The vertical component of reaction force at $O$ does not depend on $\alpha$
The horizontal component of reaction force at $O$ is equal to $W$ for $\alpha = 0.5$
The tension in the rope is $2 W$ for $\alpha = 0.5$
The rope breaks if $\alpha > 1.5$

Solution:

Since

O Q = O P

\therefore \angle P = \angle Q = 45^{\circ}

At equilibrium

R_y + \frac{T}{\sqrt{2}} = W + \alpha W

.....(1)

and

R_x = \frac{T}{\sqrt{2}}

.....(2)

Torque about point '

O

W \frac{L}{2} + \alpha W L = \frac{T}{\sqrt{2}} L

\therefore T = \sqrt{2} \left( \frac{W}{2} + \alpha W \right)

.....(iii)

\therefore R_x = \frac{T}{\sqrt{2}} = \left( \frac{W}{2} + \alpha W \right)

Therefore for

\alpha = 0.5

R_x = \frac{W}{2} + \alpha W = \frac{W}{2} + 0.5 W

R_x = W

i.e., the horizontal component of reaction force at, 0 ,

R_x = W

for

\alpha = 0.5

Now torque about point

P

T_y L = W \frac{L}{2}

\Rightarrow R_y = \frac{W}{2}

The vertical component of reaction force at

O

does not depend on

\alpha

As per question, rope can sustain a maximum tension of

2 \sqrt{2} W

\therefore 2 \sqrt{2} W = \sqrt{2} \left( \frac{W}{2} + \alpha W \right)

\Rightarrow 2 = \frac{1}{2} + \alpha \quad \therefore \alpha = \frac{3}{2}

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