Particles of weight 5 N, 1 N, x N and 6 N are placed at the points (2, q), (−7, q), (3, 3) and (9, 1), respectively - Leaving Cert Applied Maths - Question 6 - 2013

Next, to find the value of q, we use the formula for the y-coordinate of the center of gravity:

rac{W_1 y_1 + W_2 y_2 + W_3 y_3 + W_4 y_4}{W_1 + W_2 + W_3 + W_4} = ext{C.G.}_y

Where:

• $W_1 = 5$, $y_1 = q$
• $W_2 = 1$, $y_2 = q$
• $W_3 = x$, $y_3 = 3$
• $W_4 = 6$, $y_4 = 1$
• $ext{C.G.}_y = 3$

Setting up the equation:

rac{5(q) + 1(q) + 9(3) + 6(1)}{12 + 9} = 3

Solving gives:

$$5q + q + 27 + 6 = 3(21) \ ightarrow 6q + 33 = 63 \ ightarrow 6q = 30 \ ightarrow q = 5$$

Thus, the value of q is 5.

To find the center of gravity of the remaining lamina, we first calculate the area of triangle ABC:

ext{Area}_{ABC} = rac{1}{2} imes ext{base} imes ext{height} \ ext{Area}_{ABC} = rac{1}{2} (24)(18) = 216

Next, we find the center of gravity (C.G.) of triangle ABC:

C.G. of A, B, and C is calculated using:

ext{C.G.} = rac{(0 + 0 + 24)}{3}, rac{(0 + 18 + 0)}{3} = (8, 6)

Then, we calculate the area of the circle with radius 15:

ext{Area}_{circle} = rac{ ext{Area}_{circle}}{2} = rac{ ext{Area}_{circle}}{ ext{Area}_{BC}} = rac{rac{22}{7}(15)^2}{2} = rac{706 imes 86}{2} = 1571

Now, we represent the remaining lamina as the area of triangle ABC minus the area of the circle. The coordinates of the center of gravity can be determined by balancing the moments:

ext{and for } y: \ ext{C.G.}_{remaining} = rac{490 imes 86(y) + 706 imes 86(6) - 216(6)}{490 imes 86}

Upon solving, you get:

• For x, approximately $x = 13.76$
• For y, approximately $y = 10.32$

Thus, the coordinates of the center of gravity of the remaining lamina are approximately (13.76, 10.32).