Question

Below figure shows a rigid bar hinged at A and supported in a horizontal position by two vertical identical steel wires. Neglect the weight of the beam. The tension $T_1$ and $T_2$ induced in these wires by a vertical load P applied as shown are

• A. $T_1=T_2=\frac{P}{2}$
  
• B. $T_1=\frac{Pal}{(a^2+b^2)},T_2=\frac{Pbl}{(a^2+b^2)}$
  
• C. $T_1=\frac{Pbl}{(a^2+b^2)},T_2=\frac{Pal}{(a^2+b^2)}$
  
• D. $T_1=\frac{Pal}{2(a^2+b^2)},T_2=\frac{Pbl}{2(a^2+b^2)}$
  

Answer 1

The correct option is B $T_1=\frac{Pal}{(a^2+b^2)},T_2=\frac{Pbl}{(a^2+b^2)}$


The free body diagram
$R_A$ is the reaction force at A.
Let the rigid bar attains equilibrium at an angle $\theta$ from horizontal positon.
$\Delta l_2$ = change in length of steel wire carrying tension $T_2$.
$\Delta l_1$ = change in length of steel wire carrying tension $T_1$.

$\therefore \Delta 2=\frac{T_{2}L}{AE},\Delta 1=\frac{T_{1}L}{AE}$
$\therefore tan\theta =\frac{\Delta l_2}{b},=\frac{\Delta l_2}{a}or\Delta l_1=\frac{b}{a}\Delta l_1$
or $\frac{T_{2}L}{AE}=\frac{b}{a}\times \frac{T_{1}L}{AE}or{T}_2=\frac{b}{a}{T}_1$

At equilibrium summation of all moments about point A will be zero.
$\therefore \sum M_A=T_{2}b+T_{1}a-Pl=0$
$\therefore T_{2}b+T_{1}a=Pl$
$\left(\frac{b}{a}{T}_1\right)b+T_{1}a=Pl$
$T_1{b}^2+T_1{a}^2=Pal$
$T_1=\frac{Pal}{a^2+b^2}$
$\therefore T_2=\frac{b}{a}\times T_1$
$T_2\times \frac{b}{a}\times \frac{Pal}{a^2+b^2}=\frac{Pbl}{a^2+b^2}$