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Particles of weight 3 N, 1 N, 4 N, and 2 N are placed at the points $(p, q)$, $(1, p)$, $(q, 4)$, and $(0, 3)$ respectively - Leaving Cert Applied Maths - Question 6 - 2021

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Particles-of-weight-3-N,-1-N,-4-N,-and-2-N-are-placed-at-the-points-$(p,-q)$,-$(1,-p)$,-$(q,-4)$,-and-$(0,-3)$-respectively-Leaving Cert Applied Maths-Question 6-2021.png

Particles of weight 3 N, 1 N, 4 N, and 2 N are placed at the points $(p, q)$, $(1, p)$, $(q, 4)$, and $(0, 3)$ respectively. The co-ordinates of the centre of gravi... show full transcript

Worked Solution & Example Answer:Particles of weight 3 N, 1 N, 4 N, and 2 N are placed at the points $(p, q)$, $(1, p)$, $(q, 4)$, and $(0, 3)$ respectively - Leaving Cert Applied Maths - Question 6 - 2021

Step 1

Find the value of p

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Answer

To find the value of pp, we can calculate the X-coordinates of the center of gravity using the formula:

xcg=(weightixi)weightix_{cg} = \frac{\sum (weight_i \cdot x_i)}{\sum weight_i}

In this case, the weights are given as follows: 3N, 1N, 4N, and 2N at coordinates (p,q)(p, q), (1,p)(1, p), (q,4)(q, 4), and (0,3)(0, 3). The total weight is:

3+1+4+2=10N3 + 1 + 4 + 2 = 10 N

Setting up the equation for the X-coordinate:

0.5=3p+11+4q+010-0.5 = \frac{3p + 1 \cdot 1 + 4q + 0}{10}

Simplifying this gives:

5=3p+1+4q-5 = 3p + 1 + 4q 3p+4q=6[1]3p + 4q = -6 \\ [1]

Now, solving for Y-coordinate similarly:

1.5=3q+1p+43101.5 = \frac{3q + 1p + 4 \cdot 3}{10}

This simplifies to:

15=3q+p+1215 = 3q + p + 12 3q+p=3[2]3q + p = 3 \\ [2]

Now, we solve the system of equations [1] and [2]:

  1. From [2], we can express p=33qp = 3 - 3q and substitute it into [1]:

3(33q)+4q=63(3 - 3q) + 4q = -6

This simplifies to: 99q+4q=69 - 9q + 4q = -6 5q=15-5q = -15 q=3q = 3

Substituting back to find pp: p=33(3)p=2p = 3 - 3(3) \\ p = 2

Step 2

Find the value of q

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Answer

We previously found that p=2p = 2. Now substituting pp into equation [2]:

3q+2=33q + 2 = 3

3q=13q = 1 q=13q = -\frac{1}{3}

Thus, the values are:

  • p=2p = 2
  • q=3q = -3.

Step 3

Find Coordinates of Points B, C, E and F

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Answer

Assuming point A at (0,0)(0, 0), we first calculate the coordinates of point B. Since AB=25cm|AB| = 25 cm and AC=14cm|AC| = 14 cm, we can use the Pythagorean theorem for BDBD:

  1. B(7,24)B(7, 24) (based on the given dimensions)
  2. C(14,0)C(14, 0)
  3. Midpoint DD will be rac{(0 + 14)}{2}, rac{0 + 24}{2} = (7, 12).
  4. Let Fextat(7,24)F ext{ at } (7, 24) by midpoint of BDBD.
  5. E(7,12)E(7, 12) being related to D.

Step 4

Calculate the coordinates of the center of gravity of the remaining shape

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Answer

To calculate the center of gravity of the remaining shape after removing circles S1S_1 and S2S_2, we first compute the areas:

  • Area of triangle ABC=12×14×24=168ABC = \frac{1}{2} \times 14 \times 24 = 168.
  • Area of S1=πr12=π(32)=9πS_1 = \pi r_1^2 = \pi (3^2) = 9\pi.
  • Area of S2=πr22=π(22)=4πS_2 = \pi r_2^2 = \pi (2^2) = 4\pi.

Thus, the area remaining is: 1689π4π168 - 9\pi - 4\pi

Next, we compute center of gravity:

  • Center of gravity before cutting out circles is at (7,31)(7, 31) when height is 12 cm, for overall geometry taking into account: 7.31=(16813π1689π)7.31 = \left(\frac{168 - 13\pi}{168 - 9\pi}\right).

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