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

Given:
Mass of the stick = $m_1=200g$
Mass of the small object = $m_2=20$
Length of the string = $l=1\mathrm{m}$
As the system is in equilibrium, we have:
${\tau }_{total}=0\left(\mathrm{about}\mathrm{O}\right)$
$\left(T_1\times r_1\right)-\left(T_2\times r_2\right)-\left(m_{1}g\times r_3\right)=0$

$$\begin{gathered} \Rightarrow T_1\times 0.7-T_2\times 0.3-2\times 0.2\times g=0 \\ \Rightarrow 7T_1-3T_2=3.92...\left(i\right) \end{gathered}$$
Now, we have:
Total upward force = Total downward force
$$\begin{gathered} T_1+T_2=m_{1}g+m_{2}g \\ =0.2\times 9.8+0.02\times 9.8 \\ \Rightarrow T_1+T_2=2.156...\left(ii\right) \end{gathered}$$



Solving equations (i) and (ii), we get:
$$\begin{gathered} {\mathrm{T}}_1=1.038\mathrm{N}\approx 1.04\mathrm{N}; \\ {\mathrm{T}}_2=1.18\approx 1.12\mathrm{N} \end{gathered}$$