6. (a) Particles of weight 5 N, 8 N, 3 N and 1 N are placed at the points (4, 1), (-3, p), (p, q) and (15, 4), respectively - Leaving Cert Applied Maths - Question 6 - 2015

Now, to find the value of q, we can use the formula for the y-coordinate of the centre of gravity:

$\bar{y} = \frac{\sum (w_iy_i)}{\sum w_i}$

Given:

• Weights: 5 N, 8 N, 3 N, 1 N
• Positions: (4, 1), (-3, p), (p, q), (15, 4)

Substituting into the formula:

$2 = \frac{5(1) + 8(-3) + 3(q) + 1(4)}{17}$

Calculating:

$2 = \frac{5 - 24 + 3q + 4}{17}$

Simplifying:

$2 = \frac{3q - 15}{17}$

Multiplying both sides by 17:

$34 = 3q - 15$

Adding 15 to both sides:

$49 = 3q$

Thus,

$q = \frac{49}{3} \approx 16.33$

For the quadrilateral lamina, we apply the same principle:

To find the total area and the centre of gravity, we'll calculate the areas:

1. For triangle ABC: $\text{Area}_{ABC} = \frac{1}{2} \times 18 \times 3 = 27$ The centroid for triangle ABC is: $\left( \frac{0 + 18 + 6}{3}, \frac{0 + 0 + 3}{3} \right) = (8, 1)$

2. For triangle ACD: $\text{Area}_{ACD} = \frac{1}{2} \times 9 \times 6 = 27$ The centroid for triangle ACD is: $\left( \frac{0 + 6 + 0}{3}, \frac{0 + 3 + 9}{3} \right) = (2, 4)$

3. For the entire quadrilateral ABCD: $\text{Area}_{ABCD} = 54$ To find coordinates (x, y) of centre of mass of the quadrilateral:

   $54(x) = 27(8) + 27(2)$ From this, we find: $x = 5$

   Now, for y: $54(y) = 27(1) + 27(4)$ Thus, $y = \frac{27 + 108}{54} = 2.5$

The co-ordinates of the centre of gravity of the lamina are (5, 2.5).