00:01
Hello students, let's solve this question here.
00:04
For cot 3, the mass m3 is equal to 50 gm and tension t3 will be m into g which is equal to 50 into 10 raise to minus 3 into 10.
00:21
Therefore, tension t3 value is 0 .5 n.
00:26
Let's draw the free body diagram i .e.
00:30
Tension t1, t2 and t3 acting in opposite directions.
00:40
So, considering the cartesian sign at equilibrium, we can write t1 of x is equal to t1 into cos of 30 degree it will be equal to 0 .866 t1 and t1 of y is equal to t1 into sin 30 degree it will be equal to 0 .5 t1.
01:09
Now, t2 is equal to t2 into cos 60 degree and the value is 0 .5 t2 and t2 of y is equal to minus t2 into sin of 30 degree it will be 0 .5 t1.
01:32
Now, t3 will be equal to t3 equal to 0 .5 n.
01:42
And t3 of y will be 0.
01:46
Now, at equilibrium the net force acting on the ring will be 0.
01:50
Therefore, along the x direction fnet at x is equal to 0.
01:59
T1 of x plus t2 of x plus t3 of x is equal to 0 i .e.
02:09
0 .866 t1 plus 0 .5 t2 minus 0 .5 is equal to 0.
02:18
Let's mark it as equation number 1...