Since OQ=OP ∴‌‌∠P=∠Q=45∘ At equilibrium Ry+‌
T
√2
=W+αW .....(1) and Rx=‌
T
√2
.....(2) Torque about point ' O ' W‌
L
2
+αWL=‌
T
√2
L ∴‌‌T=√2(‌
W
2
+αW).....(iii) ∴‌‌Rx=‌
T
√2
=(‌
W
2
+αW) Therefore for α=0.5 Rx=‌
W
2
+αW=‌
W
2
+0.5W or Rx=W i.e., the horizontal component of reaction force at, 0 , Rx=W for α=0.5 Now torque about point P TyL=W‌
L
2
⇒‌‌Ry=‌
W
2
The vertical component of reaction force at O does not depend on α As per question, rope can sustain a maximum tension of 2√2W ∴‌‌2√2W=√2(‌