Question

A uniform rod of length $4L$ and mass $M$ is suspended form a horizontal roof by two light strings of length $L$ and $2L$ as shown. Then the tension in the left string of length $L$ is:

• A. $\frac{Mg}{2}$
• B. $\frac{Mg}{3}$
• C. $\frac{3}{5}Mg$
• D. $\frac{Mg}{4}$

Answer 1

The correct option is A $\frac{Mg}{2}$

$AC=4L$
$BC=L$
$AB=\sqrt{A{C}^2-B{C}^2}$
$=\sqrt{{\left(4L\right)}^2-L^2}$
$=\sqrt{16L^2-L^2}$
$=\sqrt{15L^2}$
$AB=\sqrt{15}L$
Also, $\cos \theta =\frac{AB}{AC}=\frac{AE}{AD}$
$=\frac{\sqrt{15}L}{4L}=\frac{AE}{2L}$
$AE=\frac{\sqrt{15}}{4}\times 2L$
$AE=\frac{\sqrt{15}}{2}L$
taking moment at $A$
$T_1\times AB=Mg\times AE$
$T_1\times \sqrt{15}L=Mg\times \frac{\sqrt{15}}{2}L$
$T_1=\frac{Mg}{2}$