Connected blocks on a surface with friction

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Homework Help Overview

The discussion revolves around a system of two blocks, A and B, connected by load cells on a horizontal surface, with a force applied to one of the load cells. The blocks have different coefficients of friction, and the scenario includes the effects of acceleration and potential impact with a wall.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting at each load cell and how they change when the system hits a wall. There are questions about the application of forces and the role of friction in the system's behavior.

Discussion Status

Some participants have provided insights into the forces acting on the load cells and the implications of acceleration. There is ongoing exploration of how additional forces, such as those from a vibrating wall, might affect the system.

Contextual Notes

Participants note the complexity introduced by varying forces and the need for free body diagrams to clarify the interactions within the system. The original poster has indicated constraints in editing the initial problem statement, which may affect the clarity of the discussion.

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This is not a homework problem rather it is a discussion that I have been having with friends. *Your thoughts and feedback would be appreciated.

Homework Statement


Blocks A and B are on a horizontal surface and connected to one another by a load cell.
Block A has mass m1 and Block B has mass m2.
Each block has difference coeff of static and dynamic friction.
Load cells are connected to the left and right of the blocks.
A right horizontal force F1 is applied to the*
All blocks accelerate at the same rate.
The applied force is much greater than the frictional force.

Diagram

-(#)- = load cell

-(1)-A-(2)-B-(3)-

Questions.

1) What are the forces at each load cell.
2) Assume that the system eventually hits a wall but F continues to be applied. *What are the forces a each load cell after impact.

-(1)-A-(2)-B-(3)-WALL

The Attempt at a Solution


1)*
The force at location (1) is:

u1*N1+u2*N2


Where u represents friction coeff and N is normal force.
The force at location (2) is:
u2*N2

The force at location (3) is simply zero.

In the case where*
The force at location (2) is*

2)*
Once we hit the wall we have a change in the reactionary forces.
The reactionary forces cancel out and the force throughout the entire chain is F.
Position 1 is F, position 2 is F, position 3 is F.
 
Last edited:
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Thoughts?.
 
Some things are not clear to me. Where exactly is the force acting on? Is it acting on load cell 1? Is it pushing load cell 1 (which is connected to the rest of the system of course) to the right?
 
Hello. Thank you for your response. The force is applied to load cell numb one and is pushing right.
 
I typed the question up in another application and just noticed that it did not properly paste all of the information. I can no longer edit the original post so I will provide the information below.

1. Homework Statement

Blocks A and B are on a horizontal surface and connected to one another by a load cell.
Block A has mass m1 and Block B has mass m2.
Each block has difference coeff of static and dynamic friction.
Load cells are connected to the left and right of the blocks.
A constant right horizontal force F1 is applied to load cell (1)
All blocks accelerate at the same rate as the load cells are rigid connections.
The applied force is much greater than the frictional force.

Diagram

-(#)- = load cell

Force-->-(1)-A-(2)-B-(3)-


Questions.

1) What are the forces at each load cell?
2) Assume that the system eventually hits a wall but F continues to be applied. What are the forces a each load cell after impact?

Force-(1)-A-(2)-B-(3)-WALL

3. The Attempt at a Solution
1)
The force at location (1) is:

u1*N1+u2*N2 as this is the force require to move the blocks. Since there is a net force they will be accelerating to the right. Assume the static and dynamic friction coeff is the same to keep things simple.


Where u represents friction coeff and N is normal force.
The force at location (2) is:
u2*N2

The force at location (3) is simply zero.

In the case where the system is against a wall...
The force at location (2) is:

2)
Once we hit the wall we have a change in the reactionary forces.
The reactionary forces cancel out and the force throughout the entire chain is a constant F.
Position 1 is F, position 2 is F, position 3 is F.
 
Last edited:
The way of solving this is to draw free body diagrams for all of the objects and apply Newton's laws.

If we take the load cells to be similar to rigid rods, then the compression forces on those rods would correspond to the readings on the load cells.

Your answer for question (1) is correct if the whole system is traveling at constant speed. But since you say the applied force F is much greater than the total frictional force, the system has to travel with acceleration. If you apply Newton's laws for each object, you'll see that the compression forces (readings) that can be calculated for the load cells are different when there is an acceleration.

For question (2), your readings for the load cells for the case when the system is stationary against a wall are correct.
 
Omoplata,

Thank you, this all makes perfect sense. If I may throw one more variable into the mix.

Assume I apply both a left horizontal force to load cell three and the fixed wall moves slowly moves back and forth with a sinusoidal x component. F1 is greater than F2 and both are much greater than the frictional forces.

The load cells and blocks start under compression. Are the forces at each load cell difference as the wall vibrates?
 
Is the force F2 the reaction force between the wall and load cell three, or is it another force that someone applies on load cell three?

Either way, all you have to do is draw free body diagrams for all the objects and apply Newton's laws. Try it, and post the results here.
 
It is the force between the wall and load cell three. I will try free body diagrams as suggested.
 

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