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
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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