Physics - Mechanics: Applications of Newton's Second Law (15 of 20) sliding block combination

[@schrodie56]

I also meant to say that these KZbin lectures are a very valuable resource for any student of physics. Many thanks.

[@ravinapanangalage2761]

Professor Michel, these are incredible set of lessons. Thanks for your effort and time. Just a small clarification. When you consider the whole system 2 friction forces in the system between m1 and m2, they are equal and opposite direction. Won't they cancelled out each other ?

[@MichelvanBiezen]

No, they are additive, since each acts on one of the two blocks which move in opposite directions. (therefore each friction force acts against the direction of motion for each block)

[@ravinapanangalage2761]

@@MichelvanBiezen : Thanks got it. Thanks for the quick reply.

[@asmaalobaidli8093]

I finally understand physics, thanks a lot !

[@charlesbromberick4247]

Nice problem - I can think of a few variations, and suspect they will appear in other videos. Thanks, I´m sure these are helping many struggling students.

[@daniyalravandeh5223]

And about the Tension which is a part of question the answer would be as follows: T = m3*(g - a) = 10*(9.8 - 5.75)= 40.50 N

[@MichelvanBiezen]

+Daniyal Ravandeh Make that m3 (not m1)

[@daniyalravandeh5223]

+Michel van Biezen You're right, sorry about that and thank you ;)

[@tobeynguyen2613]

+Michel van Biezen Is tension the same in both strings? Why or why not? Thanks!

[@daniyalravandeh5223]

+Tobey Nguyen I guess that it's not the same. On the other side (left of mass 2) tension would be as follow: To simplify the problem, assume that we have the m1=2kg a=5.75 m/s^2 Mu=0.3 So the tension on both sides of pulley on the left is equal to tension in cable which pulls mass 1 (m1): T = (m1 x a) + (Mu x m1 x g) = 2 x 5.75 + 0.3 x 2 x 9.8 = 17.38 N

[@DanishJaved]

Hey Professor. Great video, as always. I wanted to ask though if it would be possible to solve this particular problem if we were to apply the 2nd law for each block individually and work from there?

[@MichelvanBiezen]

Yes you can solve it that way, but it would be more difficult and time consuming.

[@albeetheone5752]

Professor van Biezen, questions about the tensions in the cables in different segments between the two pulleys. What is the tension force T2 right side in the cable on the right side of m2? What is the tension force T2 left side in the cable on the left side of m2??? What is the tension force T1 in the cable on the left side of m1???

[@grishproduction1243]

Thank you so much for explaining that the friction force is actually acted twice!

[@BSnicks]

I have been so confused with this example. From the calculations, it seems that there is only friction between m1 and m2. But in the comments, you answered that there is also friction between m2 and the table.

[@MichelvanBiezen]

In this example there is NO friction between m2 and the table, only between m1 and m2. But the friction force acts on both m1 and m2 and therefore you must count it twice.

[@nch8180]

During calculating net force,why is the second friction force negative ? I thought it would be positive due to the acceleration direction.

[@MichelvanBiezen]

Friction force 2 opposes the motion of m1. Friction force 1 opposes the motion of m2.

[@shadazmi5402]

this is actually something useful! thanks a million

[@schrodie56]

You say at about 2mins in that there is a force of friction applied to m2 i.e. Ffrict 2 to the right however you draw the frictional force going to the left; so did you mean to say that there is a frictional force applied to m2 going to the left?

[@MichelvanBiezen]

Yes you are correct. I "said" to the right and drew the friction force to the left. I should have said: "to the left".

[@reyna5443]

sir, you said there is a friction force to the right but you drew it to the left. m1's motion is to the left so the friction force should be to the right?

[@MichelvanBiezen]

Just like with tension, it depends on the object you are referencing. For the friction force between the 2 sliding blocks: For the bottom block moving to the rigth, the friction force acts towards the left. For the top block moving to the left, the frction force acts towards the right.

[@joeyborja423]

Question on the frictional forces Ffr1 and Ffr2. If they are opposing the motion of the whole system but themselves are opposing each other, is it wrong to surmise that, in one way, one is opposing and one is aiding the movement since they seem to cancel out and therefore treat them non-existent? Almost done with this series and so enjoying it to the max! Again, you and your wife are the best!

[@MichelvanBiezen]

In this case it helps to draw a free body diagram around each of the 2 blocks to help determine how each of the forces (and friction forces) affect each block and the system.

[@tharmabalanthuva5701]

Sir if we use the relative acceleration.

[@christianbonifacio6751]

Sir, is the tension in both string equal?

[@MichelvanBiezen]

The tension in the right string is: T = m1 g - m1 a and the tension in the left string is: T = m3 a + m3 g mu

[@anteater2536]

i am confused, how come the mass of m2 can be omitted if the right string is pulling the m2 to the right? is it correct that one of the m1 is supposed to be m2 in the equation??

[@MichelvanBiezen]

+Roy Carroyo Since there is no friction between m2 and the table, there is no contribution to the net force from m2. The weight of m2 does not contribute to the friction between m1 and m2.

[@anteater2536]

oh i see, thank you sir.

[@md.mominulislam5068]

Dear sir, i am confused as the m2 acts as a surface for m1. so friction force to be right side

[@MichelvanBiezen]

Yes, this is a confusing problem and that is why I put on this video, so we can learn how to deal with it. There is no friction between m2 and the table. There is only friction between m1 and m2. The direction of the friction force is always in the opposite direction of the motion of the object without the friction. There are two friction forces. One acting against m1 and the other acting against m2. m1 moves to the left, so the friction force against the motion of m1 is to the right. m2 moves to the right, so the friction force against the motion of m2 is to the left. The both oppose the motion of the whole system.

[@1zero53]

Is the accelaration of each block same or does all block actually have different accelaration?

[@MichelvanBiezen]

The magnitude of the acceleration of each block must be the same since they are all connected.

[@bggbbdg5625]

Biezen should mention that there is no friction between m2 and the table otherwise it is not solvable.

[@MichelvanBiezen]

There is friction between m2 and the table (coefficient of friction = 0.3)

[@khairulmarin8290]

But sir right after sometimes there will be no connection between M2 and M1... But u are calculating the frictional force as if it will be there for the whole operation.....if you could enlight me about that....would be very helpful

[@MichelvanBiezen]

Only for the portion where they are in contact like in the figure

[@khairulmarin8290]

What would change if the friction was there for whole time

[@naeemghafori5046]

Sir, why the videos getting more darker ?

[@MichelvanBiezen]

+Naeem Hakimi We had to learn a lot about making videos including sound and lighting. Those problems are now solved, but still remain with our early videos.

[@erc333]

What about the tension on m2 and m1 ...?

[@justinbeiber9290]

Reuel Cuellar because they are opposing each other like t1=t2 so they will subtract each other

[@raspberry765]

T1= 40.6N, T2= 17.4N - T1 being the rope connecting m3 to m2 and T2 being the rope connecting m1 and m2.

[@penguinzz6746]

I still don't get it why these friction forces oppose the acceleration.. Ur videos are still awesome sir.

[@MichelvanBiezen]

Think of it like this. Will pushing the brakes of your car accelerate the car?

[@drtamiz]

Aren't the frictions on the boxes internal and should not be included in the acceleration equation?

[@MichelvanBiezen]

When you rub 2 blocks together, does the friction oppose your motion?

[@rameshkumarsharma1593]

I Think that F net = m3*g-m1*g-(m1+m2)*g*0.3. Please coment whether I am correct or wrong. Thanking you Sir.

[@MichelvanBiezen]

Since there is no friction between the table and m2, the equation in the video is correct.

[@phuzo3320]

Sir the frictional forces are acting in opposite directions how could u take it to be in same direction while calculating net force... Plz explain

[@Simoos7]

I have the same question.. I don't understand this point..

[@saadtahir416]

What about the friction between surface and m1

[@MichelvanBiezen]

m1 is only in contact with the block below it.

[@exodia45]

hello sir how are you? i have a question imagine there is no any friction so Fnet = (W3-W2-W1) and with friction it would be Fnet = (W3-W2-W1-2(W1.mu)) am i right?!

[@MichelvanBiezen]

If you are referring to this problem, why would Fnet (without friction) be equal to W3 - W2 - W1 ?

[@exodia45]

@@MichelvanBiezen you are right fnet=m.a and fnet=w3 only without friction between any masses and table or each other but while existing friction between m1and m2 it should be fnet=m.a and then fnet=w3-m1.g.mu-m1.g.mu if there is friction between table and m2 it would be fnet = w3-m2.g.mu-2(m1.g.mu) is that right?!

[@MichelvanBiezen]

Fnet would be W3 - 2m1g mu - (m1+m2) g mu

[@successfulnarcissisticman9140]

Hey teacher I want to know why there's no friction directed to the left at the point of contact of m2 and the surface

[@MichelvanBiezen]

+papa thug In this example there is no friction between the surface and m2 The purpose of this example is to show the friction between moving blocks

[@successfulnarcissisticman9140]

Michel van Biezen Oh Ok I should've known . Thanks

[@adikymanyamemambolle657]

Thanks for this video sir. Now, the two frictional forces are opposing each other i.e one is to the right and the other to the left. Why do they both have the same signs when you apply N2?

[@MichelvanBiezen]

Both friction forces oppose the acceleration of the whole SYSTEM. Fnet = m total * a And Fnet = all the forces aiding the acceleration - all the forces opposing the acceleration. (They both oppose the acceleration).

[@adikymanyamemambolle657]

Thank you sir.

[@mhireteab1900]

@@MichelvanBiezen But the first friction is in the direction of the acceleration, we should add not subtract

[@rahultiwari9003]

sir how can we know that all the masses have the same acceleration ???? and also the friction forces are to the right as well as to the left so shouldn't they cancel each other

[@MichelvanBiezen]

+Rahul Tiwari All the masses are connected to each other. Each friction force acts on an object. If they don't act on the same object they do not cancel

[@rahultiwari9003]

Sir please tell me more thing I am much confused in such problems bcz I have seen that when there are multiple pulley system and there are multiple masses then the acceleration can be different so how can I know the relationship between those accelerations

[@naveednaiemi3979]

Sir if there was friction between surface and m2, would we use Meo*N2=Meo*(m1g+m2g)?

[@MichelvanBiezen]

Yes, the friction force would be mu * (m1g + m2g).

[@procrastmh]

+Michel van Biezen does that mean that there would be 3 frictional forces?

[@callistersmaifwani413]

Yes that's what it will be

[@jonasakarlsson2036]

Hey, u never write out "Tension" in the a=f/m, do the tension cancel or what? And isn't a1=0,5a2 in this problem, so if u would ask the acceleration of m1, would it be half the answer? Or why isnt it 0,5 here?

[@MichelvanBiezen]

+Jonasa Karlsson The tension on the strings are internal to the system and not needed to solve the problem, unless you want to use the technique of free body diagrams.

[@jonasakarlsson2036]

Yeah I usually do the free body diagram, because on tests if u just write in the correct numbers imedialty they suspect cheating which is sad!.

[@SSSJ0014]

Hi professor Should Fk2=N2*mu ? Or are we missing another friction force w/ m2 & the table, which will be Fk=N2*mu

[@MichelvanBiezen]

The video is correct.

[@itanibhim]

Hi sir, there is three forces acting downward that causes from m1 and m2 but you mentioning only two normal forces acting upward. Shouldn't be there another normal force? Thanks for posting this video. I am really enjoying and recalling the courses I had long time ago.

[@MichelvanBiezen]

Bhim Itani It is correct as is. The key is to figure out where the forces are acting and what the magnitude is at each location.

[@itanibhim]

Michel van Biezen Thanks

[@doodelay]

Why did you subtract mg1*fr twice instead of having the equation be mg3 - mg1*fr - (mg2+mg1*fr)? It's as if block 2 doesn't exist. Edit: OOOH I think I see, block 2 in a way doesn't resist the pull at all simply because it has no friction right?

[@MichelvanBiezen]

I didn't quite follow that, but the video is correct.

[@pavitra3819]

Just wanted to know when will we study this I school like which grade or something

[@MichelvanBiezen]

pavitra shadvani This is a typical 1st year college physics problem.

[@pavitra3819]

Thnx

[@ihteshamkhan4696]

Isn't there a friction force between m2 and the table?

[@MichelvanBiezen]

In this example there is no friction between m2 and the table

[@ihteshamkhan4696]

Michel van Biezen thank you.

[@vinceroiallendial5353]

why did we minus the friction force i thought opposite direction will be cancelled?

[@MichelvanBiezen]

Applying these "general rules" without thinking through the problem can lead to incorrect answers. We have to look at each friction force and determine how it affects the system. In both cases each of the friction forces will remove energy out of the system and thus we have to subtract both of them.

[@callistersmaifwani413]

Sir Isn't't the acceleration 5.6 to the accuracy? 🤔... Ur videos are life saver 😭 I love what u are doing here sir much love ❤️

[@MichelvanBiezen]

5.75 is the correct number. Thanks for checking. 🙂 Glad you find the videos helfpul.

[@obakengatom_sweetbrotherof1081]

i think it is fn= 9.8(3+2) for block m2

[@MichelvanBiezen]

That is the reaction force of the table pushing back against m2.

[@obakengatom_sweetbrotherof1081]

sorry sir what is the normal force of block m2

[@chiragsaxena7251]

Sir what is the easiest method to find the tension?

[@MichelvanBiezen]

This playlist will show you how: PHYSICS 17 TENSION AND WEIGHT

[@ilyassourahou7839]

why their is 2 friction forces and the movement goes on one side (sorry for my english and keep going prof)

[@MichelvanBiezen]

Think about how each friction force acts on each block. The friction force between the 2 blocks acts against the motion of each block, and since they move in opposite directions, the friction force acts in one direction for one block and in the other direction for the other block.

[@ilyassourahou7839]

@@MichelvanBiezen thanks sir

[@petereziagor4604]

Awesome

[@MichelvanBiezen]

Glad you are enjoying these.

[@obakengatom_sweetbrotherof1081]

answer 5.56

[@navodadesilva]

thanks

[@visitorsig]

Hey you never told how to get the tension I need that lol

[@whyareyou1764]

once you found the acceleration all you need to do is plug into the simplest equation, for this one it will be for block 3 T1=(m3g)-(m3ay)

[@lakshmisanthosh8429]

This video is little blur

[@MichelvanBiezen]

Yeah...this is one of the 1st videos we ever made. Since Mike is a physics major and i, a math major, neither one of us have any experiences in making videos. So please be understanding when viewing our ollder videos. Whenever I see one of our older videos, I just hang my head in shame, want re-make those videos, and pretend those videos never happened.

[@pavitra3819]

Just wanted to know when will we study this I school like which grade or something

[@daphenomenalz4100]

Now, (if you are in 11th now)😂

[@pavitra3819]

Just wanted to know when will we study this I school like which grade or something

[@HorizonSpeed26]

11th ,12th maybe 10th.

[@pavitra3819]

Just wanted to know when will we study this I school like which grade or something