(a) Particles of weight 2 N, 3 N, 4 N and 5 N are placed at the points (4,3), (−2,−3), (−5,6) and (−4,−7), respectively. Find the co-ordinates of the centre of gravity of the system. (b) A circular lamina with centre p and with point q on its circumference has the triangular portion with vertices p, q and r removed. The co-ordinates of the points are p(0,0), q(8,0) and r(4,6) respectively. Find the co-ordinates of the centre of gravity of the remaining lamina.

Leaving Cert Applied Maths Centre of Gravity: (a) Particles of weight 2 N, 3 N

(a) Particles of weight 2 N, 3 N, 4 N and 5 N are placed at the points (4,3), (−2,−3), (−5,6) and (−4,−7), respectively - Leaving Cert Applied Maths - Question 6 - 2013

Question 6

(a) Particles of weight 2 N, 3 N, 4 N and 5 N are placed at the points (4,3), (−2,−3), (−5,6) and (−4,−7), respectively. Find the co-ordinates of the centre of gravi... show full transcript

Worked Solution & Example Answer:(a) Particles of weight 2 N, 3 N, 4 N and 5 N are placed at the points (4,3), (−2,−3), (−5,6) and (−4,−7), respectively - Leaving Cert Applied Maths - Question 6 - 2013

Step 1

Find the co-ordinates of the centre of gravity of the particle system

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Answer

To find the co-ordinates of the center of gravity (ar{x}, ar{y}) of the particles, we use the formulas:

$ar{x} = rac{ ext{Σ(weight} imes ext{x-coordinate})}{ ext{Σweights}}$ $ar{y} = rac{ ext{Σ(weight} imes ext{y-coordinate})}{ ext{Σweights}}$

Calculating the total weights:

Total weight = 2 + 3 + 4 + 5 = 14 N

Now calculate the moments:

Moment about the x-axis:
- Particle 1: 2 N at (4, 3) → Moment = 2 × 4 = 8
- Particle 2: 3 N at (−2, −3) → Moment = 3 × (−2) = −6
- Particle 3: 4 N at (−5, 6) → Moment = 4 × (−5) = −20
- Particle 4: 5 N at (−4, −7) → Moment = 5 × (−4) = −20

Calculating $ar{x}$ : $ar{x} = rac{8 - 6 - 20 - 20}{14} = 1$

Now for the y-axis moments:

Moment about the y-axis:
- Particle 1: 2 N at (4, 3) → Moment = 2 × 3 = 6
- Particle 2: 3 N at (−2, −3) → Moment = 3 × (−3) = −9
- Particle 3: 4 N at (−5, 6) → Moment = 4 × 6 = 24
- Particle 4: 5 N at (−4, −7) → Moment = 5 × (−7) = −35

Calculating $ar{y}$ : $ar{y} = rac{6 - 9 + 24 - 35}{14} = -1$

Thus, the center of gravity for part (a) is at (1, -1).

Step 2

Find the co-ordinates of the centre of gravity of the remaining lamina

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Answer

For part (b), we first need to find the area of the triangular region pqr:

The area of triangle pqr can be calculated using the formula: $ext{Area} = rac{1}{2} imes ext{base} imes ext{height}$ Taking base = 8 units (from q to p) and height = 6 units (from r to line pq), we get: $ext{Area}_{pqr} = rac{1}{2} imes 8 imes 6 = 24$
The center of gravity of triangle pqr is given by: $ext{c.g.} = igg( rac{(0 + 8 + 4)}{3}, rac{(0 + 0 + 6)}{3} igg) = igg( rac{12}{3}, rac{6}{3} igg) = (4, 2)$

For the lamina:

The area of the circular lamina is: $ext{Area}_{ ext{lamina}} = ext{π}(8^2) = 64 ext{π}$

Now for the remaining area:

Remaining area = Total area - Area of triangle = $64 ext{π} - 24$

Finding the coordinates for remaining lamina:

Assume the c.g. for the remaining lamina is (x, y): Using moments about the center: $ext{Σ(areas)} = 64 ext{π}x - 24(4)\ ext{Σ(areas)} = 64 ext{π}y - 24(2)$ $e x t Σ (a re a s) = 64 e x t π x - 24 (4) e x t Σ (a re a s) = 64 e x t π y - 24 (2)$ Rearranging these:
- For x:

ightarrow -0.54$$

For y:

ightarrow -0.27$$ Therefore, the co-ordinates of the centre of gravity of the remaining lamina is approximately (-0.54, -0.27).

(a) Particles of weight 2 N, 3 N, 4 N and 5 N are placed at the points (4,3), (−2,−3), (−5,6) and (−4,−7), respectively - Leaving Cert Applied Maths - Question 6 - 2013

Worked Solution & Example Answer:(a) Particles of weight 2 N, 3 N, 4 N and 5 N are placed at the points (4,3), (−2,−3), (−5,6) and (−4,−7), respectively - Leaving Cert Applied Maths - Question 6 - 2013

Find the co-ordinates of the centre of gravity of the particle system

Find the co-ordinates of the centre of gravity of the remaining lamina

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