Question

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at the 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass ‘m’ is suspended from the rod at the 160 cm mark as shown in the figure. Find the value of ‘m’ such that the rod is in equilibrium. (g=10 m/s²)

• A. \(\frac{1}{12}\) kg
• B. \(\frac{1}{2}\)kg
• C. \(\frac{1}{2}\)kg
• D. \(\frac{1}{2}\)kg

Answer 1

The Correct Option is A

To determine the value of the unknown mass \( m \) such that the rod is in equilibrium, we apply the principle of moments. For the rod to be in equilibrium, the clockwise moments about the pivot must equal the counterclockwise moments.

First, define the moments about the pivot (at 40 cm).

1. The weight of the rod acts at its center of mass, which is at 100 cm from the left end. The distance from the pivot (40 cm mark) is \( 100 \, \text{cm} - 40 \, \text{cm} = 60 \, \text{cm} \). The force due to the mass of the rod is \( 0.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 5 \, \text{N} \). So, the moment is \( 5 \, \text{N} \times 60 \, \text{cm} = 300 \, \text{Ncm} \) (counterclockwise).
2. For the 2 kg mass at 20 cm, the distance from the pivot is \( 40 \, \text{cm} - 20 \, \text{cm} = 20 \, \text{cm} \). The force is \( 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \). Thus, the moment is \( 20 \, \text{N} \times 20 \, \text{cm} = 400 \, \text{Ncm} \) (counterclockwise).
3. The unknown mass \( m \) at 160 cm is \( 160 \, \text{cm} - 40 \, \text{cm} = 120 \, \text{cm} \) from the pivot. The force is \( m \times 10 \, \text{m/s}^2 \). The moment is therefore \( m \times 10 \times 120 \, \text{cm} = 1200m \, \text{Ncm} \) (clockwise).

According to the principle of moments for equilibrium:

\(400 - 300 = 1200m\)

\(100 = 1200m\)

Solving for \( m \):

\(m = \frac{100}{1200} = \frac{1}{12} \, \text{kg}\)

\(m = \frac{1}{12} \, \text{kg}\)