Particles of weight 7 N, 1 N, 6 N, and p N are placed at the points (5, p), (-3, p), (p, q), and (10, 7) respectively - Leaving Cert Applied Maths - Question 6 - 2018

Now to find the value of q, we can use the formula for the coordinates of the center of gravity (CG) in the y-direction:

$CG_y = \frac{ \sum (m_i y_i) }{ \sum m_i }$

Using the coordinates of gravity:

$q = \frac{ 7 \cdot p + p \cdot q + 6 \cdot 7 }{ 14 + p }$

Substituting p = 2:

$q = \frac{ 7(2) + 2q + 42 }{ 14 + 2 } = \frac{ 14 + 2q + 42 }{ 16 }$

This leads to:

$16q = 56 + 2q \Rightarrow 16q - 2q = 56 \Rightarrow 14q = 56 \Rightarrow q = 4$

To find the coordinates of the center of gravity of the remaining lamina, we need to calculate the areas and coordinates:

1. Area of Triangle ABC: $Area_{ABC} = \frac{1}{2} \times base \times height = \frac{1}{2} \times 12 \times 15 = 90$ The coordinates of the center of gravity for triangle ABC are: $(\frac{12}{3}, \frac{15}{3}) = (4, 5)$

2. Area of Triangle ADC: $Area_{ADC} = \frac{1}{2} \times 12 \times 6 = 36$ The coordinates of the center of gravity for triangle ADC are: $(\frac{12}{3}, \frac{6}{3}) = (2, 2)$

3. Area of the remaining lamina: $Remaining Area = Area_{ABC} - Area_{ADC} = 90 - 45 = 45$

4. Now calculate the center of gravity: With contributions from the remaining area:

   • In the x-direction: $x = \frac{45(4) - 36(2)}{45} = 6$
   • In the y-direction: $y = \frac{45(5) - 36(7)}{45} = 3$

Thus, the coordinates of the center of gravity of the remaining lamina are (6, 3).