# A Bead Of Mass m is attached to one end of a spring of natural length R And spring constantK = (√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)
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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 • 114
law of motion is every action and opposite reaction

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Hi
K = ?3 + 1
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EASY

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MYIO
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