Determining the Angle Formed at the Middle of the String in a Pulley-block system

Question:

Determine tha angle formed at the middle of the string where a block of mass $${\sqrt2 \ m}$$ is hanging as shown below where the pulleys are smooth and the string used is inextensible.

Solution:

Free body diagram of the blocks of different masses:

Equilibrium at the mid-point:

Equilibrium refers to the situation when the net external force on the system is zero,Central block is in equilibrium.

Vertical force:

The vertical component of the tension must balance its weight.

The vertical component of the tension T from bothsides is

$${2T\cos\theta}$$

The weight of the central block is

$${\sqrt{2} \ mg}$$

The vertical force balance each other so that

$${2T\cos\theta=\sqrt{2} \ mg}$$

Horizontal force:

The horizontal components of the tension cancel out each other since there is no horizontal movement.

Determining the value of angle formed:

From the vertical force balance equations

$${2T\cos\theta=\sqrt{2} \ mg}$$

$${2 mg \cos \theta=\sqrt{2} \ mg}$$

$${\cos\theta=\frac{\sqrt{2}}{2}}$$

$${\cos\theta =\frac{1}{\sqrt2}}$$

$${\theta = 45^\circ}$$

Conclusion:

The angle formed at the middle of the string is 45^o

(total angle between the two segment is 45^o+45^o=90^o )