System of equilibrium, why won't the normal force apply?

AI Thread Summary
The discussion centers on the absence of a normal force in the analysis of a ring in equilibrium under tension and weight. Participants clarify that while the ring does contact the string, the normal force is considered internal and thus excluded from the equilibrium equations. The tension in the string and the weight of the ring are the primary forces to consider, as the normal force exerted by the string on the ring cancels out with the force exerted by the ring on the string. A free body diagram approach is emphasized to illustrate the forces acting on the system. Ultimately, understanding the system's boundaries and the nature of internal versus external forces is crucial for solving the problem correctly.
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Homework Statement



208670-ec244440bee12f7c3ec022a2b6b33841.jpg

Homework Equations

The Attempt at a Solution


I've resolved all the relevant forces (tension, weight, normal force), but upon looking at the mark scheme, there is no normal force to be resolved.
Forgive me if I'm missing out on something obvious, but shouldn't there be a normal force opposing the weight?
 

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SmamMir said:

Homework Statement



208670-ec244440bee12f7c3ec022a2b6b33841.jpg

Homework Equations

The Attempt at a Solution


I've resolved all the relevant forces (tension, weight, normal force), but upon looking at the mark scheme, there is no normal force to be resolved.
Forgive me if I'm missing out on something obvious, but shouldn't there be a normal force opposing the weight?
Show your work, please. Draw the forces exerted on the ring.
 
ehild said:
Show your work, please. Draw the forces exerted on the ring.
Re-did as neatly as I could.
Negating the normal normal force would allow me to easily find out the tension (as it should be?), but I can't quite grasp the logic behind it.
 

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Are you looking for something besides string tension? The string tension can be found using a free body diagram at joint R.
 
PhanthomJay said:
Are you looking for something besides string tension? The string tension can be found using a free body diagram at joint R.
No, just the string tension.
I tried, and accounted for the normal force, but the mark scheme doesn't. My only query is: why?
 

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SmamMir said:
Re-did as neatly as I could.
Negating the normal normal force would allow me to easily find out the tension (as it should be?), but I can't quite grasp the logic behind it.
What do you call R? there are no other forces but the tension in the string and the weight of the ring.
 
ehild said:
What do you call R? there are no other forces but the tension in the string and the weight of the ring.
The normal force?
My only question is, why won't there be a normal force? The ring does come in contact (rests on) with the ring.
 
SmamMir said:
The normal force?
My only question is, why won't there be a normal force? The ring does come in contact (rests on) with the ring.
The ring comes in contact with the string, and both pieces of the string act on it with the tension. The horizontal components of the tension forces cancel, and the vertical components add, and that balances the weight of the ring.
 
You may be thinking that normal force is the bearing force of the string in contact with the ring. That is a valid (although not usual) way of looking at it. But when you draw a free body diagram of the ring, you have the weight force and the tension forces acting on it. The "normal" force is internal in that diagram, and thus excluded in the equilibrium equations. You only look at external forces and internal forces that are "cut" free from the system. To find the 'normal force' you need to look at a different free body diagram which is a bit hard to find. The normal force ends up being equal to the weight of the ring. But don't include it in your original diagram!
 
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  • #10
Think of it this way. You are asked to find the tension in a string which is assumed massless. OK, then consider as your system the ring plus a little piece of the string that loops around the ring and has two ends on either side. The only forces acting on this system are the two tensions at each end of the string that is now the system and the weight of the ring. The "normal" force exerted by the string on the ring is equal and opposite to the "normal" force exerted by the ring on the string, so they cancel out. Remember that you can choose as your system anything you wish and, as long as all forces are accounted for properly and the correct equations are obtained from the FBD, who's to say that one system is more appropriate than another?
 
  • #11
Now I understand the OP's problem.
The ring with the string can be considered as two systems. One the ring with the small piece of the string in contact with it (the red spot)
The other one is the small piece and the two branches of the string.
The forces acting on the ring are its weight and the normal force from the red piece of string. N-W=0
The forces acting on the small red piece are the tension forces and the normal force from the ring. 2Ty-N=0
Adding up these equations, N cancels.

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