Stacked blocks pulled horizontally

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Two blocks of equal mass on a horizontal surface experience friction when a horizontal force is applied to the bottom block. The discussion revolves around the conditions under which each block begins to move, emphasizing that static friction must be overcome for motion to occur. It is concluded that both blocks will start moving simultaneously once the applied force exceeds a certain threshold, specifically when the force surpasses the maximum static friction acting on the bottom block. As the force increases further, the acceleration of the bottom block can exceed that of the top block due to the differing frictional forces acting on each. Understanding the mechanics of static and dynamic friction is crucial to solving the problem accurately.
  • #31
Here's another question. What if the lower block had an external force towards right, with friction between two blocks remaining, but no friction between lower block and ground.
 
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  • #32
judas_priest said:
Are my FBDs correct? Where am I going wrong?

You didn't show any FBD, did you?, so how can they be correct or incorrect. Show the two FBD, now you know well the Forces acting on the two blocks.

judas_priest said:
Here's another question. What if the lower block had an external force towards right, with friction between two blocks remaining, but no friction between lower block and ground.

This is or more or less the same question as OP. Solve the first one and put the value of μground = 0.
 
  • #33
darkxponent said:
You didn't show any FBD, did you?, so how can they be correct or incorrect. Show the two FBD, now you know well the Forces acting on the two blocks.



This is or more or less the same question as OP. Solve the first one and put the value of μground = 0.

I've shown my FBDs 3 times now. You've even responded to them.
 
  • #34
I'm going to post the diagram with forces for the question I just asked (i.e What if the lower block had an external force towards right, with friction between two blocks remaining, but no friction between lower block and ground.). Let me know if I'm doing it right.
 
  • #35
judas_priest said:
I've shown my FBDs 3 times now. You've even responded to them.

You have posted just one attachment to this thread, that is the diagram of the question. I haven't find any attachment of Free Body Diagrams.
 
  • #36
Here you go. The friction acting on the top block is the friction between top block and lower block, and I've also showed it's reaction on the lower block. And there's an external force. You can't complain about me not showing about FBD anymore
 

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  • Stacked blocks_bottom_pull.png
    Stacked blocks_bottom_pull.png
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  • #37
judas_priest said:
Here you go. The friction acting on the top block is the friction between top block and lower block, and I've also showed it's reaction on the lower block. And there's an external force. You can't complain about me not showing about FBD anymore

This is not the proper FBD. You have to draw two FBDs, one for the upper block and one for the lower one. And you must show all forces acting on the object.
 
  • #38
All the forces necessary are shown. Here I'm going to describe the vertical forces for you.
N (due to upper block) + Mg downwards for the lower block

N (due to lower block ) = Mg for the upper block.

This thread has been going on endlessly with almost no point being made. Please help me if you can tell me where I'm going wrong. Else, I'll just try it on my own.

I'm bad with paint.
 
  • #39
judas_priest said:
All the forces necessary are shown. Here I'm going to describe the vertical forces for you.
N (due to upper block) + Mg downwards for the lower block

N (due to lower block ) = Mg for the upper block.

This thread has been going on endlessly with almost no point being made. Please help me if you can tell me where I'm going wrong. Else, I'll just try it on my own.

I'm bad with paint.

I was just about to say that.

If you are new to drawing FBDs, i think this will help https://en.wikipedia.org/wiki/Free_body_diagram

Okay i don't if i am going against the rules but i will form the equations for you.

Upper object: f1 = m1a1

Lower object: F-f1-f2 = m2a2

Is this what you are getting.
 
  • #40
darkxponent said:
I was just about to say that.

If you are new to drawing FBDs, i think this will help https://en.wikipedia.org/wiki/Free_body_diagram

Okay i don't if i am going against the rules but i will form the equations for you.

Upper object: f1 = m1a1

Lower object: F-f1-f2 = m2a2

Is this what you are getting.
I'm not new. I'm bad at mechanics. It's the area in physics I'm trying to improve in the vacations I have. Rest all is pretty fine I'd say. I built my interest in science after mechanics was taught. Trying to smoothen it up now.

Also, both the masses are same. What is f2? There is no friction between the ground and lower block. I'm trying to clarify about the latter question. that is :

The coefficient of static friction for the top contact surface is greater than zero and the bottom contact surface is frictionless. There's an external force pulling the lower block.
 
  • #41
judas_priest said:
I'm not new. I'm bad at mechanics. It's the area in physics I'm trying to improve in the vacations I have. Rest all is pretty fine I'd say. I built my interest in science after mechanics was taught. Trying to smoothen it up now.

Also, both the masses are same. What is f2? There is no friction between the ground and lower block. I'm trying to clarify about the latter question. that is :

The coefficient of static friction for the top contact surface is greater than zero and the bottom contact surface is frictionless. There's an external force pulling the lower block.

For the latter question, there is no friction at ground so f2 = 0. For OP f2 ≤μN
 
  • #42
darkxponent said:
For the latter question, there is no friction at ground so f2 = 0. For OP f2 ≤μN

But you wrote F-f1 = ma1.

Shouldn't it be F + f1 ma1?
Because the reaction friction force and the external Force are in the same direction
 
  • #43
judas_priest said:
But you wrote F-f1 = ma1.

Shouldn't it be F + f1 ma1?
Because the reaction friction force and the external Force are in the same direction

No they are in opposite direction. See what i wrote in post 20

darkxponent said:
I did not talk about circuits anywhere!. I am talking mechanics only. At the surface between the blocks, the Lower block has tendency to move forward, but as is the property of friction, which does exist between the two blocks, Friction opposes the motion of the lower block, so it applies a force on lower block backward, means that the upper block exerts a force on the lower block backwards. Then Newtons Third Law comes into play and the lower block exerts an equal amount of Force on the upper block in the opposite direction that is forward direction. This Friction f1 will be static until the two surfaces don't move relative to each other.

BTW i wrote F-f1 = ma2! and f1 =ma1, if you take f2 as zero.
 

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