b. Why did you take 3cos(30)cos(30)?
What is the component of the weight of ring R downwards?
A small smooth ring R, of mass 0.6 kg, is threaded on a light inextensible string. One end of the string is attached to the fixed point A and the other end is attached to a ring B, of mass 0.2 kg, which is threaded on a fixed rough horizontal wire which passes through A (see diagram (excuse the obviously unequal angles)). The system is in equilibrium, with B about to slip and with the part AR of the string making an angle of 60° to the wire.
a. Explain, with reference to the fact that ring R is smooth, why the part BR of the string is inclined at 60° to the wire.
b. Show that the normal contact force between B and the wire has magnitude 5 N.
c. Find the coefficient of friction between B and the wire.
For a. I was unsure of what wording to use exactly, and answered the following:
a. Ring R being smooth and ring B being on the same horizontal plane as fixed point A, ring R will have slipped to the midpoint of the string: angles ABR and RAB are therefore equal.
Angle RAB being given as 60°, this means that the part BR of the string is inclined at 60° to the wire.
For b. however I have several problems. Firstly, of how the 6N weight of R is distributed over both string segments (RB and AR). I would assume that the each segment carries half the load, but I’m unsure of how to prove this. Secondly, even taking 3cos30 N (half of R’s weight resolved for the slope of the string) as the tension in the string segment BR, I end up with normal contact force = 2 (weight of B) + 3cos30(cos30) = 4.25 N
Any suggestions?