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 can be determined using a Newton scale. By suspending the metre stick and measuring the force required to support it, the student could record the weight as 1.2 N.

The centre of gravity of the metre stick is not at the 50.0 cm mark due to its non-uniform distribution. The weight is not evenly distributed along its length, causing the centre of gravity to shift to the 50.4 cm mark.

The student knew that the metre stick was in equilibrium when it remained stationary and horizontal after adjusting the applied forces. This indicates that the net force was zero and the moments about any axis were balanced.

To calculate the net force, sum the forces acting on the metre stick:

$F_{net} = F_{up} - F_{down} = 4.5 + 3.0 + 5.7 - 2.0 - 4.0 = 10.2 - 10.2 = 0 ext{ N}$

To find the total clockwise moment:

$C.T.M = (4.5 ext{ N} imes (50.4 - 26.2)) + (3.0 ext{ N} imes (50.4 - 38.3)) = 4.5 imes 24.2 + 3.0 imes 12.1 = 108.9 + 36.3 = 145.2 ext{ N cm}$

The anti-clockwise moment can be calculated as follows:

$A.C.T.M = 2.0 ext{ N} imes (50.4 - 11.5) + 5.7 ext{ N} imes (50.4 - 70.4) = 2.0 imes 38.9 + 5.7 imes 20.0 = 77.8 + 114.0 = 191.8 ext{ N cm}$

The net force is zero, which supports the first condition of equilibrium. The total moments calculated also cancel each other out, confirming that the system is in equilibrium, with clockwise moments balancing the anti-clockwise moments.