the larger one👌👌
What Dave did is correct. The frictional force is always opposite to the applied force that's why has negative sign and the sine component of the weight is positive because it has the same direction with the applied force
Concepts are simply built...Thanks Sir..
thank you so much sir, your excellent explanation helped me a lot today
Professor Dave Explains , The Classic Vintage Intro.
Professor Dave is my life saver . While my physics professor destroy it😆
How do we decide whats cos and sin at 6:00
@@suzuki12334 at my channel looke
@@newtube553 ua-cam.com/video/TC23wD34C7k/v-deo.html heres what i thought but ty
@@suzuki12334 The easiest way to decide what needs to be sine and cosine, is to inspect the limiting cases. When theta = 0, the sine = 0, and the cosine = 1. Vice-versa happens, when theta = 90 degrees. With your intuition, think about what will realistically happen when the angle is zero, and when the angle is 90 degrees. If the force in question should also be zero when theta =0, then it is most likely going to be sine. If the force in question should be its full amount when theta=0, then it should be cosine.
You might have a tangent instead of sine, so also check the other limit. See if it should be cosine, when theta=90 degrees, to conclusively determine that it should be sine.
thank you i really understand friction so easy
I fucking love the intro its catchy as fuck😂😂🎉
7:05
No, THANK YOU!!!!!!!
Keanu reeves
If you were not on UA-cam I’d be having a lot more trouble with physics 😅
highly recommended☺
could u explain more about the process of coldwelding that also catalyze static friction?
Very precise
Super! You made the subject simple! Thanks
thank you for explaining clearly
This model works very well for hard surfaces. I wonder how the contact area and other factors start to be important when analyzing friction in deformable surfaces like rubber or human tissue? Would be interesting to know :)
I love you and your channel so much dude
HOLY SHIT you just taught 45 mins in 7 mins and I actually understood everything
You're the best
That fact that I understand friction more by watching him than by listening to my teachers lecture
How did you get so good at solving, what did you do when you were a student... I want tipsssss as many as the s' s i put
I think he has used calculator. How can someone know the values of sin 35 and cos 35 and use their decimal values in an equation? The equation becomes long and we have multiply big numbers. I think giving angles like 30,45 or 60 is good because their values are easy to work with at the time of calculation.
But Sir how we will know what is the value of cos 35° and sin 35°?
Hey man you are the legend of science...
I love that intro song..
Thank you for illustrating perfectly
Thank you. It helped a lot .
@professor_dave_explains do you still stands by the statement that rougher surfaces have more friction?
True and false -
1. The limiting force of static friction depends on area and contact and independent of material
2. The limiting force of kinetic friction depends upon area of contact and independent of material
3. The limiting force of kinetic friction is independent of area of contact and it depends on material
4. The limiting force of static friction is independent of area and contact and depends on material.
5. Both limiting friction and static friction depends on area of contact and nature of material
Can work done against non conservative forces get stored in the form of potential energy?
oh man, good question. i can't think of how that would work but i'm sure there must be some example.
Professor Dave Explains the Induced magnetic field in a coil is a non conservative field. Still work is stored in the form of magnetic potential energy. Is it the same thing? Thanks :)
Can't. That work will be wasted mostly as heat. This is what the meaning of "nonconservative"
It is very clear that, any conservative force must be accompanied with a kind of potential energy. But Nonconservative forces are not. By the way, conservative force is defined by the equation of F = dU/dx. Where dU is the differential of potential energy. Work done against a conservative force will result an increase in the kind of potential energy and vise versa. Work done by a nonconservative force will definitely waste the energy which is not reversible as dU. Thank you Sir.
Excuse me. At 6:30 is there a mistake in the formula?
force is a vector so it has magnitude and direction ( net force could be (negative mg sin cita) + f friction or mg sin cita - f friction )
If I wind a rope around a spool body, how many winds do I need to get a frictional grab where it can not slip? Thanks
6:58 I almost got it right tho, I just forgot to calculate the sin of Its Normal Force
which I didn't understand before the answer was shown.
To be fair, mass is not required to solve the check comprehension problem. Net force equals Fnet = mgsin(α) - μmgcos(α) = mg(sin(α)-μcos(α)). Now a = Fnet/m therefore masses cancel out and we have final formula a = g(sin(α)-μcos(α)).
COOL SUBARU CLIP! really piqued my interest. are you a subaru driver, dave?????
Prof Dave. At :58 s. This is not correct. You state that "every surface has different coefficient." No single surface has a coefficient of friction. Mu is a comparison between two surfaces, never just one. One surface, alone, does not have friction.
Hello. I'm just confuse because I read the book named "Conceptual Physics" that says that Increasing contact points doesn't add to the friction. Can someone clarify thanks
I'm not positive but I think it has more to do with the force of gravity on the contact points than the points themselves. If I have two boxes of the same size on the ground (therefore both boxes have equal contact points with the ground) and one weighs more, the normal force will be greater, and therefore the frictional force will also be greater.
Brother your intro very nice😂❤❤
2:44 I was unable to understand this point. Can someone explain me, please?
It’s Newton‘s 3rd law, every action has an equal reaction. However, every surface has a limit on how much force it can offer an equal reaction to. Once that is exceeded, the object moves. Hope that clears it up :)
@@Silberstern-bd9jx Thanks!! I somehow cleared my point sometime ago cz I had an exam. But THANKS for your kind help!! 😭🖤
Thank you, Prof Dave!
One quick question,
As you mentioned Fnet = Ff +mg sinθ
But in the last example
The solution is Fnet = mg sin 35° - Ff
I want to know why it is negative (or you subtract Ff in stead of add)
Is it because mg sin 35° and Ff is in opposite direction so that you make it negative?
Thank you very much for your effort and I am looking forward to complete the whole series of 43 videos!!:D
As you know, F=MA
And the block travels downwards with acceleration "a".
Frictional force always opposes the motion of a surface over another surface.
So as the frictional force increases, the acceleration of the block decreases.
Therefore, the friction force will be negative because it affects the acceleration.
And when acceleration is affected, the force will also be affected.
This means Fnet = mg sin 35° - Ff.
@@maynur1 Thank you so much for your explanation!! Now I understand more about it!:DDD
Can anyone explain why add? 6:27
now I understood the difference between static and kinetic. whoa!!
Anyone from India 🇮🇳
Me
Hi bro
Yo! I'm
Hi bro
Anyone else confused about the equations at the end? At 6:35 it says Fnet=Ff + mg sin(theta) . . .
yet at 6:56, it said Fnet= mg sin(theta) - Ff
So which is it?! Do I add or subtract the frictional force?!
Bucciarati is physics a jojo reference?? I’d assume you’d subtract since friction goes the opposite direction of applied force
@@barakatalhinai3541 physics is definitely a jojo reference. think about Pucci’s stand power to change the physics of gravity 😮 and thanks for the answer!
most welcome Dave
Sir why are we subtracting friction forces from mgsin35 to have friction net force instead of adding to satisfy the static condition of forces in the x- axis
Friction is in another direction it must be taken away
Wow thanks alot ,it was really well explained
Thankyou Professor Dave!
Great explanation - thanks
I have a question: Why is static friction used to walk forward? I know our foot pushes backwards, and the friction lets us go forward, but i thought that static friction only resisted the initiation of motion. Is there an another definition for static friction or am i just not understanding this concept correctly?
It is only locally that static friction opposes the initiation of motion. Locally at the point of contact between two objects. The static friction in this example, is what stops your foot from moving backwards as you push it backwards while trying to walk. If you were on a truly frictionless surface, your foot would slide backwards and you would run in place if you tried to walk.
Think of the Earth's surface a "work mirror". You push backward on the ground, by pushing your foot backward while it is in contact with the ground. By Newton's third law, the ground also pushes forward on your foot with an equal and opposite force. The force on the ground by your foot, cannot practically do any measurable amount of work on the Earth, because of the Earth's inertia is so large that its motion due to this force is immeasurably small. Instead what happens, is that the ground reflects this force back to you, through its third law pair of this force, as the ground applies a forward force to your foot, while your body moves.
which is correct, is it @6.31 : net force = kinetic friction + mg sin theta or @6.53 : net force = mg sin theta - kinetic friction
confused help
yes atomic is the root of the constact known as friction. its all part to understand as a whole like an ecosystem. so u can build with perfect natural law in harmony and then we can multiply this to sustain energy. but all this is to be equated in order by function amd presence naturally according to purpose relative to said application. its constant. like anything u cant have just peices and parts but as a whole .
u are really good teacher I was having a test and u saved me tnx to GOD
Someone can help explain where does 112.4 N comes?
mg sin 35° = 20 (9.8) sin 35°
Here value of sin 35 is irrational. He has taken it till 4 to 5 decimal places I think which is 0.5735. So we have
20 (9.8)(0.573)= 112.4(approx)
Actually the problem with this question is that how we can find the values of trigonometric ratios of angles like 35°.
But why does static friction push back exactly as much as I push the object? And why is there a maximum? Why is dynamic friction always weaker?
Nice explained.
I don't think I was taught half this in school.
Did you record the first seven mins on one take? I didn't notice any frame splices... ( or was I fooled with great editing?) 🙂
Editing skills!
Helpful :) thank you.
ya really
I gave my life for Physics