In an experiment to verify the laws of equilibrium, the centre of gravity and the weight of a metre stick were found - Leaving Cert Physics - Question 1 - 2016

The weight of the metre stick was found using a newton balance or weighing scales. The mass of the stick was measured, and the gravitational force was calculated by multiplying the mass by the acceleration due to gravity (approximately 9.81 m/s²).

The upward forces were measured using newton balances, which indicated the forces acting upwards due to the applied weights. The downward forces were determined based on known weights attached to the stick plus the weight of the stick itself.

The centre of gravity is not at the 50.0 cm mark because the stick may be unbalanced, possibly due to a hole in one side or an uneven weight distribution.

To calculate the net force, we sum the upward and downward forces:

• Upward force: $3.9 ext{ N} + 4.1 ext{ N} = 8.0 ext{ N}$
• Downward force: $2 ext{ N} + 3 ext{ N} + 2 ext{ N} + 1.1 ext{ N} = 8.1 ext{ N}$
• Net vertical force: $0.1 ext{ N} = ext{upward} - ext{downward}$

To find the sum of the moments about the 40 cm mark:

• Moment is calculated as force times displacement:

• Clockwise moments: $(2 ext{ N} imes 0.52) + (1.1 ext{ N} imes 0.10) + (3.9 ext{ N} imes 0.04) = 1.3082 ext{ Nm}$

• Anti-clockwise moments: $(2 ext{ N} imes 0.24) + (4.1 ext{ N} imes 0.1) = 1.3 ext{ Nm}$

• Sum of moments: $1.3082 ext{ Nm} - 1.3 ext{ Nm} = 0.01 ext{ Nm}$

For the system to be in equilibrium:

1. The net vertical force must equal zero, which is confirmed as we calculated $0.1 ext{ N}$ (this indicates a small imbalance).
2. The sum of moments about a point, in this case, the 40 cm mark, should also equal zero, and our calculation of $0.01 ext{ Nm}$ suggests a slight imbalance, further verifying the system's equilibrium state.