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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.
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Solution
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
\[T_1 = 1 . 038 N \approx 1 . 04 N; \]
\[ T_2 = 1 . 18 \approx 1 . 12 N\]
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