Mechanics question please help

Question

A particle Y of mass 2m connected to a fixed point X by a light inextensible string. It is also connected to a particle Z of mass m by a light inextensible string which passes over a light pulley Q, which is at the same horizontal level as X. The system is in equilibrium in a vertical plane, with string XY making an angle G with the horizontal and string XY is perpendicular with the part of the string YQ. Find the exact value of the tension in string XY and the value of G.

Answer 1

Hi Orawan:

 

    Same setup as the last problem, just with some different variable names.  But the key is still setting up Newton's 2nd law on the positions of Particle Z and Particle Y.  At Particle Y, in particular, this involves setting up the horizontal and vertical equilibrium expressions for Newton's 2nd law, to solve for two unknowns.  In this case, the unknowns are the tension in string XY (T_xy) and the angle G between XY and the horizontal.

 

The tension in the string over the pulley is mg.  Make sure you understand why this is.  We can talk about it further online if you have questions about it.

 

Then, the horizontal and vertical equilibrium equations from Newton's 2nd law for Particle Y are:

 

T_xycos(G)-mgcos(90-G) = 0

T_xysin(G)+mgsin(90-G)-2mg = 0

 

Consulting your diagram, you need to make sure you understand how these equations are set up, based on the orientation of all the forces acting at Particle Y.

 

You can then solve these two equations for the two unknowns, T_xy and G.  The solutions involve some trig ratio identities, a Pythagorean trig identity, and algebra.  My solutions are:

 

G = 60^o or π/3 rad

T_xy = mgtan(60^o) ~ 1.73mg

 

If you want to go into more detail about the steps involved here, we can set up an online session to follow the steps more closely, with drawings.  Again, no guarantees about the absolute correctness of my algebra, but the technique is definitely the right path.