Suppose the masses are equal for the two blocks and that we call the mass M. Lets also suppose that the friction between the two blocks and the friction between the floor and the bottom block is the same and lets call that mu. Is it fair to then say that the acceleration of the bottom block is a=((mu*3Mg)-F)/M
Thank you for this great video! Here is a question: you use one value mu for coefficient of friction. If we made a distinction between static friction and kinetic friction, would it be correct to use the value of mu static to first find all the friction forces (assuming that m1 is just about to begin sliding), but then to use mu kinetic in the expression for the acceleration of m1?
If you are not sure if the system will overcome the static friction and begin to move, then yes, you need to calculate the forces using the static coefficient of friction. However if it is assumed that the system will accelerate, then you only need to use the kinetic coefficient of friction.
for the final answer, why isn't the force aiding equivalent to F3 (mass 2)? isn't that the force that opposes the other two forces motion? or is it because we are only talking about m1 in this problem?
I'm confused by 6:28, when determining the direction of F f3. Relative to the A block the B block is moving to the right, and as friction opposes the direction of motion F f3 should be moving to the left not to the right? Clarification would be appreciated, thanks
F fr3 is equal in magnitude to the tension in the string that keeps m2 from moving. If F fr3 were directed to the left, there would be no tension in the string and block m2 would be pushed to the left. Therefore F fr 3 is directed to the right.
Michel van Biezen Hm I got it. By the way I really love you. You're a great lecturer. You're the only person who help me to understand the hardest topic in physics. I am really grateful to you. I wish I had such a lecturer. Thanks for all the videos.
@@MichelvanBiezen Actually I missed my coaching class and was finding it difficult to understand concept ..Then I came across you and it clarified Thank you 😀😀
2:48 Excuse me sir, but how we assume that there’s no friction, then we find the friction, I got a little confused. May you explain this to me sir? Thank you so much 😊
In order to find the direction of the friction force, you ask yourself the question: ""which way would the object move if there was no friction?". Once you answer that, the direction of the friction will be in the opposite direction.
Hello, I'm study for an exam on the 10th knowing my lecturer he will be using tension, could you possibly give me a brief explanation of how to take tension of m1 and m2 into account please?? Would the tension2 on m2 be equal and opposite of mg and m1 tension1 be equal and opposite the tension2 and during the calculations would we add the tension 1 to friction depending which way the block is accelerated if at all, also... Would you use tension 2 as an aid or opposition and this depending on direction deduct tension2 from tension1 or would you need to go the equation route using the correct signs and find the aiding and opposing forces that way?? Thank you
F can be any value as long as it is greater than the two friction forced combined. The acceleration is given in terms of F. (F is not given and cannot be found, so you pick any value for F and then you can find the acceleration)
@@MichelvanBiezen I think T = mue1 × ( m1 + m2 ) g And f1 = mue 1 (m1 + m2 ) g And f2 = mue 2 m2 g Be cause I have done other similar/ same question where there was 3 block system and the upper block is tied to a string .
This is a great problem. It really touches a lot of the basic mechanics concepts in one nice package.
But for Indians it is very simple. Because I think level of my country questions is high. This problem is simplest one
It will become a little bit more difficult if we remove that rope
Your an angel😭😭😭😭😭❤️❤️❤️
This video saved me, thank you!
Continuously I've been watching the friction playlists from video no-01 to 11
& my boringness is 0%.
I'm really of ur fan sir♥️♥️
We are glad you like them.
Fr3 should cancel out Fr2 since it is equal in magnitude but opposite in direction and since it is in the direction of the applied force
Suppose the masses are equal for the two blocks and that we call the mass M. Lets also suppose that the friction between the two blocks and the friction between the floor and the bottom block is the same and lets call that mu. Is it fair to then say that the acceleration of the bottom block is a=((mu*3Mg)-F)/M
Take the final answer, replace m1 with m, replace m2 with m, replace mu1 with mu and mu2 with mu and you will get the answer you are looking for.
Prof...my prof you are the best!!!!
Thank you
Thank you soo much! You have no idea how your videos are helping me😭❤❤ perfect explanations
Glad the videos are helping.
m2 cannot move, therefore only m1 can be accelerated by F
I am going to ask you this qsn, but you had already answered. Thanks!
The dog also wants to lecture :-} thank you for the video
Thank you for this great video! Here is a question: you use one value mu for coefficient of friction. If we made a distinction between static friction and kinetic friction, would it be correct to use the value of mu static to first find all the friction forces (assuming that m1 is just about to begin sliding), but then to use mu kinetic in the expression for the acceleration of m1?
If you are not sure if the system will overcome the static friction and begin to move, then yes, you need to calculate the forces using the static coefficient of friction. However if it is assumed that the system will accelerate, then you only need to use the kinetic coefficient of friction.
Or have you divided for only m1 because there is a rope (=T)?
thanks for helping me out 😊
God bless you!
Amazing
Thank you!
for the final answer, why isn't the force aiding equivalent to F3 (mass 2)? isn't that the force that opposes the other two forces motion? or is it because we are only talking about m1 in this problem?
The net force on m1 equals F minus the friction force on the bottom and minus the friction force at the top of the block.
If m1 above m2 then any force exerts in m1 and make it accelrate by a then The force in m2 will be m2 a
Right ?
If you switch the positions of m1 and m2 you would also switch m1 and m2 in all the equations shown.
It was great. thank you sooo much
I'm confused by 6:28, when determining the direction of F f3. Relative to the A block the B block is moving to the right, and as friction opposes the direction of motion F f3 should be moving to the left not to the right? Clarification would be appreciated, thanks
F fr3 is equal in magnitude to the tension in the string that keeps m2 from moving. If F fr3 were directed to the left, there would be no tension in the string and block m2 would be pushed to the left. Therefore F fr 3 is directed to the right.
Thank you for amazing videos sir
You’re a blessing
you are the definition of a king
No, not a king. Just a simple man with lots of faults, (just ask my wife). :)
simp
thanks sir.
You are welcome.
“…and that’s how it’s done” *drops mic
We do it in "style" :)
I couldn't understand why we didn't take total mass in the denominator when we calculate acceleration.?
Can you explain please?
m2 cannot move, therefore only m1 can be accelerated by F
Michel van Biezen Hm I got it. By the way I really love you. You're a great lecturer. You're the only person who help me to understand the hardest topic in physics.
I am really grateful to you.
I wish I had such a lecturer.
Thanks for all the videos.
@@MichelvanBiezen even if m2 cannot move, wouldn't its weight play a role in trying to move m1?
To find a, you have divided for m1. But must not you divided for m1+m2?
Which object is accelerating?
The dog was wandering why the friction on m2 was not an aiding force to the moving box m1?
The direction of the friction force is opposite to the direction of motion of the object if there was no friction.
The concept of considering N1=[m1g+m2g] is clear but why we haven't considered the gravitational force of block 1 as (m1+m2)g ???
You have to take each block separately. The weight of block 1 is m1g and the weight of block 2 is m2g. The normal force pushes back against both,
Thank you sir
Really helpful
Glad it helped
@@MichelvanBiezen Actually I missed my coaching class and was finding it difficult to understand concept ..Then I came across you and it clarified
Thank you 😀😀
thanks
Without the tension, is the maximum force that can be applied to mass1 given byl to (m1+m2)ug?
That is correct.
is force friction 2 & 3, newton's third law pair ?
You can think of it like that.
2:48
Excuse me sir, but how we assume that there’s no friction, then we find the friction, I got a little confused.
May you explain this to me sir?
Thank you so much 😊
In order to find the direction of the friction force, you ask yourself the question: ""which way would the object move if there was no friction?". Once you answer that, the direction of the friction will be in the opposite direction.
@@MichelvanBiezen
Thank you so much sir.
Perfect 🤩
Glad you liked it. 🙂
Nice
Glad it was helpful.
Hello, I'm study for an exam on the 10th knowing my lecturer he will be using tension, could you possibly give me a brief explanation of how to take tension of m1 and m2 into account please?? Would the tension2 on m2 be equal and opposite of mg and m1 tension1 be equal and opposite the tension2 and during the calculations would we add the tension 1 to friction depending which way the block is accelerated if at all, also... Would you use tension 2 as an aid or opposition and this depending on direction deduct tension2 from tension1 or would you need to go the equation route using the correct signs and find the aiding and opposing forces that way?? Thank you
There are a number of videos on tension in this playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS
thank you very much :)
Nice job man
Thanks!
The direction of f3 should be towards the left side, right?
No, Ffr 3 acts to the right as shown in the video.
Why there is f3?
It is the friction force acting on m2 caused by N2
That dog lol
thanks from india for jee
Welcome to the channel!
How to find F? That's my question😭😭
F can be any value as long as it is greater than the two friction forced combined. The acceleration is given in terms of F. (F is not given and cannot be found, so you pick any value for F and then you can find the acceleration)
@@MichelvanBiezen okay thank you so much sir
i wish its was turkish :(
I've never seen physics done so wrong...
What made you think like that?
Sorry I think this is wrong
The video is correct. Thanks for checking.
@@MichelvanBiezen I think T = mue1 × ( m1 + m2 ) g
And f1 = mue 1 (m1 + m2 ) g
And f2 = mue 2 m2 g
Be cause I have done other similar/ same question where there was 3 block system and the upper block is tied to a string .
I think there will be no f3