Question

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

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

Answer 1

Answer: (B)

$\frac{Mg}{1+2{\cos }^3\theta }$

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 $P$,
$T_2+2T\cos \theta =Mg...\left(i\right)$
$PR=a$
$\therefore PQ=PS=a\sec \theta$
If $\delta l_{(}1)$ and $\delta l_2$ be the elongations in the wires $PR$ and $PQ$ (or $PS$) respectively then
$\delta l_2=\delta l_1\cos \theta$ (from geometry)
If $A$ and $Y$ be the cross-sectional area and Young's modulus of each of the three wires, then
$\frac{T}{AY}\left(a\sec \theta \right)=\frac{T_2}{AY}\left(a\cos \theta \right)\Rightarrow T=T_2{\cos }^2\theta ...\left(ii\right)$
Solving equations (i) and (ii) we get $T_2=\frac{Mg}{1+2{\cos }^3\theta }$