Prooving a system will travel up to 180 degrees

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Homework Help Overview

The discussion revolves around proving that a system can travel freely up to 180 degrees, involving a counterweight (m2) and a mass of a pan (m1 = 3 kg). The original poster has provided a free body diagram (FBD) and some equations related to the forces acting on the system.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to analyze the conditions under which the system will rotate freely, questioning the validity of their equation relating the weights and distances involved. Other participants seek clarification on the problem statement, the forces in the diagram, and the meaning of various terms and variables used in the equations.

Discussion Status

The discussion is ongoing, with participants providing clarifications and raising questions about the setup and the equations presented. There is a focus on understanding the components of the system and the relationships between them, with no consensus reached yet.

Contextual Notes

Participants note difficulties in interpreting the provided diagram and equations, highlighting potential ambiguities in the problem setup, such as the definition of terms and the pivot point of the system.

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Homework Statement


To prove that system will travel freely upto 180 degree
m2 is counterweight and m1 is mass of pan (=3kg)
i have attched the fbd or another link http://www.imagebam.com/image/54bb5a394377595

Homework Equations


m1(h + a sin θ) g x = m2 y h g
m2 = 9.13 kg

The Attempt at a Solution


we are stuck at the following equation
Pwg ≥( m1 x^2 + m2 y^2) sin2 θ / [ x(h-xsin2 θ)]
LHS is weight of pan and arm and if LHS>RHS, system will rotate freely
is this correct? can you pls explain it properly because it will clear my concepts.
Thanks.
 

Attachments

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Hello, welcome to PF :smile:

Could you clarify the problem statement a bit more? The diagram is difficult to read, I see no forces, what is what, what is the pivot point, what is a, what is P w (pr Pw ?), etc.
 
a is the length of bearing housing(the middle block which is the reference axis) hence a sin θ is added to h for balancing. my issue is the second equation. i cannot understand the RHS side.LHS is simply the wt of pan and tray.(ρwg).
can you help me with the equation on RHS
( m1 x^2 + m2 y^2) sin2 θ / [ x(h-xsin2 θ)]
Thanks a lot for a warm welcome in this wonderful community.
 
Still lost -- or rather not "in":

The thing on the right is a pan. What is a pan ? The thing in the middle is a bearing housing. Interesting. Does it pivot around some axis ? What axis ?

If the LHS was the weight of the pan and arm and now is the weight of the pan and tray, how come m1 doesn't feature in it ? What tray ?
What is s ?

What does the equation m1(h + a sin θ) g x = m2 y h g represent ? I see something in kg m3 /s2

What is ##\theta## ? What's the blue horizontal line ? And the black sloping line just underneath ?

THe fat black lines to L and R ? The thin black lines (where the h are mentionsed) ?

What is dangling from the bearing housing ? Where is the reference point for a ?

Are lengths measured in meters ?

What has to rotate 180 degrees ? Doesn't it bum into the fat line on the left ?
 

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