A uniform pencil AB weighing 40g can be balanced horizontal on a knife edge at 2cm from end A when a mass 60g is hung from This end what is the length of pencil

Answer 1

To determine the length of the uniform pencil AB given the conditions of the problem, we can use the principle of moments (torque).

1. Assign Variables:

   • Let \( L \) be the length of the pencil.
   • The weight of the pencil (40g) acts at its midpoint, which is at \( \frac{L}{2} \) from end A.
   • The 60g mass is hung from end A, which is 2 cm from the pivot (knife edge).

2. Calculate Moments:

   • Moment due to the weight of the pencil is calculated about the knife edge:
     • The distance from the knife edge to the center of the pencil is \( \frac{L}{2} - 2 \) cm.
     • Moment (Torque_1) due to the pencil's weight = Weight × Distance = \( 40 , \text{g} \times \left( \frac{L}{2} - 2 \right) \).
   • Moment due to the 60g weight:
     • Since the 60g mass is 2 cm from the knife edge, the moment (Torque_2) due to the mass = \( 60 , \text{g} \times 2 , \text{cm} = 120 , \text{g cm} \).

3. Setting Up the Equation: Since the system is in equilibrium, the moment around the pivot from the pencil should equal the moment from the 60g mass: \[ 40 \left( \frac{L}{2} - 2 \right) = 120 \]

4. Solve for L: Divide both sides by 40: \[ \frac{L}{2} - 2 = 3 \] Add 2 to both sides: \[ \frac{L}{2} = 5 \] Multiply by 2: \[ L = 10 , \text{cm} \]

Therefore, the length of the pencil AB is \( 10 , \text{cm} \).