One end of a horizontal uniform beam of weight 𝑊 and length 𝐿 is hinged on a vertical wall at point O and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q, at a height 𝐿 above the hinge at point O. A block of weight $$\alpha W$$ is attached at the point P of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of ($$2\sqrt{2}$$)W. Which of the following statement(s) is(are) correct?

The beam OA is uniform, horizontal and of length $$L$$. It is hinged at the wall at $$O$$.
The light, inextensible rope is tied at the free end $$A$$ of the beam and at the point $$Q$$ which is vertically above $$O$$ by the same distance $$L$$.
Hence the rope is of length $$\sqrt{L^{2}+L^{2}} = L\sqrt{2}$$ and it makes an angle of $$45^{\circ}$$ with both the horizontal and the vertical.

Let

• $$W$$ = weight of the beam (acts at its mid-point $$L/2$$ from the hinge).
• $$\alpha W$$ = weight of the block hung at the point $$P$$ situated at a distance $$x$$ from the hinge.
• $$T$$ = tension in the rope.
• $$H$$ = horizontal component of the hinge reaction, taken positive towards the right (along +x).
• $$V$$ = vertical component of the hinge reaction, taken positive upwards (along +y).

Step 1 : Equilibrium of forces

Horizontal balance: $$H + \big(-T\cos 45^{\circ}\big)=0 \;\;\Longrightarrow\;\; H=\dfrac{T}{\sqrt{2}} \quad -(1)$$

Vertical balance: $$V + T\sin 45^{\circ} -W-\alpha W =0 \;\;\Longrightarrow\;\; V = W(1+\alpha)-\dfrac{T}{\sqrt{2}} \quad -(2)$$

Step 2 : Equilibrium of moments about the hinge O (anticlockwise positive)

The moment of the tension about $$O$$ is produced by the perpendicular distance $$L$$ of its line of action: $$\tau_{T}=+\,L\;T\sin 45^{\circ}= \dfrac{L\,T}{\sqrt{2}}.$$ The clockwise moments due to the two weights are

$$\tau_{W}= -\,W\left(\dfrac{L}{2}\right), \qquad \tau_{\alpha W}= -\,\alpha W\,x.$$

Setting the net moment equal to zero gives

$$\dfrac{L\,T}{\sqrt{2}} - W\left(\dfrac{L}{2}\right) - \alpha W x = 0 \;\;\Longrightarrow\;\; T = \sqrt{2}W\!\left(\dfrac12 + \alpha\,\dfrac{x}{L}\right) \quad -(3)$$

Step 3 : Vertical reaction

Substituting the value of $$T/\sqrt{2}$$ from (3) into (2):

$$\dfrac{T}{\sqrt{2}} = W\!\left(\dfrac12 + \alpha\,\dfrac{x}{L}\right)$$

$$\Longrightarrow\;\; V = W\!\left[1+\alpha-\dfrac12 - \alpha\,\dfrac{x}{L}\right] = W\!\left[\dfrac12 + \alpha\!\left(1-\dfrac{x}{L}\right)\right] \quad -(4)$$

The quantity $$\alpha$$ will disappear from expression (4) only when the bracket multiplying it is zero, i.e.

$$1-\dfrac{x}{L}=0 \Longrightarrow x = L.$$

This means the block must be hung from the extreme end $$A$$ of the beam (which is indeed the position of point $$P$$ in the figure). With this condition one obtains

$$V = \dfrac{W}{2},$$

which is a constant, completely independent of the value of $$\alpha$$. Hence statement A is correct.

Step 4 : Verifying the remaining statements

Putting $$x=L$$ and $$\alpha =0.5$$ in (3):

$$T = \sqrt{2}W\!\left(\dfrac12 + 0.5\right)= \sqrt{2}W.$$

Therefore from (1)

$$H = \dfrac{T}{\sqrt{2}} = W.$$

The horizontal reaction, however, acts towards the right (positive) while the numerical value quoted in the option carries no sign. Since direction is an essential part of a vector component, the statement is not accepted as correct.

For $$\alpha =0.5$$ we have already obtained $$T=\sqrt{2}W\approx1.414\,W$$, not $$2W$$. Hence statement C is wrong.

The rope can withstand a maximum tension $$T_{\text{max}}=2\sqrt{2}\,W$$. From (3) with $$x=L$$, the tension is

$$T = \sqrt{2}W\!\left(\dfrac12+\alpha\right).$$

Equating $$T=T_{\text{max}}$$ gives

$$\sqrt{2}W\!\left(\dfrac12+\alpha\right)=2\sqrt{2}W \;\;\Longrightarrow\;\; \alpha = 1.5.$$ For $$\alpha =1.5$$ the rope is exactly at its limiting tension; it will start snapping the instant $$\alpha$$ exceeds this value, i.e. for $$\alpha \ge 1.5$$. Because the option claims breakage only when $$\alpha > 1.5$$ (excluding the equality case), it is not strictly correct.

Thus, the only statement that is wholly correct is

Option A which states: “The vertical component of reaction force at O does not depend on $$\alpha$$.”