00:01
This problem we have given a panata of mass 8 kg is 8 kg is attached to a rope of negligible mass that is strung between the tops of two vertical poles.
00:15
This is left pole and this is right pole the distance between these two poles is 2 meter and this right pole is given it is 0 .5 meter higher than the top of the left pole so this is 0 .5 meter and it is given that the penanta is attached to the rope at a horizontal position halfway between the two poles that means on this line and a vertical distance one meter below the top of the left pole so so here is suppose finanta this distance is 1 meter and this distance is also 1 meter like this this rope is connected this distance is 1 .5 meter also sorry 1 meter because it is half way between the two poles so it is 1 meter now we can find this angle theta theta is equal to 45 degree right we have to find the tension in each part of the rope due to weight of the finata right so we know that suppose tension in this ring is t1 and this is t2 and suppose this angle is alpha okay this distance is 1 meter so this total distance is 1 .5 meter so so we can find this alpha angle also.
02:26
We know that 10 alpha will be equal to 1 .5 divided by 1.
02:33
So alpha will be equal to 10 inverse 1 .5.
02:38
So this is equal to 10 inverse 1 .5 which is equal to 56 .3 degree.
02:54
This alpha is 56 .3 degree.
02:57
We know there this penanta is in equilibrium position.
03:01
So the net force in horizontal direction should be equal to 0.
03:06
Net force in horizontal direction should be equal to 0.
03:10
So the horizontal component of t1, which is t1 cos theta, should be equal to t2, cos alpha...