Finding the Stretch in a Spring for a 2D Equilibrium Problem

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    2d Equilibrium
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Homework Help Overview

The discussion revolves around a 2D equilibrium problem involving a rod and a spring. Participants are exploring the forces acting on the rod and how to properly represent these forces in a free body diagram (FBD). The problem includes considerations of spring stretch as a function of the angle of the rod.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the rod, particularly the normals and the force in the spring. There are attempts to clarify the orientation of these forces and how they relate to the FBD. Questions arise about the unstretched and stretched lengths of the spring and how to calculate the amount of stretch based on the position of the rod.

Discussion Status

The discussion has progressed with participants providing hints and guidance regarding the representation of forces in the FBD and the relationship between the angle of the rod and the spring's stretch. Multiple interpretations of the spring's behavior and the forces involved are being explored, with some participants expressing confusion about specific aspects of the problem.

Contextual Notes

There is an ongoing discussion about the unstretched length of the spring, which is noted as unknown and unneeded for the calculations at hand. Participants are encouraged to focus on the height differences related to the spring's stretch as the rod moves from vertical to an angle.

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Homework Statement



Here is the problem
2a14yx.jpg


Homework Equations



Sum of the forces in x, y and moments

The Attempt at a Solution



I am having troubles drawing a good FBD.
 
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What forces act on the rod?

Hint: Find the spring stretch as a function of angle.
 
Doc Al said:
What forces act on the rod?

Hint: Find the spring stretch as a function of angle.

I know there are 2 normals that act on the bar (from the collars) but which way are the normals acting?
 
Which way do "normals" usually act? :wink:
 
Doc Al said:
Which way do "normals" usually act? :wink:

perpendicular, so the normal at the bottom is vertical and the normal at the top is horizontal?
 
Oblivion77 said:
perpendicular, so the normal at the bottom is vertical and the normal at the top is horizontal?
Exactly.
 
Ok, so where does the force in the spring come along in the FBD?
 
Oblivion77 said:
Ok, so where does the force in the spring come along in the FBD?
You tell me. How is the spring oriented? Which way does it push?
 
Would the force in the spring be tension going straight up?
 
  • #10
Oblivion77 said:
Would the force in the spring be tension going straight up?
Yep. You should have a good handle on that FBD now.
 
  • #11
Doc Al said:
Yep. You should have a good handle on that FBD now.

Thanks for the help!
 
  • #12
I am having some trouble seeing the stretched and un-stretched lengths of the spring. When the bar is vertical it looks like the un-stretched length is 0.
 
  • #13
Oblivion77 said:
I am having some trouble seeing the stretched and un-stretched lengths of the spring. When the bar is vertical it looks like the un-stretched length is 0.
The unstretched length of the spring is unknown and unneeded. What you do need is how much the spring is stretched when the bar is at an angle. When the bar is vertical the spring is unstretched, thus the the amount of stretch is zero.

Hint: When the bar is at some angle α, how far down has the end of the bar pulled the spring compared to when the bar was vertical?
 
  • #14
Is it something like 5sinα?
 
  • #15
Oblivion77 said:
Is it something like 5sinα?
Something like that--but not that. :wink:

Locate the right triangle whose hypotenuse is the bar. What's the height of that triangle? That's the height of the right end of the bar. Compare that to its height when vertical. That will tell you the amount the spring has been stretched.
 
  • #16
Ok, the hypotenuse is 5. So the adjacent side(vertical) is 5Cosα and opposite(horizontal) is 5Sinα. So is 5Cosα the stretched?
 
  • #17
Oblivion77 said:
So is 5Cosα the stretched?
No, it's the height of the spring end measured from the horizontal fixed bar. Compare that to where the spring was when the angled bar was vertical (and the spring unstretched). The difference in height will tell you the amount the spring was stretched.
 
  • #18
I still can't see what is happening. I am not good at spring questions. I know F=k(L-Lo). but earlier you said the un-stretched length was not needed.
 
  • #19
Oblivion77 said:
I know F=k(L-Lo). but earlier you said the un-stretched length was not needed.
You need L-Lo, but not Lo.

Do this. Find the height of spring end above the fixed horizontal bar when the 5m bar is vertical. (That should be easy!) Then find its height when the 5m bar is at an angle. (You've already done that.) The difference between those two heights is the amount of stretch in the spring (which is all that L-Lo is) when the bar moves from vertical to some angle.
 
  • #20
Doc Al said:
You need L-Lo, but not Lo.

Do this. Find the height of spring end above the fixed horizontal bar when the 5m bar is vertical. (That should be easy!) Then find its height when the 5m bar is at an angle. (You've already done that.) The difference between those two heights is the amount of stretch in the spring (which is all that L-Lo is) when the bar moves from vertical to some angle.

Thanks for all the help! I solved it now.
 

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