One end of a horizontal uniform beam of weigh | vedclass

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Question

(A) Let the angle the rope makes with the horizontal beam be $	heta$. Since the height $OQ = L$ and the length $OP = L$,$	an 	heta = \frac{L}{L} =

Vedclass · Accepted Answer

$A, B, D$

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(A) Let the angle the rope makes with the horizontal beam be $	heta$. Since the height $OQ = L$ and the length $OP = L$,$	an 	heta = \frac{L}{L} = 1$,so $	heta = 45^{\circ}$.Resolving forces in the vertical direction at the hinge $O$:$R_y + T \sin 45^{\circ} = W + \alpha W$$R_y + \frac{T}{\sqrt{2}} = W(1 + \alpha) \quad . . . (i)$Resolving forces in the horizontal direction at the hinge $O$:$R_x = T \cos 45^{\circ} = \frac{T}{\sqrt{2}} \quad . . . (ii)$Taking torque about point $O$ for the beam:$W \left(\frac{L}{2}ight) + (\alpha W) L = (T \sin 45^{\circ}) L$$\frac{W}{2} + \alpha W = \frac{T}{\sqrt{2}}$$T = \sqrt{2} W \left(\frac{1}{2} + \alphaight) \quad . . . (iii)$From $(ii)$ and $(iii)$:$R_x = W \left(\frac{1}{2} + \alphaight)$. For $\alpha = 0.5$,$R_x = W(0.5 + 0.5) = W$. Thus,statement $(B)$ is correct.From $(iii)$,if $\alpha = 0.5$,$T = \sqrt{2} W (0.5 + 0.5) = \sqrt{2} W$. Thus,statement $(C)$ is incorrect.For the vertical reaction $R_y$,from $(i)$:$R_y = W(1 + \alpha) - \frac{T}{\sqrt{2}} = W(1 + \alpha) - W(\frac{1}{2} + \alpha) = W(1 - 0.5) = 0.5 W$. Since $R_y$ is independent of $\alpha$,statement $(A)$ is correct.The rope breaks if $T > T_{\max} = 2\sqrt{2} W$:$\sqrt{2} W (\frac{1}{2} + \alpha) > 2\sqrt{2} W$$\frac{1}{2} + \alpha > 2 \implies \alpha > 1.5$. Thus,statement $(D)$ is correct.

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