Why Is Static Friction Acting on Block B?

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Homework Help Overview

The discussion revolves around the forces acting on two blocks, A and B, which are pressed against a wall. The problem involves understanding the static friction forces between the blocks and the wall, as well as the implications of Newton's third law in this context.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the nature of the forces acting on the blocks, particularly questioning why the static friction force acting on block A is relevant to block B. There are discussions about the normal force and the conditions under which static friction applies.

Discussion Status

Participants are actively engaging with the concepts of friction and Newton's laws, with some expressing confusion about the relationships between the forces. There is a mix of correct and incorrect understandings being clarified through the dialogue, with no explicit consensus reached yet.

Contextual Notes

Some participants express uncertainty about the definitions and roles of the forces involved, particularly regarding the frictional forces and their interactions. There is also mention of the coefficients of friction and their applicability in the scenario presented.

gracy
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Homework Statement

;Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is
2BLOCKS.png
[/B]

Homework Equations

:[/B]
f 1(static friction force acting on block A)=20 N=mass of A multiplied by g.

The Attempt at a Solution

:
f2.png

Here f 2=f1+ m g B= 120 N
The only thing I don't understand is why f1 is acting on block B?[/B]
 
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gracy said:

Homework Statement

;Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is
View attachment 81959 [/B]

Homework Equations

:[/B]
f 1(static friction force acting on block A)=20 N=mass of A multiplied by g.

The Attempt at a Solution

:View attachment 81960
Here f 2=f1+ m g B= 120 N
The only thing I don't understand is why f1 is acting on block B?[/B]
Because action and reaction are equal and opposite. If B is exerting a force f1 up on A, A must be exerting a force of the same magnitude down on B.
 
The coefficient of [static] friction is a limit on how high the frictional force can go before the objects that are in contact start slipping. It may or may not be the applicable limit in this case. First one needs to decide whether the blocks are sliding or are sticking.

A coefficient of friction [static or dynamic] relates the tangential force of friction to the perpendicular force with which the surfaces are pressing on one another. This perpendicular force is usually called the "normal force". The word "normal" in this case means "perpendicular", not "ordinary".

With this in mind, what is the normal force between block B and the wall?
 
jbriggs444 said:
The word "normal" in this case means "perpendicular", not "ordinary".
Oh,this I know of course.:smile:
 
haruspex said:
Because action and reaction are equal and opposite. If B is exerting a force f1 up on A, A must be exerting a force of the same magnitude down on B.
But ,f1 is not exerted by A alone,it's a force of friction.It is due to contact between both A and B.Newton's third law will be applicable here also?
 
gracy said:
But ,f1 is not exerted by A alone,it's a friction force.It is due to contact between both A and B.Newton's third law will be applicable here also?
As far as A is concerned, it is just an external force exerted on it. Doesn't matter whether it's friction or, or a nail, or magnetic...
Similarly for B. Yes the action and reaction law applies.
 
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haruspex said:
Yes the action and reaction law applies.
Oh,this I didn't know of course.:smile:
 
Thanks jbrigs444 and haruspex .
 
Then f 1(static friction force acting on block A) should also be =f2 +mA multiplied by g
 
  • #10
gracy said:
Then f 1(static friction force acting on block A) should also be =f2 +mA multiplied by g
But if it is true we will not get the same answer for f2 i.e 120N.
 
  • #11
gracy said:
Then f 1(static friction force acting on block A) should also be =f2 +mA multiplied by g
Why?
 
  • #12
For the same reason i.e Newton's third law.
 
  • #13
gracy said:
For the same reason i.e Newton's third law.

Show your work. Why does Newton's third law require that f2 = f1 + mAg?
 
  • #14
Because of Newton's third law f2 will also act on block A But in downward direction because f2 is acting on B upward.So for equilibrium of block A.
So f1=f2+mA g
 
  • #15
gracy said:
Because of Newton's third law f2 will also act on block A But in downward direction because f2 is acting on B upward.So for equilibrium of block A.
So f1=f2+mA g
That is a completely incorrect understanding of Newton's third law. F2 acts between block B and on the wall. It does not act on block A at all.
 
  • #16
jbriggs444 said:
F2 acts between block B and on the wall. It does not act on block A at all.
Oh,sorry.I misunderstood f2.My bad.I thought f2 is force of friction between the two blocks experienced by Block A.
I labeled that f2 =friction between wall and block B but forgot.Really sorry.
 
Last edited:
  • #17
gracy said:
.I thought f2 is force of friction between the two blocks experienced by Block A.
Even this is wrong.But now I am clear .I know force of friction between the two blocks experienced by Block A is f1 acting in downward direction.
 
  • #18
My ideas and thoughts are all jumbling.I think,I should sleep now.
 

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