Question

A uniform meter stick having a mass 400g is suspended from the fixed supports through two vertical light strings of equal lengths fixed at the ends. A small object of mass 100g is put on the stick at a distance of 60cm from the left end. Calculate the tension in the two strings.(g=9.8$m/{s^2}$)

Answer 1

Hint: In such types of questions we calculate moments about a point and balance the clockwise and counterclockwise torques as the system is in equilibrium.

Formula used – $\sum {{M_A} = 0} $

Complete step-by-step answer:
Let the tension in the left string be ${T_1}$ and the right string be${T_2}$.
Three forces are acting on the meter stick, i.e. ${T_1}$​,${T_2}$​ and mg, Where the mass of the small object is m. Since the rod does not rotate so that the net torque is zero and the system is in translational balance, we have the condition that the sum of torques around any point on the stick must be zero.
We can choose any point we want to do our torque analysis, let's choose the left end of the stick because the torque ${T_1}$ ​will be zero.
$({T_1} \times {r_1}) - ({T_2} \times {r_2}) - ({m_1}g)({m_2}) = 0$
${m_1}$​=400g=0.4kg,
${r_1}$​=60cm=0.6m,
${r_2}$=1−0.6=0.4m, l=1m,

${m_2}$$\sum {{M_A} = 0} $ ​=100g=0.1kg
$ \Rightarrow {T_1} \times 0.6 - {T_2} \times 0.4 - 0.4 \times 0.1 \times 9.8 = 0$
$ \Rightarrow 0.6{T_1} - 0.4{T_2} = 0.392.........(i)$
The two tensions must support the total weight of the system, so we have
${T_1} + {T_2} = 0.4 \times 9.8 + 0.1 \times 9.8 = 4.9....(ii)$
Multiply eqn(ii) with 0.4 and add with eqn(i), we get
$
  0.6{T_1} - 0.4{T_2} = 0.392 \\
  \dfrac{{0.4{T_1} + 0.4{T_2} = 1.96}}{{ \Rightarrow {T_1} = 2.35N}} \\
 $
Substitute the value of ${T_1}$in eqn (ii) we get,
$ \Rightarrow {T_2} = 2.55N$
Hence the tensions in the two strings are 2.35N and 2.55N.

Note: In such types of questions, we need to take care in selecting the point about which we calculate the next moment as such points should be selected to eliminate as many variables as possible.