hi Michel. why can't we get the maximum magnitude of the friction force in both masses then add them and make the maximum force that will not make them slipping equals or greater than them. but when i try it it gave me different value. thanks :)
One simple question, the coeficient of friction 2 is static or dynamic? I assume is static and thats is why this problem is tricky. Normally with a static coefficient We tend to consider that there is not acceleration and here there is due to the movement of the down block. Another thing that confused me is the fact that the heavy block is the upper one which is smaller in size in the drawing, I couldn´t solve the problem until I realize that. Now I think I got it, unless you tell me that the coeficient 2 is dynamic, which would drive me crazy. Anyway, your work is great and you are one of the few online teachers who have exercises at a university level (Khan academy is great but the level of its exercises is lower) Keep up the great job and thanks from Spain!
It is usually figured out by context. The problem states that there is no slipping between m1 and m2. Thus by default u2 is static. The 2 blocks accelerate, thus u1 must be kinetic.
Hi Michel. Are you sure the free body diagrams of the normal force are accurate? I think there is an additional force, because if we were to cancel out the forces, we should really be at F_net_y = 0, as the system would not be accelerating down. Please let me know. Thank you very much.
Yes, they are correct. Keep in mind that only Ffr1 has an affect on the acceleration. Ffr2 and Ffr3 only ensure that the top block does not slip relative to the bottom block.
Hi Michel, I have a similar problem to solve except the bottom block is being pulled up an 10 degree incline. How can I incorporate this into my problem?
When looking at the friction between 2 moving blocks where one is on top of the other, you must determine how the friction affects EACH block separately. Since the blocks move in opposite directions, the associated friction forces will be in opposite directions.
So I was wondering if there was a force applied to the bottom block to the right and friction between the two blocks...was the second friction still gonna point to the right?
I should have looked at it closer. You are correct that they are moving in the same direction in this example. (I was thinking of another example). But the principle holds the same. The friction between the 2 blocks acts in opposite directions relative to each block.
The friction between the 2 blocks: For the top block the friction acts to the left (opposes the force pulling the top block). For the bottom block the friction acts to the right, since it is the only force that can pull the bottom block to the right.
since we want to calculate Fmax that can be done on m2 without causing the two masses to seprate ,so can we say that Fmax is equal or less than the fractional force F3. Because when F max is bigger than( F3 ) ,the m2 starts to accelerate and will separate from m1 . Thank you A lot .
Michel van Biezen ok but the magnitude of f3 is (29.4) fo F max should be equal or less than f3 or (29.4 ) not as solved in vedio (52.3) Thank you so much .
yes. a = 4.57 m/sec^2 is the acceleration of m1. which means that if F is greater than 29.4 N, m2 will slide relative to m2. And the force required to acceleration both masses at 4.57m/sec^2 would be 52.3 N.
What if the coefficient of friction is zero on the ground plane? Would it mean there is no force? Or do you just disregard multiplying by the coefficient?
@@MichelvanBiezen Cool thanks! Makes sense it would have zero frictional force as a result for that part of the equation. By the way your video helped break down the free body diagram in a much simpler way than I've seen on MIT and made it a lot easier for me to understand. Thanks so much Michel!
If the force on the upper box is increased such that the upper box begins to move relative to the lower box, then the coefficient between the two boxes would decrease (kinetic friction vs static friction) and the upper box would accelerate at a hight rate than the lower box.
we found the maximum force 52.3 N but the maximum friction force between these two block equal (5*9.8*0.6=29.4 N) then if we apply the maximum 52.3N force that larger than the friction force between these two blocks then how they will remain together?
what happens when F=0.000001 ? Acording to your formula Ffr = μ*m*g, the body would have to start moving in the direction of that friction force, since the only force opposing it would be that of magnitude 0.000001 and that's negligible ....
Hi Michel I am having some issues with dynamics questions involving incline plane and pulley problems, mass A sliding down plane pulling mass B up the vertical side by pulley system with mechanical advantage 2:1.When checking solutions my accelerations are off because they select down as positive for mass B making the 2T in cable negative even though it aids its acceleration upwards. Is this an error thank you.
Michel van Biezen Thank you for reply, I think I can see what is going on it is a problem similar to ones you have done except there is a mechanical advantage taken place with the pulleys on the inclined plane. There is a kinematic equation for rope length 0=2aA+3aB, its a 2:3 mechanical advantage set up. The coordinate system used for acceleration is not intuitive in that the chosen positive direction for each mass is opposite, however this seems to correspond to rope equation -2aA=3aB in that if mass A is moving negative direction B is increasing length and vice versa.The question is from Kraige 6th dynamics 3/25. I appreciate this might be asking a bit much in these conditions, however thanks for informative videos.
The key to understanding the mechanical advantage with pulleys is that if you have for example a 2:1 advantage, the force required to pull the object is only half its weight (when hanging), but the distance you must pull is twice the distance the object moves.
It is a dragging force originally caused by F(max) but its acting indirectly thru M2 pulling against its binding friction with M1. Its simply Fmax conducted thru M2 & when it arrives at the contact surface it acts on M1 in the direction of Fmax. If that frictional force didnt exist, M1 would not be dragged along with M2.
It depends on how hard you pull. I you pull gently, the top block would move with the bottom block without slipping. If you pull hard enough the top block would slip.
The force calculated equals the friction force between m1 and m2 PLUS the force required to accelerate m2. Note that if you calculate both forces they add up to the 52.3 N.
Michel van Biezen professor . Can't we find Fmax by making it equal to friction force 3 which is occur between m1 and m2? Because in other video you find Fmax like that . Here is the video. ruclips.net/video/3Dn1Ze8maEc/видео.html But when I tried, it didn't give the same result :( so what is the problem?
glad you got that audio problem fixed.. your videos are very helpful. keep up the great work sir
hi Michel. why can't we get the maximum magnitude of the friction force in both masses then add them and make the maximum force that will not make them slipping equals or greater than them. but when i try it it gave me different value. thanks :)
One simple question, the coeficient of friction 2 is static or dynamic? I assume is static and thats is why this problem is tricky. Normally with a static coefficient We tend to consider that there is not acceleration and here there is due to the movement of the down block. Another thing that confused me is the fact that the heavy block is the upper one which is smaller in size in the drawing, I couldn´t solve the problem until I realize that. Now I think I got it, unless you tell me that the coeficient 2 is dynamic, which would drive me crazy. Anyway, your work is great and you are one of the few online teachers who have exercises at a university level (Khan academy is great but the level of its exercises is lower) Keep up the great job and thanks from Spain!
It is usually figured out by context. The problem states that there is no slipping between m1 and m2. Thus by default u2 is static. The 2 blocks accelerate, thus u1 must be kinetic.
Difficult and tricky video :D...Thank you so much sir. I am tired I will give it a new try hopefully soon. Blessing to you and your family.
Thank you.
espceially the second friction force, very confusing. i still dont get it how...
Hi Michel. Are you sure the free body diagrams of the normal force are accurate? I think there is an additional force, because if we were to cancel out the forces, we should really be at F_net_y = 0, as the system would not be accelerating down. Please let me know. Thank you very much.
Yes, they are correct. Keep in mind that only Ffr1 has an affect on the acceleration. Ffr2 and Ffr3 only ensure that the top block does not slip relative to the bottom block.
Hi Michel, I have a similar problem to solve except the bottom block is being pulled up an 10 degree incline. How can I incorporate this into my problem?
Take a look at the incline problems. That should give you a good idea how to go about it.
Why don't we consider the internal frictional forces whilst calculating the accn?
We considered all the friction forces. Which one are you referring to?
A greater thanks for Thanks for video.
You are welcome. 🙂
why is the second frictional force on the bottom block not pointing to the left just like we did in the previous activities?
When looking at the friction between 2 moving blocks where one is on top of the other, you must determine how the friction affects EACH block separately. Since the blocks move in opposite directions, the associated friction forces will be in opposite directions.
@@MichelvanBiezen So they are actually moving in opposite directions? I thought they are both moving in to the right?
So I was wondering if there was a force applied to the bottom block to the right and friction between the two blocks...was the second friction still gonna point to the right?
I should have looked at it closer. You are correct that they are moving in the same direction in this example. (I was thinking of another example). But the principle holds the same. The friction between the 2 blocks acts in opposite directions relative to each block.
The friction between the 2 blocks: For the top block the friction acts to the left (opposes the force pulling the top block). For the bottom block the friction acts to the right, since it is the only force that can pull the bottom block to the right.
since we want to calculate Fmax that can be done on m2 without causing the two masses to seprate ,so can we say that Fmax is equal or less than the fractional force F3. Because when F max is bigger than( F3 ) ,the m2 starts to accelerate and will separate from m1 .
Thank you A lot .
Yes, the force (F) cannot be bigger than Ffr3 so that there will not be any slipping between m1 and m2.
Michel van Biezen
ok but the magnitude of f3 is (29.4) fo F max should be equal or less than f3 or (29.4 ) not as solved in vedio (52.3)
Thank you so much .
yes. a = 4.57 m/sec^2 is the acceleration of m1. which means that if F is greater than 29.4 N, m2 will slide relative to m2. And the force required to acceleration both masses at 4.57m/sec^2 would be 52.3 N.
Very nicely explained sir. Thank you
What if the coefficient of friction is zero on the ground plane?
Would it mean there is no force? Or do you just disregard multiplying by the coefficient?
Work out the problem in exactly the same way, just put a zero for that coefficient of friction.
@@MichelvanBiezen Cool thanks! Makes sense it would have zero frictional force as a result for that part of the equation.
By the way your video helped break down the free body diagram in a much simpler way than I've seen on MIT and made it a lot easier for me to understand.
Thanks so much Michel!
Hi,, Prof. What if the upper box is pulled at higher force or vice versa will the acceleration Of the lower box remain moving forward?
If the force on the upper box is increased such that the upper box begins to move relative to the lower box, then the coefficient between the two boxes would decrease (kinetic friction vs static friction) and the upper box would accelerate at a hight rate than the lower box.
we found the maximum force 52.3 N but the maximum friction force between these two block equal (5*9.8*0.6=29.4 N) then if we apply the maximum 52.3N force that larger than the friction force between these two blocks then how they will remain together?
I know I'm 2 years late but this is not a static problem, the block is moving. You can see that there is acceleration. Have a nice day.
@@mihadbinislamtanim6267isnt the acceleration for the bottom block.he is saying they will not slip
Thank you so much! You're literally a life saver
great lesson
Thanks. Glad you liked it. 🙂
sir i think there is a mistake
a = (fk2 - fk1) / m1 + m2
i can explain why
"am i wrong"
The video is correct.
sir,
If we neglect the friction between m1 and the surface
What will be the acceleration of each block?
I tried it and i got a1
@@oussamabaadj9355 I agree with you
Can't thank you enough!
Glad this was helpful.
what happens when F=0.000001 ? Acording to your formula Ffr = μ*m*g, the body would have to start moving in the direction of that friction force, since the only force opposing it would be that of magnitude 0.000001 and that's negligible ....
The friction force will adjust so that it will not exceed the force pulling the block.
I mean since the bottom block is moving to the right?
See comment above.
Hi Michel I am having some issues with dynamics questions involving incline plane and pulley problems, mass A sliding down plane pulling mass B up the vertical side by pulley system with mechanical advantage 2:1.When checking solutions my accelerations are off because they select down as positive for mass B making the 2T in cable negative even though it aids its acceleration upwards.
Is this an error thank you.
If you can write out the text of the problem, I may be able to figure it out.
Michel van Biezen Thank you for reply, I think I can see what is going on it is a problem similar to ones you have done except there is a mechanical advantage taken place with the pulleys on the inclined plane. There is a kinematic equation for rope length 0=2aA+3aB, its a 2:3 mechanical advantage set up. The coordinate system used for acceleration is not intuitive in that the chosen positive direction for each mass is opposite, however this seems to correspond to rope equation -2aA=3aB in that if mass A is moving negative direction B is increasing length and vice versa.The question is from Kraige 6th dynamics 3/25. I appreciate this might be asking a bit much in these conditions, however thanks for informative videos.
The key to understanding the mechanical advantage with pulleys is that if you have for example a 2:1 advantage, the force required to pull the object is only half its weight (when hanging), but the distance you must pull is twice the distance the object moves.
Michel van Biezen Thank You
whats the purpose of force friction 3?
It is a dragging force originally caused by F(max) but its acting indirectly thru M2 pulling against its binding friction with M1.
Its simply Fmax conducted thru M2 & when it arrives at the contact surface it acts on M1 in the direction of Fmax.
If that frictional force didnt exist, M1 would not be dragged along with M2.
What happens if we pull m1 instead of m2?
It depends on how hard you pull. I you pull gently, the top block would move with the bottom block without slipping. If you pull hard enough the top block would slip.
Can i use the same formula as your example when we pull m1? I want to find maximum force to make m2 slides
didn't the max force calculated exceeds the max static friction and causes block 2 slips? that's the only part i'm confused about.
The force calculated equals the friction force between m1 and m2 PLUS the force required to accelerate m2. Note that if you calculate both forces they add up to the 52.3 N.
Michel van Biezen professor . Can't we find Fmax by making it equal to friction force 3 which is occur between m1 and m2? Because in other video you find Fmax like that . Here is the video.
ruclips.net/video/3Dn1Ze8maEc/видео.html
But when I tried, it didn't give the same result :(
so what is the problem?
That problem is a static situation. Nothing is moving. Here the blocks are moving.
Now I understood everything. Thanks for your answer
this is my sign to drop out lol
But these types of examples can really help in understanding specific concepts. 🙂
😭
Glad you liked it. 🙂
@@MichelvanBiezen I will take. EUEE this year and I'm trying hard to change my scores but I can't I don't get it