Solve Physics Question on Ring Equilibrium w/ Masses & Friction

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The discussion revolves around a physics problem involving two rings, A and B, with different masses and coefficients of friction, connected by a string with two hanging particles. The key points include establishing that the angles alpha and beta are equal due to the system being in equilibrium, and the need to derive the normal reactions at both rings in terms of mass and gravitational force. Participants are clarifying the forces acting on the particles, specifically the balance of tension, friction, and weight. There is also a focus on determining which ring is on the verge of sliding based on the frictional forces involved. The conversation emphasizes the importance of correctly setting up the equations of motion for accurate analysis.
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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 , T_1\cos \alpha + \mu W=W --- 1

For particle Q , T_2\cos \beta+\mu W=W ---2

Consider horizontal componenets , T_1\sin \alpha=T_2\sin \beta ---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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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 , T_1\cos \alpha + \mu W=W --- 1

For particle Q , T_2\cos \beta+\mu W=W ---2

Consider horizontal componenets , T_1\sin \alpha=T_2\sin \beta ---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 ?
 
What's W?
 
diazona said:
What's W?

W=mg , the gravity pull
 
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?
 
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 .
 
any other response ?
 
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