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Question: A uniform metre stick having mass 400 g is suspended from the fixed supports through two vertical li...

A uniform metre stick having mass 400 g is suspended from the fixed supports through two vertical light strings of equal lengths fixed at the ends. A small object of mass 100 g is put on the stick at a distance of 60 cm from the left end. Calculate the tensions in the two strings. (g = 10 m/s²)

Answer

The tensions in the two strings are 2.4 N and 2.6 N.

Explanation

Solution

Explanation of the solution:

  1. Identify Forces and Positions:

    • Mass of metre stick (M) = 400 g = 0.4 kg. Its weight (W_stick = Mg) acts at its center (50 cm mark).
    • Mass of small object (m) = 100 g = 0.1 kg. Its weight (W_object = mg) acts at 60 cm from the left end.
    • Tension in left string (T₁) acts at the 0 cm mark.
    • Tension in right string (T₂) acts at the 100 cm mark.
    • Given g = 10 m/s².

  2. Calculate Weights:

    • W_stick = 0.4 kg × 10 m/s² = 4 N
    • W_object = 0.1 kg × 10 m/s² = 1 N

  3. Apply Translational Equilibrium (Sum of vertical forces = 0): The upward forces (T₁ + T₂) must balance the downward forces (W_stick + W_object).

    T1+T2=Wstick+WobjectT_1 + T_2 = W_{stick} + W_{object}

    T1+T2=4 N+1 NT_1 + T_2 = 4 \, \text{N} + 1 \, \text{N}

    T1+T2=5 N(Equation 1)T_1 + T_2 = 5 \, \text{N} \quad \text{(Equation 1)}

  4. Apply Rotational Equilibrium (Sum of torques = 0): Choose the left end (0 cm mark) as the pivot point to simplify calculations (T₁ creates no torque about this point).

    • Torque due to W_stick (clockwise): Wstick×(distance from pivot to center)W_{stick} \times (\text{distance from pivot to center})

      =4 N×0.5 m=2 Nm= 4 \, \text{N} \times 0.5 \, \text{m} = 2 \, \text{Nm}

    • Torque due to W_object (clockwise): Wobject×(distance from pivot to object)W_{object} \times (\text{distance from pivot to object})

      =1 N×0.6 m=0.6 Nm= 1 \, \text{N} \times 0.6 \, \text{m} = 0.6 \, \text{Nm}

    • Torque due to T₂ (counter-clockwise): T2×(distance from pivot to right end)T_2 \times (\text{distance from pivot to right end})

      =T2×1 m=T2 Nm= T_2 \times 1 \, \text{m} = T_2 \, \text{Nm}

    For rotational equilibrium, sum of clockwise torques = sum of counter-clockwise torques:

    2 Nm+0.6 Nm=T2 Nm2 \, \text{Nm} + 0.6 \, \text{Nm} = T_2 \, \text{Nm}

    2.6 Nm=T2 Nm2.6 \, \text{Nm} = T_2 \, \text{Nm}

    T2=2.6 N(Equation 2)T_2 = 2.6 \, \text{N} \quad \text{(Equation 2)}

  5. Solve for T₁: Substitute the value of T₂ from Equation 2 into Equation 1:

    T1+2.6 N=5 NT_1 + 2.6 \, \text{N} = 5 \, \text{N}

    T1=5 N−2.6 NT_1 = 5 \, \text{N} - 2.6 \, \text{N}

    T1=2.4 NT_1 = 2.4 \, \text{N}

The tensions in the two strings are 2.4 N and 2.6 N.