JEE Mains Mechanics (Equilibrium) 2

Above, you can see the forces acting on each end of one of the sections of the string. It is important to note that I denoted the forces going downwards $mg$ because those are their values, not because the weight of the balls are actually acting on those points. You can also note that the tension $T_2$ is equal in both ends, because it is said that is a light string.

We can now write Newton's second law for both ends as

Eq. 1: $T_{1}\sin\theta_{1}-T_{2}\sin\theta_{2}-mg=0$

Eq. 2: $T_{2}\cos\theta_{2}-T_{1}\cos\theta_{1}=0$

Eq. 3: $T_{2}\sin\theta_{2}-mg=0$

Note the sum of forces on the horizontal direction of the lower end has not been written. Replacing the value of $mg$ taken from Eq. 3 on Eq. 1 and re-writing Eq.1 and 2 we obtain

$T_{1}\sin\theta_{1}=2T_{2}\sin\theta_{2}$

$T_{1}\cos\theta_{1}=T_{2}\cos\theta_{2}$

If we divide these equations we get

$\tan\theta_{1}=2\tan\theta_{2}$

Since $\tan\theta_{1}=2$ , then $\tan\theta_{2}=1$ , which of course gives the answer of $\theta_{1}=45$ degrees.

Apply horizontal force equilibrium to the point where the leftmost ball is suspended from. The horizontal components of the tensions in the two l-length segments are equal. Then apply vertical force equilibrium to the point where the second ball is suspended from. The vertical component of the tension in the diagonal l-length segment equals the weight of the ball. Using this with vertical force equilibrium of the point where the leftmost ball is suspended from, the vertical component of the tension in the leftmost l-length segment equals twice the weight of the ball. Therefore, the second l-length segment has the same horizontal component, but half the vertical component of the leftmost l-length segment. Therefore, the tangent of the angle it makes is half the tangent of the other segment, so theta2 = arctan(2/2) = arctan(1) = 45 degrees.

BREAK THE VECTORS TO COMPONENTS IN I CAP AND J CAP