To find the forces acting in a system in equilibrium

AI Thread Summary
The discussion revolves around resolving forces in a system in equilibrium, specifically focusing on the calculations involving a ring and a rod system. The initial confusion stemmed from incorrectly accounting for the torque produced by the weight of rod BC when analyzing the forces acting on the ring. Upon realizing that the system should be considered as a whole, including both the ring and rod BC, the correct torque was identified, leading to the resolution of the problem. The participants noted that as the angle between the bars increases, the horizontal reaction forces and necessary friction at the ring also increase significantly. This highlights the importance of understanding the mechanics involved in such systems for accurate calculations.
gnits
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Homework Statement
To find the forces acting in a system in equilibrium
Relevant Equations
moments
Could I please ask where I have gone wrong with my reasoning in the following question:

nowp.JPG


The answers in given in the book are:

(1/2)W tan(Ө)
W
vertically
(1/2)W tan(Ө) horizontally

Here is my diagram:

now.png

Considering the system as a whole:

(In the text below "Ya" and "Xa" are the forces at the hinge at A)

Resolving vertically gives:

Ya + R = 2W + ω (call this "equation 1")

Resolving horizontally gives:

F = Xa (this agrees with the book answer)

Taking moments about C gives:

W * a sin(Ө) + W * 3a * sin(Ө) = Ya * 4a * sin(Ө)

which gives:

Ya = W (this agrees with the book answer)

And so from "equation 1" we have that R = W + ω (call this "equation 2")

Now, considering the ring alone and taking moments about B gives:

R * 2a * sin(Ө) = ω * 2a * sin(Ө) + F * 2a * cos(Ө)

which gives:

R * tan(Ө) = ω * tan(Ө) + F

So using "equation 2" gives:

F = W * tan(Ө)

But book answer is F = (1/2) * W * tan(Ө)

Where have I reasoned wrongly?

Thanks for any help.
 
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gnits said:
Now, considering the ring alone and taking moments about B gives:

R * 2a * sin(Ө) = ω * 2a * sin(Ө) + F * 2a * cos(Ө)
You have forgotten the moment (torque) produced by BC's weight.
 
Steve4Physics said:
You have forgotten the moment (torque) produced by BC's weight.
Thanks very much for that. Indeed, if I include the torque produced by BC's weight everything works out.

This torque is given by W * a sin(Ө)I guess I am having trouble justifying to myself, why, when I am considering forces on the ring only, I need to include W due to the weight of BC in this way. If I imagine a diagram with the ring only then, if anything, I would label the force on the ring due to BC as being an arrow pointing vertically downwards from the ring of magnitude W. But then it's moment about B would be be

W * 2a * sin(Ө)

not W * a * sin(Ө) as I use above to get the right answer?

How am I thinking of this wrongly?

*EDIT* I think that my confusion is one of visualization. I think that, "mentally", I should be thinking in terms of considering "the rod BC and the ring" as the entity which I am considering in isolation, rather than just "the ring", then indeed I would use W * a sin(Ө)
 
Last edited:
You have a mechanism that would tend to relocate itself to the point of lowest potential energy, where bars AB and BC would become both vertical if nothing would be holding ring C in its position.
AB would simply rotate, while BC must rotate and move, while following the circular trajectory of point B on one end and a horizontal trajectory on the other end.
 
Thanks all for the replies. They have helped me to solve to problem.
 
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gnits said:
Thanks all for the replies. They have helped me to solve to problem.
As the angle between the bars increases, the magnitude of the horizontal reaction at pivot A and the necessary friction force at ring C dramatically increases.
At 180 degrees, the value of those forces will be infinite.

Please, see this practical application of this principle of increasing mechanical advantage:
https://roperescuetraining.com/physics_angles.php

You are welcome. :)
 
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