a) Particles of weight 4 N, 5 N, 3 N and 2 N are placed at the points (11, 5), (p, q), (-4, 1) and (7, p), respectively - Leaving Cert Applied Maths - Question 6 - 2009

Next, we find the value of q using the equation for the y-coordinate of the centre of gravity:

rac{4(5) + 5(q) + 3(1) + 2(p)}{14} = q

Substituting p = 2 into the equation:

rac{20 + 5q + 3 + 4}{14} = q

Simplifying:

rac{27 + 5q}{14} = q

Multiplying both sides by 14:

$27 + 5q = 14q$

Rearranging gives:

$27 = 14q - 5q$

$27 = 9q$

Thus, we get:

$q = 3$

To find the centre of gravity of the remaining lamina after removing triangle ade, we first calculate the areas:

Area of quadrilateral abcd:

$ext{Area}_{abcd} = (12)(8) = 96$

Centre of gravity of quadrilateral abcd:

$c.g = (6, 4)$

Area of triangle aed:

$ext{Area}_{aed} = \frac{1}{2}(12)(8) = 36$

Centre of gravity of triangle aed:

$c.g = (7, 2)$

Total area of the lamina after removing triangle aed:

$ext{Area}_{lamina} = 96 - 36 = 60$

Using the formula for centre of gravity:

For the x-coordinate:

$x = \frac{(60)(x_{abcd}) - (36)(x_{aed})}{60}$

Substituting the values:

$x = \frac{(60)(6) - (36)(7)}{60} = 5.4$

For the y-coordinate:

$y = \frac{(60)(y_{abcd}) - (36)(y_{aed})}{60}$

Substituting the values:

$y = \frac{(60)(4) - (36)(2)}{60} = 5.2$

Thus, the co-ordinates of the centre of gravity of the remaining lamina are (5.4, 5.2).