What is the net force on the system without friction?

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Homework Help Overview

The discussion revolves around determining the net force on a system of blocks without friction. Participants are analyzing the forces acting on each block and attempting to derive equations of motion based on their diagrams and assumptions about acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to write equations based on the forces acting on each block, including tensions and weights. There is discussion about the correctness of the equations and the signs used. Some participants question whether the acceleration is the same for all blocks and how to proceed with solving for it.

Discussion Status

There is a mix of attempts to clarify the setup and derive equations. Some guidance has been offered regarding checking the equations and considering the system as a whole. Multiple interpretations of the forces and their relationships are being explored, but there is no explicit consensus on the next steps.

Contextual Notes

Participants are working under the assumption that there is no friction and are trying to reconcile their equations with the physical setup described. There is uncertainty regarding the correct application of forces and the resulting equations.

canicon25
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Homework Statement



determine the acceleration of the system. there is no friction on the surface
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Homework Equations


Fnet=ma

The Attempt at a Solution



Left block
Ft=9.81+a1

Middle block
Ft=Fs-7a2
Right block
Fs=5a3+49.1

Not sure how to proceed with this one. What is equal to what?
 
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The first thing to do is draw the forces acxting on each object on the diagram.
The 1kg mass has weight acting down and a tension (T1) acting up. The 5kg mass has a weight acting down and a tension (T2) acting up.
The 7kg mass has the 2 tensions acting on it.
You should be in a better position to write down the resultant force equations with this diagram
 
i did make diagrams, i just didn't post them.

Ft = T1
Fs = T2

the equations i wrote are from the diagrams

left
T1-mg=ma
T1-9.81=a
T1=9.81+a
right
T2-mg=ma
T2-49.1=5a
T2=49.1+5a

The middle block would be

-T1-(-T2)=ma
-T1+T2=ma
-T1+T2=7a

i am assuming the acceleration is the same for all? can it be solved for acceleration this way? please help.
 
Yes, your method is correct. The individual equations look okay, though it can be tricky to get all the signs right. Check your work by thinking of the whole system as one 13 kg mass with forces 5g and -1g so net of 4g to the right. Use F = ma on that.
 
canicon25 said:
right
T2-mg=ma
T2-49.1=5a
T2=49.1+5a

Would it be T2 - mg ?
 

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