A uniform metre stick of mass 200 g is suspended from the ceiling through two vertical strings equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance 70 cm from the left end. Find the tensions in the two strings.

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Solution

As the system in the translational equilibrium $T_{1} + T_{2} = 2 + 0.2 = 2.2$
The rod does not rotate so the net torque is zero
Finding torque about centre of strings

$0.2 \times 0.2 + T_{1} \times 0.5 = T_{2} \times 0.5$

$(T_{2} - T_{1}) \times 0.5 = 0.04$

$(T_{2} - T_{1}) = \frac{0.04}{0.5}$

$\frac{4}{100} \times \frac{10}{5} = 0.08$

$2 T_{2} = 2.28$

$T_{2} = 1.14$ N

$T_{1} = 2.2 - 1.14$

$= 1.16 N$

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