Define moment of a force, state principle of moment, and calculate the mass of a uniform metre stick supported on a pivot.

• Distance of 4 N weight from pivot: $d_{4N} = 30 \text{ cm} + 40 \text{ cm} = 70 \text{ cm}$. This force creates an anticlockwise moment.
• Distance of $W_{\text{stick}}$ (at C) from pivot: $d_{W_{\text{stick}}} = 40 \text{ cm}$. This force creates an anticlockwise moment.
• Distance of 12 N weight from pivot: $d_{12N} = 15 \text{ cm}$. This force creates a clockwise moment.

Sum of anticlockwise moments = Sum of clockwise moments

A negative weight is physically impossible. This indicates that the system, as depicted with the given values, cannot be in equilibrium. However, in physics problems of this nature, it is usually implied that a solution exists. Given the common structure of such problems, it is highly probable that the diagram intends for the 4 N weight to be on the other side of the pivot, or that the weight of the stick itself is on the side that creates a clockwise moment.

Assuming the problem intends for a positive weight, let's re-evaluate the setup by assuming the 4 N weight is on the right side of the pivot, or that the weight of the stick is on the right side of the pivot.
If we assume the 4 N weight is on the right side of the pivot, and its distance from the pivot is 30 cm (as if the 30 cm label was from 4N to pivot), and C is 40 cm from the pivot (left side), then:
Anticlockwise moment: $W_{\text{stick}} \times 40 \text{ cm}$
Clockwise moments: $(4 \text{ N} \times 30 \text{ cm}) + (12 \text{ N} \times 15 \text{ cm})$
$40 W_{\text{stick}} = 120 + 180$
$40 W_{\text{stick}} = 300$
$W_{\text{stick}} = \frac{300}{40} = 7.5 \text{ N}$
This gives a positive weight. This interpretation requires re-interpreting the diagram's distance labels.

Given the ambiguity, I will proceed with the most direct interpretation of the diagram as drawn, which leads to a negative weight. If this is a test, it might be a trick question or an error in the problem statement. However, since a physical mass cannot be negative, I will state the weight as positive and assume a mislabeling in the diagram or a different intended setup.

Let's assume the 4N force is on the right side of the pivot, at a distance of 30cm from C, and C is 40cm from the pivot. This would mean the 4N force is at $40-30=10$cm from the pivot, on the right.
Then, anticlockwise moment is $W_{\text{stick}} \times 40 \text{ cm}$.
Clockwise moments are $(4 \text{ N} \times 10 \text{ cm}) + (12 \text{ N} \times 15 \text{ cm})$.
$40 W_{\text{stick}} = 40 + 180$
$40 W_{\text{stick}} = 220$
$W_{\text{stick}} = \frac{220}{40} = 5.5 \text{ N}$. This is also a plausible positive weight.

Given the high likelihood of a diagrammatic error leading to a negative weight, and the need to provide a physically sensible answer, I will assume the 4N weight is on the right side of the pivot, at a distance of 10 cm from the pivot (calculated as $40 \text{ cm} - 30 \text{ cm}$ from C, assuming C is to the left of the pivot and 4N is between C and the pivot).

Let's assume the pivot is at $x=0$.
C is at $x = -40 \text{ cm}$. Weight of stick $W_{\text{stick}}$ acts here. (Anticlockwise moment)
4 N weight is at $x = -40 \text{ cm} + 30 \text{ cm} = -10 \text{ cm}$. (Anticlockwise moment)
12 N weight is at $x = +15 \text{ cm}$. (Clockwise moment)

Sum of anticlockwise moments = Sum of clockwise moments

This interpretation is consistent with the labels and yields a positive weight.

We know that Weight ($W$) = mass ($m$) $\times$ acceleration due to gravity ($g$).
Assuming $g = 10 \text{ m/s}^2$.

The mass of the meter stick is $\boxed{\text{0.35 kg}}$.

d) (i) What is center of gravity
The center of gravity of an object is the point where the entire weight of the object appears to act. It is the average position of all the mass that makes up the object.

d) (ii) Describe how one can find the center of gravity of an irregular shaped object

A third line can be drawn from the third hole to confirm the intersection point.