- Home
- Standard 11
- Physics
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 ?
$(A)$ 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 $2 W$ for $\alpha=0.5$
$(D)$ The rope breaks if $\alpha>1.5$
$A,B,D$
$A,B,C$
$A,B$
$A,D$
Solution
$R _{ y }+\frac{ T }{\sqrt{2}}= W +\alpha W$ $. . . . . . (i)$
$R _{ x }=\frac{ T }{\sqrt{2}}$ $. . . . . . (ii)$
Taking torque about ' $O$ '
$W \frac{\ell}{2}+\alpha W \ell=\frac{ T }{\sqrt{2}} \ell$
$T =\sqrt{2}\left(\frac{ W }{2}+\alpha W \right)$ $. . . . . . (iii)$
$R _{ x }=\frac{ T }{\sqrt{2}}=\left(\frac{ W }{2}+\alpha W \right)$
Taking torque about $P$
$R _{ y } \ell= W \frac{\ell}{2}$
$R _{ y }=\frac{ W }{2}$
$\text { when } T = T _{\max }$
$2 \sqrt{2} W =\sqrt{2}\left(\frac{ W }{2}+\alpha W \right)$
we get $\alpha=\frac{3}{2}$
