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

Click For Summary

Discussion Overview

The discussion revolves around the dynamics of two blocks being pushed on a floor with friction, focusing on the forces acting on each block and the implications of those forces on their motion. Participants explore concepts related to Newton's laws, frictional forces, and the conditions under which the blocks may or may not accelerate.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that when a force F is applied to block M1, block M2 will experience a force transmitted through M1, but the magnitude of this force will be less than F.
  • Others argue that the presence of friction between the blocks and the wall affects the net forces acting on them, particularly when the blocks are not accelerating.
  • A participant questions why the force between the blocks equals F when they are pushed against a wall, while it is less than F in other scenarios.
  • There is a discussion about the role of friction in both accelerating and non-accelerating cases, with some participants suggesting that friction must be accounted for in the equations of motion.
  • One participant emphasizes the importance of drawing free-body diagrams to visualize the forces acting on the blocks.
  • Another participant expresses uncertainty about how to incorporate friction into the equations and seeks guidance on the approach.
  • There is mention of static friction balancing gravitational forces, with participants discussing the implications of this balance on the motion of the blocks.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact nature of the forces acting on the blocks or the role of friction in different scenarios. Multiple competing views remain regarding how to apply Newton's laws and the implications of friction on the blocks' motion.

Contextual Notes

Limitations include the potential for missing assumptions regarding the system's configuration, the dependence on definitions of forces, and unresolved mathematical steps related to the application of Newton's laws in different contexts.

gracy
Messages
2,486
Reaction score
83
WS4c.JPG

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?
APPLIED.jpg

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.
 
Physics news on Phys.org
gracy said:
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.
 
Than why in this case

WHY.png

There is friction between the two blocks as well as between the block and wall.
 
gracy said:
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.
 
Doc Al said:
In this case, the blocks are not accelerating
This seems to me the result ,not the reason.
 
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
proxy.php?image=http%3A%2F%2Fwww.phys.ttu.edu%2F%7Erirlc%2FWS4c.JPG
 
Last edited by a moderator:
gracy said:
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.

gracy said:
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
proxy.php?image=http%3A%2F%2Fwww.phys.ttu.edu%2F%7Erirlc%2FWS4c.JPG
What applies in both cases are Newton's laws.
 
Last edited by a moderator:
Doc Al said:
Not sure what you are looking for.
ok.I can point out.
Doc Al said:
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.
 
gracy said:
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.)
 
  • Like
Likes   Reactions: gracy
  • #10
Doc Al said:
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?
 
  • #11
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.
 
  • Like
Likes   Reactions: gracy
  • #12
gracy said:
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.
 
  • #13
So,you mean acceleration will be same for both the blocks.
 
  • #14
But you did not mention anything about friction in the accelerating case" (and taking the horizontal axe oriented to the right).
 
  • #15
gracy said:
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.
 
  • #16
mattt said:
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.
 
  • #17
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?
 
  • #18
mattt said:
Do you know Newton laws, applied to one particle and to system of particles?
Yes.As far as I know,I know.
 
  • #19
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.
 
  • #20
mattt said:
Try to do it, and tomorrow I'll come back here again to see.
OK.I will try.Thanks again.
 
  • #21
mattt said:
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?
 
  • #22
gracy said:
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.
 

Similar threads

Find the acceleration of the system (2 blocks sliding on a table)
  • · Replies 23 ·
Replies
23
Views
3K
How much force does the 20 kg block exert on the 10 kg block?
  • · Replies 17 ·
Replies
17
Views
2K
Undergrad Friction force has different effect than Applied Force?
  • · Replies 9 ·
Replies
9
Views
3K
Undergrad Force applied to 2 connected mass.
  • · Replies 10 ·
Replies
10
Views
2K
Undergrad Question about Momentum vs. Kinetic Energy vs. Deforming In Collisions
  • · Replies 2 ·
Replies
2
Views
2K
When Friction the block m2 start to move?
  • · Replies 11 ·
Replies
11
Views
2K
Graduate Compressive force between objects
  • · Replies 2 ·
Replies
2
Views
3K
Basic Forces and Motion Problem (friction)
  • · Replies 2 ·
Replies
2
Views
1K
Coefficient of Friction when Normal Force is Reduced [HS]
  • · Replies 5 ·
Replies
5
Views
2K
High School Calculating Force Between Two Blocks on a Frictionless Surface
  • · Replies 2 ·
Replies
2
Views
2K