Particles of weight 4 N, 10 N, P N, and 7 N are placed at the points (3, 8), (p, -6), (4, q), and (p, p) respectively - Leaving Cert Applied Maths - Question 6 - 2019

For the y-coordinate of the center of gravity, we apply a similar formula:

$y_{cg} = \frac{\sum m_iy_i}{\sum m_i}$

Substituting the known values:

$q = \frac{4 \cdot 8 + 10 \cdot (-6) + P \cdot q + 7 \cdot p}{21 + P}$

Given that $p = 1$, we have:

$22q = -21 + q$

Solving this gives:

$22q + q = -21$ $q = -1$

The center of gravity for the quadrilateral lamina can be determined by finding the centroid of each triangle formed by the vertices

1. Triangle AEF:

   • Area (A): 54
   • Centroid: (4.5, 3)

2. Triangle DFC:

   • Area (A): 27
   • Centroid: (6, 8)

3. Triangle EBC:

   • Area (A): 36
   • Centroid: (11, 4)

4. Total Area:

   • Area ABCD: 117

Using the area to weight these: $x_{cg} = \frac{54 \cdot 4.5 + 27 \cdot 6 + 36 \cdot 11}{117}$

Calculating: $x_{cg} = 6.85$

For y-coordinate: $y_{cg} = \frac{54 \cdot 3 + 27 \cdot 8 + 36 \cdot 4}{117}$

Calculating: $y_{cg} = 4.46$

Thus, the co-ordinates of the centre of gravity of the lamina are (6.85, 4.46).