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

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 $P$,
$T_2+2Tcos\theta =Mg$ $\dots \left(i\right)$
$PR=a$
$\therefore$ $PQ$ $=Ps$ $=asec\theta$
If $\delta l_1$ and $\delta l_2$ be the elongations in the wires $PR$ and $PQ$ $\left(orPS\right)$ 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
$\frac{T}{AY}$ $\left(asec\theta \right)$ $=\frac{T_2}{AY}acos\theta$ $\Rightarrow$ $T=T_{2}co{s}^2\theta$ $\dots \left(ii\right)$
Solving eqns. $\left(i\right)and\left(ii\right)$, we get $T_2$ $=\frac{Mg}{1+2co{s}^3\theta }$