Force Applied -- General doubt about pushing 2 blocks on a floor with friction

[gracy]

Force" F"is applied on M1 .Although force applied is not directly on M2,but it will also experience this force" F" via M1,right?

But I don't think it is correct.Because it implies that no matter how big the force on M1,the net force on M1 will always remain zero.And there will be unbalanced force "F" on block M2.It is like we are pushing M1 by force "F"but it is in it's place(not moving)instead M2 moves.Note that between M1 and M2 there will be frictional force ,as they are pressing against each other but it will be along the surface of contact and as there is no force to counter in y direction the frictional force will not come into picture.
I am not posting this in homework section because It is not my homework .It's my conceptual doubt.
Thanks.

 

[Doc Al]

Force" F"is applied on M1 .Although force applied is not directly on M2,but it will also experience this force" F" via M1,right?

As you seem to have realized, this is not the case. The force F is applied to M1. In turn, M1 will apply a different force on M2. That force will be less than F.

 

[gracy]

Than why in this case

There is friction between the two blocks as well as between the block and wall.

 

[Doc Al]

Than why in this case

View attachment 81954
There is friction between the two blocks as well as between the block and wall.

In this case, the blocks are not accelerating.

 

[gracy]

In this case, the blocks are not accelerating

This seems to me the result ,not the reason.

 

[gracy]

Why https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/65/65001-ae59a43002c3519191eb12aaa6fd31ab.jpg
is applicable in this case
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/65/65022-752f6be69d8fa34ccdbce044f6fd5318.jpg
but not in

 

[Doc Al]

This seems to me the result ,not the reason.

Not sure what you are looking for. Given the force F applied, you then deduce any other forces using Newton's laws.

Why https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/65/65001-ae59a43002c3519191eb12aaa6fd31ab.jpg
is applicable in this case
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/65/65022-752f6be69d8fa34ccdbce044f6fd5318.jpg
but not in

What applies in both cases are Newton's laws.

 

[gracy]

Not sure what you are looking for.

ok.I can point out.

That force will be less than F.

but in the case when blocks are pushed against wall
the force will be equal to F.Why?That's what I am looking for.

 

[Doc Al]

ok.I can point out.

but in the case when blocks are pushed against wall
the force will be equal to F.Why?That's what I am looking for.

In the first case, the blocks are accelerating. Thus, there must be a net force on each block.

In the second case, when pushed against the wall, the blocks are not accelerating. Thus, there must be no net force on them. (Which implies that the force between the blocks must equal the applied force.)

 

[gracy]

That force will be less than F.

That means for block M1 there will be less acceleration than M2 ,if F is not much larger than force on M2?

 

[mattt]

Expanding what Doc Al just said:

in "the accelerating case" (and taking the horizontal axe oriented to the right) you have:

$F = (M_1 + M_2) a$

$F-R = M_1 a$

$R = M_2 a$

R is (the magnitude of) the force the second block exerts on the first block and also it is (the magnitude of) the force the first block exerts on the second block.In the "against the wall case" you have:

$F - N = (M_1 + M_2) a = 0$

$F - R = M_1 a = 0$

$R - N = M_2 a = 0$

N is the "Normal Force" the wall exerts on the second block.

 

[mattt]

That means for block M1 there will be less acceleration than M2 ,if F is not much larger than force on M2?

No, look at my previous post.

 

[gracy]

So,you mean acceleration will be same for both the blocks.

 

[gracy]

But you did not mention anything about friction in the accelerating case" (and taking the horizontal axe oriented to the right).

 

[mattt]

But you did not mention anything about friction in the accelerating case" (and taking the horizontal axe oriented to the right).

Do you mean friction between each block and the floor? You can add that force too. Tell me if you want me to write the above equations adding the friction force between each block and the floor.

 

[gracy]

Tell me if you want me to write the above equations adding the friction force between each block and the floor.

Yes.So nice of you.But I am afraid that I would be not showing any effort if I ask you to write all equations for me.Better,you guide me how to write them then I will write them by myself.

 

[mattt]

Well, first thing I should have asked you is, what have you already studied?

Do you know Newton laws, applied to one particle and to system of particles?

 

[gracy]

Do you know Newton laws, applied to one particle and to system of particles?

Yes.As far as I know,I know.

 

[mattt]

I have to go back to work right now, but if you understood my previous post ( # 11 ) then it won't be that difficult to you to add friction forces between each block and the floor.

The best way (when you are learning) is drawing a picture and drawing in it each and every (vector) force, and then to apply Newton laws, to the whole "block1 + block2", and/or (depending on what you want) to each block singled out.

Try to do it, and tomorrow I'll come back here again to see.

 

[gracy]

Try to do it, and tomorrow I'll come back here again to see.

OK.I will try.Thanks again.

 

[gracy]

Try to do it, and tomorrow I'll come back here again to see.

I think in post #11 ,you have explained how horizontal forces add up to zero,so there is no horizontal motion .But as we see in real observations there is no motion in vertical direction also.So,here friction comes into picture.It balances forces due to gravity.
f 1(static friction force acting on block M1)=M1 multiplied by g
f 2(static friction force acting on block M2)=f1+ M2 multiplied by g
Right?

 

[Doc Al]

But as we see in real observations there is no motion in vertical direction also.So,here friction comes into picture.It balances forces due to gravity.
f 1(static friction force acting on block M1)=M1 multiplied by g
f 2(static friction force acting on block M2)=f1+ M2 multiplied by g
Right?

Looks good to me.