The diagram shows a metre stick which is suspended from its mid-point (50 cm) with three masses hanging from it - Leaving Cert Physics - Question c - 2022

To calculate the total anticlockwise moment due to the 7 N force acting at 20 cm from the midpoint:

• Moment due to 7 N force:

  $ext{Moment} = 7 imes 0.2 = 1.4 ext{ Nm}$

Thus, the total anticlockwise moment is:

$1.4 ext{ Nm}$

The law of equilibrium states that for a system to be in equilibrium, the sum of clockwise moments about any point must equal the sum of anticlockwise moments about that same point. Therefore,

$ext{Total Clockwise Moments} = ext{Total Anticlockwise Moments}$

Given that the upward force on the metre stick is 15 N, and we have established that:

$ext{Upward Force} = ext{Weight of Metre Stick} + ext{Other Forces}$

$15 = 2 + ext{Weight of Metre Stick}$

Thus, the weight of the metre stick can be calculated as follows:

$ext{Weight of Metre Stick} = 15 - 2 = 13 ext{ N}$

The assumption that the centre of gravity of the metre stick acts at the mid-point may be invalid due to several factors, including:

• The metre stick may not be uniform in mass distribution (for example, if it is chipped or has additional weights).
• Any additional objects attached to the stick could shift the centre of gravity.
• Variations in material density along the stick may lead to an uneven distribution of weight.