Mechanics question please help

Question

A particle of mass m suspended from a fixed point A by a light inextensible string which passes through a small light ring R. The ring is attached to a fixed point B, at the same horizontal level as A, by a light inextensible string of length L (where L<AB). The system hangs at rest in equilibrium. Denoting the angle between AR and the vertical by 2z, show that the angle ABR =z and find, in terms of m,g,and z, the tension in the string BR.

Answer 1

[I apologize; I realized my original solution to the previous part contained an error]

 

I was not able to determine that ABR = z if the indicated angle is 2z.  In fact, I think there is a counterexample that shows that, with the given information, that is not necessarily true.  Imagine the case where the ring hangs directly below B.  Then ABR = 90.  If that were z, and the angle between AR and the vertical were 2z, that would mean the angle between AR and the vertical would have to be 180 degrees, which it is clearly not (in fact, it is at most 90 degrees, depending on how long the string supporting the ring is).

 

If I am incorrect, I apologize.  But I think the statement that ABR=z is not necessarily true, with the information given.  Is there some other constraint? Or did you mean an angle besides ABR?  Just let me know if there is any more information.

 

That said, there is probably some position of the ring where ABR = z when the angle between AR and the vertical is 2z.

 

If we take as given that ABR = z, the second part about the tensions can still be solved.

 

Assuming ABR=z, getting the tension in BR is a matter of setting up a system of two equations with two unknowns  The two unknowns are the tension in string AR (T_AR) and the tension in string BR (T_BR).  The two equations come from Newton's 2nd law for both the x and y directions, written knowing that the system is in equilibrium (meaning a = 0).  Look at the position of the ring, and the three forces coming off of it:  T_AR, T_BR, and the weight of the mass (applied directly downward through the light, inextensible string).  With those forces, and the geometry of the diagram, Newton's 2nd law in the horizontal and vertical directions are:

 

-T_ARsin(2z) + T_BRcos(z) = 0

T_ARcos(2z) + T_BRsin(z) - mg = 0

 

I defined, in my drawing, to the right and up as positive, and put in the signs I knew for the forces from my drawing.

 

Make sure that you understand how these equations are obtained, given your drawing.  We can talk about it in more detail online, if you like.

 

After cross-substitution, and using more trig identities (specifically, double-angle identities), I obtained:

 

T_BR = 2mgsin(z)

 

If you have any questions about these setups, please let me know.  If you want to go into more detail about how to get from those starts to those ends, please let me know that, too; but do try to fill those steps in yourself first.  I am not 100% guaranteeing the algebra, as I had to do it quickly, but the technique is sound.  

 

I am available for online physics and math tutoring, if you would like to talk about these or any other physics and math topics in more detail.