Stacked blocks pulled horizontally

  • Thread starter judas_priest
  • Start date
  • Tags
    Blocks
In summary, the conversation discusses the concept of friction and its role in the movement of two blocks on a horizontal table. It is determined that the force required for both blocks to move is the same, as friction is a self-adjusting force that increases with an increase in force. The values of friction are calculated using the formula Fstatic ≤ μN and Fdynamic = μN, and it is noted that dynamic friction only occurs when there is relative movement between two surfaces. Theoretical and practical approaches are both suggested for understanding this concept.
  • #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
 

Attachments

  • Stacked blocks_bottom_pull.png
    Stacked blocks_bottom_pull.png
    367 bytes · Views: 443
Physics news on Phys.org
  • #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.
 

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
961
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
386
  • Introductory Physics Homework Help
Replies
12
Views
928
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
901
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
2K
Back
Top