What is the net force on the system without friction?

AI Thread Summary
The discussion revolves around determining the net force and acceleration of a system of blocks without friction. Participants analyze the forces acting on each block, including tension and weight, and formulate equations based on these forces. They emphasize the importance of drawing diagrams to visualize the forces and clarify relationships between tensions and masses. The consensus is that the acceleration is likely the same for all blocks, and they suggest using the entire system's mass to simplify calculations. The conversation highlights the need to ensure the correct signs in the equations for accurate results.
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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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