A student investigated the laws of equilibrium for a set of co-planar forces acting on a metre stick - Leaving Cert Physics - Question 1 - 2014

The diagram should depict the metre stick with the following forces illustrated:

• A downward force of 1.5 N at the 50 cm mark (weight of the metre stick).
• A downward force of 5 N.
• A downward force of 15 N.
• An upward force of 9 N.
• An upward force of 12.5 N.

The total upward force can be calculated by adding the upward forces:

$9 ext{ N} + 12.5 ext{ N} = 21.5 ext{ N}$

The total downward force includes the weight of the metre stick and the other downward forces:

$1.5 ext{ N} + 5 ext{ N} + 15 ext{ N} = 21.5 ext{ N}$

According to the law of equilibrium, the sum of upward forces must equal the sum of downward forces when an object is in equilibrium. In this case, both the total upward force (21.5 N) and the total downward force (21.5 N) are equal, confirming that the metre stick is in a state of equilibrium.

The anticlockwise moment is calculated using the formula:

$ext{Moment} = ext{Force} imes ext{Distance}$

For the upward forces:

• 9 N at 10 cm: $9 ext{ N} imes 0.1 ext{ m} = 0.9 ext{ Nm}$
• 12.5 N at 12.5 cm: $12.5 ext{ N} imes 0.125 ext{ m} = 1.5625 ext{ Nm}$

Total anticlockwise moments:

$0.9 ext{ Nm} + 1.5625 ext{ Nm} = 2.4625 ext{ Nm}$

The clockwise moment is also calculated using:

For the downward forces:

• 1.5 N at 50 cm: $1.5 ext{ N} imes 0.5 ext{ m} = 0.75 ext{ Nm}$
• 5 N at 5 cm: $5 ext{ N} imes 0.05 ext{ m} = 0.25 ext{ Nm}$
• 15 N at 15 cm: $15 ext{ N} imes 0.15 ext{ m} = 2.25 ext{ Nm}$

Total clockwise moments:

$0.75 ext{ Nm} + 0.25 ext{ Nm} + 2.25 ext{ Nm} = 3.25 ext{ Nm}$

For an object to be in equilibrium, the total clockwise moments must equal the total anticlockwise moments. Upon comparison, the total anticlockwise moment (2.4625 Nm) should equate to the total clockwise moment (3.25 Nm), demonstrating imbalance and indicating that adjustments may be needed to achieve true equilibrium.