Question

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 $\left(2\sqrt{2}\right)W.$ Which of the following statement(s) is (are) correct ?

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

Answer 1

Answer: (A), (B), (D)

The rope breaks if $\alpha$ $>1.5$.

The free body diagram:


At equilibrium,

$R_y+T\sin {45}^{\circ }=W+\alpha W$

$\Rightarrow R_y+\frac{T}{\sqrt{2}}=W+\alpha W...\left(i\right)$
And,

$R_x+T\cos {45}^{\circ }=0$

$\Rightarrow R_x+\frac{T}{\sqrt{2}}=0...\left(ii\right)$

Taking torque about ${}^{\prime }{O}^{\prime }$,

$W\frac{l}{2}+\alpha Wl=T\sin {45}^{\circ }\times l$

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

Using equation $\left(ii\right)$ and $\left(iii\right)$,

$R_x=-\frac{T}{\sqrt{2}}=-(\frac{W}{2}+\alpha W)$

$\Rightarrow |R_x{|}_{\alpha =0.5}=(\frac{W}{2}+0.5\times W)=W$

Hence, $\left(B\right)$ is correct option.

Taking torque about $P$,

$R_{y}l=W\frac{l}{2}$

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

As, $R_y$ depends only on $W$, hence option $\left(A\right)$ is correct.

When, $T=T_{\text{max}}$,

$2\sqrt{2}W=\sqrt{2}(\frac{W}{2}+\alpha W)$

$\therefore \alpha =\frac{3}{2}$

As, $T>T_{max}\text{for}\alpha >1.5$. So, the rope breaks if $\alpha >1.5$