Question

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.

Answer 1

Given

Mass of the stick

\[m_1  = 200  g\]

Mass of the small object = \[m_2  = 20\]

Length of the string = \[l = 1  m\]

As the system is in equilibrium, we have

\[\tau_{\text{total}}  = 0 \left(\text{about O}\right)\]

\[\left( T_1 \times r_1 \right) - \left( T_2 \times r_2 \right) - \left( m_1 g \times r_3 \right) = 0\]

\[\Rightarrow  T_1  \times 0 . 7 -  T_2  \times 0 . 3 - 2 \times 0 . 2 \times g = 0\]

\[ \Rightarrow 7 T_1  - 3 T_2  = 3 . 92 ...........(1)\]

Now, we have

Total upward force = Total downward force

\[T_1  +  T_2  =  m_1 g +  m_2 g\]

\[= 0 . 2 \times 9 . 8 + 0 . 02 \times 9 . 8\]

\[ \Rightarrow  T_1  +  T_2  = 2 . 156 ..........(2)\]

Solving equations (1) and (2), we get

\[T_1  = 1 . 038  N \approx 1 . 04  N; \]

\[ T_2  = 1 . 18 \approx 1 . 12  N\]