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Particles of weight 5 N, 2 N, 3 N and 8 N are placed at the points \((p, q), (7, p), (-2, q)\) respectively - Leaving Cert Applied Maths - Question 6 - 2019

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Particles-of-weight-5-N,-2-N,-3-N-and-8-N-are-placed-at-the-points-\((p,-q),-(7,-p),-(-2,-q)\)-respectively-Leaving Cert Applied Maths-Question 6-2019.png

Particles of weight 5 N, 2 N, 3 N and 8 N are placed at the points \((p, q), (7, p), (-2, q)\) respectively. The co-ordinates of the centre of gravity of the system ... show full transcript

Worked Solution & Example Answer:Particles of weight 5 N, 2 N, 3 N and 8 N are placed at the points \((p, q), (7, p), (-2, q)\) respectively - Leaving Cert Applied Maths - Question 6 - 2019

Step 1

Find (i) the value of p

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Answer

To determine the value of (p), we will use the formula for the centre of gravity (CG):

CGx=∑(mi⋅xi)∑mi\text{CG}_x = \frac{\sum (m_i \cdot x_i)}{\sum m_i}

Where:

  • (m_i) is the weight of each particle,
  • (x_i) is the x-coordinate of each particle.

We know:

  • Total weight = 5 N + 2 N + 3 N + 8 N = 18 N
  • The x-coordinate of CG is given as 2:

2=5(p)+2(7)+3(−2)+8(1)182 = \frac{5(p) + 2(7) + 3(-2) + 8(1)}{18}

Expanding: 2=5p+14−6+8182 = \frac{5p + 14 - 6 + 8}{18} 2=5p+16182 = \frac{5p + 16}{18}

Cross-multiplying gives: 36=5p+1636 = 5p + 16 $$ 5p = 20 \implies p = 4. $

Step 2

Find (ii) the value of q

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Answer

Now, to find the value of (q), we apply the same concept using the y-coordinate:

Given that the y-coordinate of CG is 0:

0=5(q)+2(6)+3(q)+8(−6)180 = \frac{5(q) + 2(6) + 3(q) + 8(-6)}{18}

Expanding: 0=5q+12+3q−48180 = \frac{5q + 12 + 3q - 48}{18} 0=8q−36180 = \frac{8q - 36}{18}

Cross-multiplying gives: $$ 0 = 8q - 36 \implies 8q = 36 \implies q = 4.5. $

Step 3

Find the co-ordinates of the centre of gravity for the quadrilateral lamina

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Answer

To find the centre of gravity for the quadrilateral lamina, we first confirm the areas of triangles formed:

  1. Area of triangle (abc): Area=12×base×height=12×6×6=18.\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 6 = 18. Using the centre of gravity formula: ( \text{CG}_{abc} = (2, 5) ).

  2. Area of triangle (acd): Area=12×12×9=54.\text{Area} = \frac{1}{2} \times 12 \times 9 = 54. Using the coordinates for CG: ( \text{CG}_{acd} = (6, 3). $$

Total area of the lamina = Area of (abc + acd = 18 + 54 = 72. $$

Thus, for the final coordinates: (72)(x)=54(6)+18(2)  ⟹  x=5.(72)(x) = 54(6) + 18(2) \implies x = 5. (72)(y)=54(3)+18(5)  ⟹  y=3.5.(72)(y) = 54(3) + 18(5) \implies y = 3.5.

Therefore, the co-ordinates of the centre of gravity are ((5, 3.5)).

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