A student investigated the laws of equilibrium for a set of co-planar forces acting on a metre stick - Leaving Cert Physics - Question 1 - 2007

The weight of the metre stick was determined through a weighing scale or a Newton balance that measures gravitational force in newtons. The measurement obtained confirmed the weight as 12 N.

The centre of gravity is not at the 50.0 cm mark because the metre stick is likely non-uniform or has been manufactured unevenly. This can create an imbalance, causing the center of mass to shift away from the geometric center, resulting in a centre of gravity found at 50.4 cm.

The student recognized that the metre stick was in equilibrium when it remained stationary and balanced horizontally. The adjustments in applied forces reached a point where all forces and moments acting on the stick were balanced, confirming equilibrium.

To calculate the net force, we need to sum the forces acting on the metre stick. The forces are:

• Downwards: 2.0 N (at 11.5 cm) + 5.7 N (at 70.4 cm)
• Upwards: 4.5 N (at 26.2 cm) + 3.0 N (at 38.3 cm) + 4.0 N (at 80.2 cm)

Thus, the net force is:

$F_{net} = (2.0 + 5.7) - (4.5 + 3.0 + 4.0) = 7.7 - 11.5 = -3.8 ext{ N}$

Using the clockwise moment equation:

• For forces causing clockwise moments:

1. For 2.0 N at 11.5 cm: $Moment_1 = 2.0 imes 11.5 = 23.0 ext{ Ncm}$
2. For 5.7 N at 70.4 cm: $Moment_2 = 5.7 imes (70.4 - 50) = 5.7 imes 20.4 = 116.88 ext{ Ncm}$

Thus, the total clockwise moment is:

$M_{CW} = 23.0 + 116.88 = 139.88 ext{ Ncm}$

For the upward forces creating anti-clockwise moments:

1. For 4.5 N at 26.2 cm: $Moment_1 = 4.5 imes (50 - 26.2) = 4.5 imes 23.8 = 106.71 ext{ Ncm}$
2. For 3.0 N at 38.3 cm: $Moment_2 = 3.0 imes (50 - 38.3) = 3.0 imes 11.7 = 35.1 ext{ Ncm}$
3. For 4.0 N at 80.2 cm: $Moment_3 = 4.0 imes (80.2 - 50) = 4.0 imes 30.2 = 120.8 ext{ Ncm}$

Thus, the total anti-clockwise moment is:

$M_{ACW} = 106.71 + 35.1 + 120.8 = 262.61 ext{ Ncm}$

To verify the laws of equilibrium, we check that the net force equals zero and that the total clockwise moments equal the total anti-clockwise moments.

1. Net Force: Since net force $F_{net}$ is -3.8 N, which indicates an imbalance, the stick is not in translational equilibrium. However, if considering unbalanced moments separately, it is confirmed that they are required for static equilibrium.

2. Moments:

   • Total Clockwise Moment = 139.88 Ncm
   • Total Anti-clockwise Moment = 262.61 Ncm

Thus:

• If both moments are balanced (matched by appropriate distances and forces), it indicates rotational equilibrium even in net force, broader conclusions must assess how external forces provide balancing.