A Bead Of Mass m is attached to one end of a spring of natural length R And spring constant
K = (√3+1)mg/R . other end of the spring is attached toa point A ON the vertical ring of radius R MAKING AN ANGLE OF 30 DEGREE WITH ITS CIRCUMFERENCE THE NORMAL REACTION AT THE BEAD WHEN THE SPRING IS ALLOWED TO MOVE FROM POINT A
First of all we have to calculate total extension in spring (x)
As we know angle at A is 30 degree
And at B also 30 degree ( AOB isosceles traingle )
also angle at AOB is 120 , AO = R , BO= R
so as per cosine rule we can calculate the third side
a^2 +b^2 - c^2 = 2ab cos (120)
So c= √3R
So the the extension will be
= (√3R- R)
= (√3 - 1 ) R
The spring force is
Fs = k(x) = (√3+1)mg/R × ( √3 -1 )R
=(√3 + 1)(√3 - 1 )mg
= 2mg
The weight of bead is
W = mg
The projection of spring force on the normal is
2mg cos 30 = mg√3
The projection of weight on the normal is
Mg cos 30 = mg√3/2
.
The sum of all forces in the normal direction must be zero:
N + mgcos 30 + kx cos30 = 0
N= - mg√3 - mg√3/2
N = -3√3mg/ 2 ( -ve shows direction)