Equilibrium of a uniform bar inclined at an angle

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Homework Help Overview

The problem involves a uniform bar resting in equilibrium on a smooth inclined plane, with specific angles given for the plane and the bar. The task is to analyze the forces acting on the bar and determine the reactions at the points of contact.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the two angles provided in the problem and question their relationship to the bar and the plane. There is uncertainty about the orientation of the angles and the lack of friction coefficients. Some participants suggest drawing diagrams to clarify the setup.

Discussion Status

The discussion has explored various interpretations of the problem geometry and the forces involved. Participants have shared diagrams and equations, but there remains some confusion regarding the relationships between the angles and the forces acting on the bar. There is no explicit consensus on the correct interpretation yet, but several productive lines of reasoning have been proposed.

Contextual Notes

Participants note the absence of certain variables that could aid in solving the problem, such as the size of the peg and the direction in which the angles are measured. The problem is described as poorly worded, leading to varied interpretations.

  • #31
I've got the sum of forces in y and the moments. Just don't have the sum of forces in x.

Are the ones I have correct?
 
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  • #32
I'll let you work this problem out the most obvious way for now .

There is a more elegant method possible - perhaps we'll look at that later if you are interested .
 
  • #33
Yes I'd like to know what the way method is as well?
 
  • #34
Use the diagram in post #16 as a basis for doing a clear free body diagram for the bar .

Show us the three equations that you should now have and , if possible , your final solution .
 
  • #35
I already drew the free body diagram on the basis of post 16 and attached it in one of my previous posts and I already mentioned the 2 equations I get from the forces in the y direction and the moment around A, just doing it for the x direction now and that should give me the answer.
 
  • #36
Solved! Thank you
 

Attachments

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  • #37
Good !
 
  • #38
Nidum said:
Good !
I know this is from a few months ago, but why do the normal vectors at C and A break up into components?
 
  • #39
I don't think this is right. The normal at A and C don't break up into components and what direction is the normal at A?
 
  • #40
Sorry, mind's a little messed up. nothing's wrong
 
  • #41
OK .
 
  • #42
Well done :)
 

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