6. (a) Particles of weight 4 N, 7 N, 3 N and 5 N are placed at the points $(p, 2)$, $(-6, 1)$, $(9, q)$ and $(12, 13)$, respectively - Leaving Cert Applied Maths - Question 6 - 2012

To find the value of $q$, we use a similar approach for the center of gravity in the y-coordinate:

ext{CG}_y = rac{ ext{Sum of moments about x-axis}}{ ext{Total Weight}}

The sum of the moments about the x-axis is: $4(2) + 7(1) + 3(q) + 5(13)$

Calculating this gives: $= 8 + 7 + 3q + 65$ $= 3q + 80$

Setting up the equation: rac{3q + 80}{19} = 9

Multiplying both sides by 19, we have: $3q + 80 = 171$

Thus, $3q = 171 - 80$ $3q = 91$ q = rac{91}{3} = 30.33

To find the coordinates of the centre of gravity of the remaining lamina, we first need to calculate the area of triangle ABC:

ext{Area}_{ABC} = rac{1}{2} imes ext{base} imes ext{height} = rac{1}{2} imes 36 imes 27 = 486

Next, we find the coordinates of the centroid of triangle ABC, given by:

ext{CG}_{ABC} = rac{(0+0+36)}{3}, rac{(0+27+0)}{3} = (12, 9).

Now to find the area of the circle with radius 9:

ext{Area}_{circle} = rac{1}{3} imes rac{22}{7} imes (9)^2 = 254.57

Now we can find the coordinates of the center of gravity of the lamina after subtracting the circle:

Let the coordinates be $(x, y)$. Using the formula for the center of gravity:

$ext{Remaining area} = (231 - 43)(x) = 486(12) - 254.57(9)$

Now we substitute and solve for $x$ and $y$. For $x$:

$(231 - 43)(x) = 486(12) - 254.57(9)$ $x = 15.3$

And for $y$:

$(231 - 43)(y) = 486 - 254.57(9)$ $y = 9$

Thus, the centre of gravity of the remaining lamina is approximately at coordinates $(15.3, 9)$.