Question

Three elastic wires, $PQ$, $PR$ and $PS$ support a body $P $ of mass $M$, as shown in figure. The wires are of the same material and cross-sectional area, the middle one being vertical. The load carried by middle wire is

• A. $Mg/ 1+2$ $cos^{2}$ $\theta$
• B. $Mg/ 1+2$ $cos^{3}$ $\theta$
• C. $Mg$ $cos^{2}$ $\theta$ / $1+2$ $cos^{3}$ $\theta$
• D. $Mg$ $cos \theta$/ $1+2\, cos^{3}$ $\theta$

Answer 1

Answer: (B)

Figure shows the situation when the load (body P) descends due to the stretch in the wires. The new positions of wires $PQ$ and $PS$ are shown by the dotted lines. Let $T_{1}$, $T_{2}$ and $T_{3}$ be the loads (tensions) carried by the three wires $PQ$, $PR$ and $PS$ respectively with the wires $PQ$ and $PS$ making angle $\theta$ with the vertical.
Considering the horizontal equilibrium of point $P$,
$T_{1}$ $sin\, \theta =T_{3} sin\, \theta$ $\Rightarrow $ $T_{1}$ $=T_{3}=T$ $\left(say\right)$
Considering the vertical equilibrium of point $P$,
$T_{2}+2T cos \theta=Mg$ $\quad\ldots\left(i\right)$
$PR=a$
$\therefore $$\quad$ $PQ$ $=Ps$ $=a \, sec \theta$
If $\delta l_{1}$ and $\delta l_{2}$ be the elongations in the wires $PR$ and $PQ$ $\left(or PS\right)$ respectively then
$\delta l_{2}$ $=\delta l_{1}$ $cos \theta$ $\left(\text{from geometry}\right) $
If $A$ and $Y$ be the cross-sectional area and Young’s modulus of each of the three wires, then
$\frac{T}{AY}$ $\left(a \,sec \theta\right)$ $=\frac{T_{2}}{AY} a \,cos \theta$ $\Rightarrow $ $T=T_{2} cos^{2} \theta$ $\quad\ldots\left(ii\right)$
Solving eqns. $\left(i\right) and \left(ii\right)$, we get $T_{2}$ $=\frac{Mg}{1+2 \,cos^{3}\, \theta}$