Solve Physics Question on Ring Equilibrium w/ Masses & Friction

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In summary: Same goes to the other ring .any other response ?In summary, the rings are in equilibrium due to the static friction between the ring and the rough wire. The tension in the string provides the downward force. The upward force is the gravity pull.
  • #1
thereddevils
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Homework Statement



The diagram above shows two rings , A and B with masses 2m and m respectively which are threaded through a piece of rough wire that is fixed horizontally. THe coefficient of friction between ring A and the wire is \mu whereas the coefficient of friction between ring B and the wire is 2\mu.One end of a light inextensible string is tied to ring A and the other end is tied to ring B. Two particles P and Q each with mass m are attached to the string and hang freely,with the portion od the string PQ horizontal and portions AP and BQ each making an angle of alpha and beta with the vertical.The system is in a limiting equilibrium in a vertical plane.

(1) Show that alpha=beta

(2)Find the normal reactions at ring A and ring B in terms of m and g

(3) Determine , between ring A and B , which ring is about to slide

Homework Equations




The Attempt at a Solution



(1) Since the system is in equalibrium , resultant force on both particles P and Q is zero.

Consider vertical components ,

For particle P , [tex]T_1\cos \alpha + \mu W=W[/tex] --- 1

For particle Q , [tex]T_2\cos \beta+\mu W=W[/tex] ---2

Consider horizontal componenets , [tex]T_1\sin \alpha=T_2\sin \beta[/tex] ---2

Then i will need to play around with these 3 equations, before i proceed , i would like to check if my set up is correct . THanks
 

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  • #2
thereddevils said:

Homework Statement



The diagram above shows two rings , A and B with masses 2m and m respectively which are threaded through a piece of rough wire that is fixed horizontally. THe coefficient of friction between ring A and the wire is \mu whereas the coefficient of friction between ring B and the wire is 2\mu.One end of a light inextensible string is tied to ring A and the other end is tied to ring B. Two particles P and Q each with mass m are attached to the string and hang freely,with the portion od the string PQ horizontal and portions AP and BQ each making an angle of alpha and beta with the vertical.The system is in a limiting equilibrium in a vertical plane.

(1) Show that alpha=beta

(2)Find the normal reactions at ring A and ring B in terms of m and g

(3) Determine , between ring A and B , which ring is about to slide

Homework Equations




The Attempt at a Solution



(1) Since the system is in equalibrium , resultant force on both particles P and Q is zero.

Consider vertical components ,

For particle P , [tex]T_1\cos \alpha + \mu W=W[/tex] --- 1

For particle Q , [tex]T_2\cos \beta+\mu W=W[/tex] ---2

Consider horizontal componenets , [tex]T_1\sin \alpha=T_2\sin \beta[/tex] ---2

Then i will need to play around with these 3 equations, before i proceed , i would like to check if my set up is correct . THanks


any help ?
 
  • #3
What's W?
 
  • #4
diazona said:
What's W?

W=mg , the gravity pull
 
  • #5
OK, then there's something that doesn't seem right with your equations. What forces (or components of forces) act on particle P in the vertical direction?
 
  • #6
diazona said:
OK, then there's something that doesn't seem right with your equations. What forces (or components of forces) act on particle P in the vertical direction?

this is what i had in mind , the upwards force is the limiting static friction and the tension in the string , the downward force is simply the gravity pull and since they are in equalibrium , they equal each other . Same goes to the other ring .
 
  • #7
any other response ?
 

1. What is the concept of equilibrium in physics?

The concept of equilibrium in physics refers to the state in which the net force acting on an object is equal to zero. This means that the object is either at rest or moving with constant velocity. In other words, the forces acting on the object are balanced and there is no acceleration or change in motion.

2. How do you determine the equilibrium position of a ring with masses and friction?

To determine the equilibrium position of a ring with masses and friction, you need to consider the forces acting on the ring. These include the weight of the ring, the weight of the masses attached to it, and the frictional force. You can use Newton's laws of motion and the equations for static equilibrium to calculate the position at which these forces are balanced.

3. What factors affect the equilibrium of a ring with masses and friction?

The equilibrium of a ring with masses and friction is affected by several factors, including the weight of the ring and masses, the coefficient of friction, and the distance between the ring and the support point. These factors determine the magnitude and direction of the forces acting on the ring, and thus, affect its equilibrium position.

4. Can the equilibrium position of a ring with masses and friction change?

Yes, the equilibrium position of a ring with masses and friction can change if any of the factors affecting it change. For example, if the weight of the masses increases, the equilibrium position will shift. Similarly, if the coefficient of friction or the distance between the ring and the support point changes, the equilibrium position will also change.

5. What is the significance of studying equilibrium in physics?

Studying equilibrium in physics is important because it helps us understand the conditions under which an object is at rest or moving with constant velocity. This knowledge is essential in various applications, such as designing stable structures, predicting the motion of objects, and analyzing the forces acting on systems. Additionally, understanding equilibrium is crucial in further studies of mechanics and other branches of physics.

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