What is the tension in the string in the following case:
A small smooth ring R of mass 0.1 kg is threaded on a light string. The ends of the string are fastened to two fixed points A and B. The ring hangs in equilibrium with the part of the string AR inclined at 40 degrees to the horizontal. Show that the part of RB of the string is also inclined at 40 degrees to the horizontal, and find the tension in the string.
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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.
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