
A uniform meter stick of mass 200 g is suspended from the ceiling through two vertical strings of equal lengths fixed to the ends. A small object of a mass of 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in two strings.
Answer
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Hint: Tension is defined as the pulling force transmitted axially using a string or a chain or a cable. Tension force is the force transmitted through a string. Torque is defined as a measure of force that causes an object to rotate about an axis. The direction depends on the direction of the force on the axis.

Complete step by step solution:
Mass of the stick, M = 200 gm = 0.2 kg
Mass of the object, m = 20 gm = 0.02 kg
The weight of the stick is $0.2 \times 10 = 2N$
The weight of the object is $0.02 \times 10 = 2N$
Let the tension in two strings be $T_1$ and $T_2$,
The system is in equilibrium and thus
\[{T_1} + {T_2} = 2 + 0.2 = 2.2N\_\_\_\_\_\_\_\_\_\_\_\_(1)\]
As the stick does not rotate, the net torque acting on it is zero.
But the torque about the midpoint is given by,
\[\left( {0.2 \times 0.2} \right) + \left( {{T_1} \times 0.5} \right) = \left( {{T_2} \times 0.5} \right)\]
\[ \Rightarrow 0.04 + 0.5{T_1} = 0.5{T_2}\]
\[ \Rightarrow \left( {{T_2} - {T_1}} \right) = 0.08\_\_\_\_\_\_\_\_\_\_(2)\]
Thus from equations 1 and 2, we get
\[2{T_2} = 2.28N\]
\[ \Rightarrow {T_2} = 1.14N\]
\[ \Rightarrow {T_1} = \left( {1.14 - 0.08} \right)N\]
\[ \Rightarrow {T_1} = 1.06N\]
Thus the tension in two strings, $T_1$ and $T_2$ are $1.06 N$ and $1.14 N$.
Note: 1. An object is said to be in a state of equilibrium when all the forces that are acting upon the object are balanced and thus the net force will be zero.
2. For the object to be in equilibrium, the object must have no acceleration which means that the net force and the net torque acting on the object must be zero. For the object to be in equilibrium, the sum of the torques about any point must also be equal to zero.
3. Tension can also be defined as the action-reaction pair of forces acting on it.

Complete step by step solution:
Mass of the stick, M = 200 gm = 0.2 kg
Mass of the object, m = 20 gm = 0.02 kg
The weight of the stick is $0.2 \times 10 = 2N$
The weight of the object is $0.02 \times 10 = 2N$
Let the tension in two strings be $T_1$ and $T_2$,
The system is in equilibrium and thus
\[{T_1} + {T_2} = 2 + 0.2 = 2.2N\_\_\_\_\_\_\_\_\_\_\_\_(1)\]
As the stick does not rotate, the net torque acting on it is zero.
But the torque about the midpoint is given by,
\[\left( {0.2 \times 0.2} \right) + \left( {{T_1} \times 0.5} \right) = \left( {{T_2} \times 0.5} \right)\]
\[ \Rightarrow 0.04 + 0.5{T_1} = 0.5{T_2}\]
\[ \Rightarrow \left( {{T_2} - {T_1}} \right) = 0.08\_\_\_\_\_\_\_\_\_\_(2)\]
Thus from equations 1 and 2, we get
\[2{T_2} = 2.28N\]
\[ \Rightarrow {T_2} = 1.14N\]
\[ \Rightarrow {T_1} = \left( {1.14 - 0.08} \right)N\]
\[ \Rightarrow {T_1} = 1.06N\]
Thus the tension in two strings, $T_1$ and $T_2$ are $1.06 N$ and $1.14 N$.
Note: 1. An object is said to be in a state of equilibrium when all the forces that are acting upon the object are balanced and thus the net force will be zero.
2. For the object to be in equilibrium, the object must have no acceleration which means that the net force and the net torque acting on the object must be zero. For the object to be in equilibrium, the sum of the torques about any point must also be equal to zero.
3. Tension can also be defined as the action-reaction pair of forces acting on it.
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