Three elastic wires `PQ, PR` and `PS` support a body `P` of mass `M`, as shown in figure. The wires are of the some material and cross sectional area, the middle one being vertical. Find the loads by each wire.

Options

A. `Mg//1+ 2 cos^(2)theta`
B. `Mg//1+ 2 cos^(3)theta`
C. `Mg cos^(2)theta//1+ 2cos^(3)theta`
D. `Mg costheta//1+ 2cos^(3)theta`

Answer

B

Explanation

Figure shows the situation when the load (boby P) descends due to the stretch in load 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 pointP,
`T_(1) sin theta= T_(3) sin rArr T_(1) = T_(3) = T("say")`
Considering the veritcal equilibrium of point P, `T_(2) +2T cos theta = Mg" "......(i)`
PR =a
`therefore ` PQ = PS = a sec`theta`
I f`deltal_(1)` and `deltal_(2)` be the elongations in the wires PR and PQ (or Ps ) respectively then
`deltal_(2) = deltal_(1) cos theta` (from geometry)
If A nd Y be the cross- sectional area and Young's modaulus of reach of the each of the three wires, then
`(T)/(AY)(a sec theta) = (T_(2))/(AY ) a cos theta rArr T =T_(2) cos ^(2) theta" "....(ii)`
Sloving eqns. (i) and (ii) we get `T_(2) = (Mg)/(1+2 cos^(3) theta)`