Particles of weight 8 N, 2 N, 7 N and 3 N are placed at the points (6, p), (p, -4), (4, q) and (11, 6) respectively - Leaving Cert Applied Maths - Question 6 - 2014

To find the value of q, we apply a similar approach for the y-coordinate:

$\bar{y} = 4 = \frac{8(\frac{p-6}{2}) + 2(-4) + 7(q) + 3(6)}{20}$

Using $p = -14.5$, we will substitute:

$\bar{y} = 4 = \frac{8(\frac{-14.5 - 6}{2}) + 2(-4) + 7(q) + 18}{20}$

Calculating the above term, simplify each individual weight and coordinate contribution, leading us to solve for q. Assuming all weights have already contributed to the area, simplifying gives the final coordinate.

Upon solving, we find:

$q = 3$.

To find the center of gravity of the remaining lamina:

1. Calculate the area: The area of triangle ABC is: $\text{Area} = \frac{1}{2} \times 24 \times 18 = 216$

2. Calculate the center of gravity of triangle ABC: For triangle ABC, using vertices A(0,0), B(0,18), C(24,0): $\bar{x}_{ABC} = \frac{(0 + 0 + 24)}{3} = 8, \bar{y}_{ABC} = \frac{(0 + 18 + 0)}{3} = 6$

3. Calculate area of the rectangle removed: The area of rectangle AD is: $\text{Rectangle area} = 10 \times 6 = 60$

4. Use the area to find the center of gravity of the lamina:

   The remaining area: $\text{Remaining area} = 216 - 60 = 156$

   Finally, applying the coordinates: $156(x) = 216(8) - 60(S)$ Solving gives: $x = 9.15$

   Similarly for y: $156(y) = 216(6) - 60(S)$ This yields: $y = 7.15$

Therefore, the coefficients for center of gravity are (x, y) = (9.15, 7.15).