To find the forces acting in a system in equilibrium

In summary, when considering forces on a ring only, the weight of the bar at pivot A must be taken into account.
  • #1
gnits
137
46
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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  • #2
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.
 
  • #3
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:
  • #4
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.
 
  • #5
Thanks all for the replies. They have helped me to solve to problem.
 
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Likes Steve4Physics and Lnewqban
  • #6
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. :)
 

What is the definition of equilibrium in a system?

Equilibrium in a system refers to a state where all the forces acting on an object or system are balanced, resulting in no net movement or acceleration.

What are the different types of forces that can act in a system?

The different types of forces that can act in a system include gravitational force, normal force, frictional force, tension force, and applied force.

How do you find the forces acting in a system in equilibrium?

To find the forces acting in a system in equilibrium, you must first draw a free body diagram and identify all the forces acting on the object or system. Then, use the equations of equilibrium (sum of forces in the x-direction and y-direction equal to zero) to solve for the unknown forces.

Can an object be in equilibrium if it is moving?

Yes, an object can be in equilibrium even if it is moving. This is known as dynamic equilibrium, where the object is moving at a constant velocity with all the forces acting on it being balanced.

What happens to an object in equilibrium if an external force is applied?

If an external force is applied to an object in equilibrium, the state of equilibrium will be disrupted and the object will start to accelerate in the direction of the applied force. The object will continue to move until a new state of equilibrium is reached.

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