In the problem the ring that is threaded in the string has a mass of 0.1 kg. The string is attached at the ends A and B with the bead hanging in the middle at point R

The tension is the same throughout the string. The end AR is inclined...

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In the problem the ring that is threaded in the string has a mass of 0.1 kg. The string is attached at the ends A and B with the bead hanging in the middle at point R

The tension is the same throughout the string. The end AR is inclined at an angle 40 degree to the horizontal. We can divide the tension T in this portion into a horizontal and a vertical component given by T*cos 40 and T*sin 40 respectively. As the ring is stationary there is no net force acting on it. The tension in the other end RB is also equal to T but acts in the opposite direction. Let the part RB form an angle A with the horizontal.

So we have T*cos A = T* cos 40

=> A = 40 degrees.

The part RB of the string is also inclined at 40 degrees to the horizontal.

The vertical components of the two parts add up the weight of the ring.

2*T*sin 40 = 0.1*9.8

=> T = 0.1*9.8 / 2*sin 40

=> T = 0.7623 N

**The tension in the string is 0.7623 N, and the part RB is also inclined at 40 degrees to the horizontal.**