00:01
Hello, here we have to solve a given problem.
00:03
We have to find magnitude of the tension in each cable.
00:08
So the system is at the equilibrium, that's why a vector sum of t1, t2, and mg equals to 0.
00:19
Let's introduce y and x axis, and let's project this equation on both axes.
00:30
So let's start with x -axis.
00:33
So here t2x minus t1 equals to 0 and projection onto y -axis is t2y minus mg equals to 0.
00:52
So we have to find the projection of t2 onto 2 axis so if this angle is teta 2 then this angle alpha equals to angle theta 2 which is 3802 which is 380 0 .5 degree.
01:18
So it means that the first equation becomes t2 times cosine alpha equals to t1 and the second equation becomes t2 times sine alpha equals to mg.
01:45
So let's divide the second equation by the first, sorry, the first equation, let's divide the second equation by the first equation.
01:55
It becomes sine alpha over cosine alpha equals to m g over t1 so that equals to tangent alpha so therefore t1 equals to m g over tangent alpha that is m g over tangent of the angle tetal 2 that is 27 .2 kilograms multiplied by 9 .81 meters per second squared divided by tangent of angle theta 2 which is 38 and a half degree...