00:01
Hi, here in this given problem, first of all this is the vertical rod attached with this vertical rod.
00:12
There are two strings, massless strings of equal lengths and holding a block of mass 4 .00 kilogram and this block is being rotated in a horizontal circle.
00:33
Length of each string 1 .25 meter and the gap between the ends of the string which are attached with the vertical rod.
00:46
That gap is 2 .0 meter.
00:50
So here this angle will be same.
00:57
So if you consider this right angled triangle in this right angle triangle, perpendicular of the triangle that will be same.
01:06
Be half of two means this is 1 .00 meter tension in the strings in the upper string t1 in the lower string this is t2 value of t1 is given to us as 80 newton and t2 that is missing so these tensions may be resolved into two components horizontal component for t1 horizontal component is t1 cost for the lower string t2, horizontal component is t2, cost theta and vertical components.
01:56
For the tension t1, that is t1, sine theta upward, and for t2, vertical component is downward, and that is t2, sine theta weight of the block, 4 into jit, that is also acting vertically downward.
02:16
In the right angled triangle, sine theta is given as perpendicular, which is 1 .00 meter, divided by hypotenuse 1 .25 meter, calculated to be equal to 0 .8.
02:33
So, this angle theta, that will be given as sine inverse, given by sine inverse of 0 .8, and it is calculated to be equal to 53 .1 degree.
02:50
In the first part of the problem where we have to find the value of t2 as the block is in translational equilibrium in vertical direction.
03:04
So we can say there should be no net force acting on the block in vertical direction.
03:10
Summation of fy should be zero.
03:12
It means t2 sine theta plus 4g acting vertically down.
03:21
That should be equal to t1 sine theta acting vertically up.
03:29
So t2, sine theta will be given by t1 sine theta minus 4g...