Equilibrium and Tension in Hinged Uniform Beam with Suspended Mass

To determine which statements are correct, we analyze the static equilibrium of the horizontal beam.

1. Geometry of the System

Let the hinge at point $O$ be the origin $(0,0)$.

• Point $P$, where the block is suspended, is at $(L,0)$.
• Point $Q$, where the upper end of the rope is fixed, is at $(0,L)$.

The angle $\theta$ that the rope $QP$ makes with the horizontal beam $OP$ is given by: $\tan \theta = \frac{OQ}{OP} = \frac{L}{L} = 1 \implies \theta = 45^\circ$

Thus, $\sin \theta = \cos \theta = \frac{1}{\sqrt{2}}$.

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2. Forces Acting on the Beam

The forces acting on the beam are:

1. Weight of the beam $W$, acting downward at its center of mass $\left(\frac{L}{2}, 0\right)$.
2. Weight of the block $\alpha W$, acting downward at point $P(L, 0)$.
3. Tension $T$ in the rope acting along $PQ$ at point $P(L, 0)$:
   • Horizontal component: $T_x = T \cos 45^\circ = \frac{T}{\sqrt{2}}$ (towards the left)
   • Vertical component: $T_y = T \sin 45^\circ = \frac{T}{\sqrt{2}}$ (upward)
4. Hinge reaction force at $O$:
   • Horizontal component: $R_x$
   • Vertical component: $R_y$

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3. Rotational Equilibrium

Taking the torque about the hinge $O$ to be zero ($\sum \tau_O = 0$): $\left(T \sin 45^\circ\right) \cdot L - W \cdot \left(\frac{L}{2}\right) - (\alpha W) \cdot L = 0$

Dividing by $L$: $\frac{T}{\sqrt{2}} = W \left(\frac{1}{2} + \alpha\right)$

Solving for tension $T$: $T = \sqrt{2} W \left(\frac{1}{2} + \alpha\right)$

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4. Translational Equilibrium

Horizontal Direction ($\sum F_x = 0$):

$R_x - T \cos 45^\circ = 0 \implies R_x = \frac{T}{\sqrt{2}}$

Substituting $\frac{T}{\sqrt{2}} = W \left(\frac{1}{2} + \alpha\right)$: $R_x = W \left(\frac{1}{2} + \alpha\right)$

Vertical Direction ($\sum F_y = 0$):

$R_y + T \sin 45^\circ - W - \alpha W = 0$ $R_y = W(1 + \alpha) - \frac{T}{\sqrt{2}}$

Substituting $\frac{T}{\sqrt{2}} = W \left(\frac{1}{2} + \alpha\right)$: $R_y = W(1 + \alpha) - W\left(\frac{1}{2} + \alpha\right) = \frac{W}{2}$

Notice that $R_y = \frac{W}{2}$ is a constant and does not depend on $\alpha$.

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5. Analysis of Options

• Option A: The vertical component of reaction force at $O$ is $R_y = \frac{W}{2}$, which is independent of $\alpha$.
  Statement A is correct.

• Option B: For $\alpha = 0.5$: $R_x = W \left(\frac{1}{2} + 0.5\right) = W$
  Statement B is correct.

• Option C: For $\alpha = 0.5$: $T = \sqrt{2} W \left(\frac{1}{2} + 0.5\right) = \sqrt{2} W \neq 2W$
  Statement C is incorrect.

• Option D: The maximum tension the rope can sustain is $T_{\max} = 2\sqrt{2} W$.
  The rope breaks when $T > T_{\max}$: $\sqrt{2} W \left(\frac{1}{2} + \alpha\right) > 2\sqrt{2} W$ $\frac{1}{2} + \alpha > 2 \implies \alpha > 1.5$
  Statement D is correct.

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Conclusion

The correct statements are A, B, and D.