Three particles of masses 14 kg, 21 kg,

and 69 kg are located at the vertices of an equilateral triangle of sides 7 m.

A)Find the x-coordinate of the center of gravity of this system with respect to a coordinate system with origin at the 14 kg particle and with the 21 kg particle located on the positive y-axis. Answer in units of m

B)Find the y-coordinate of the center of gravity. Units of m

x3=7•cos30=6.06 m, y3=7/2 =3.5 m.

x=(m1•x1+m2•x2+m3•x3)/(m1+m2+m3) = =0+0+69•6.06/104 =4.022 m.
y=( m1•y1+m2•y2+m3•y3)/(m1+m2+m3) = =0+21•7+69•3.5/104 =3.74 m.

To find the center of gravity of the system with respect to the given coordinate system, we need to calculate the x-coordinate and y-coordinate separately.

A) To find the x-coordinate of the center of gravity, we can use the concept of moments. The moment of a particle about an axis is given by the product of its mass and the distance between the particle and the axis. The center of gravity is the point where the sum of the moments of all the particles is zero.

Let's consider the 14 kg particle as our reference point (origin). The x-coordinate of this particle is 0 since it is the reference point. The 21 kg particle is located on the positive y-axis, which means its x-coordinate is also 0. So, we only need to find the x-coordinate of the 69 kg particle.

Since the triangle is an equilateral triangle of sides 7 m, the distance between the 14 kg and 69 kg particles is half the length of the side, i.e., 7/2 = 3.5 m. Therefore, the x-coordinate of the 69 kg particle (with respect to the origin at the 14 kg particle) is 3.5 m.

B) Finding the y-coordinate of the center of gravity follows a similar process. In this case, we need to find the y-coordinate of the 69 kg particle since the 21 kg particle is already on the positive y-axis.

The vertical distance between the 14 kg and 69 kg particles can be determined using trigonometry in an equilateral triangle. The height of an equilateral triangle is given by (sqrt(3)/2) * side length. Therefore, the vertical distance between the 14 kg and 69 kg particles is (sqrt(3)/2) * 7 m ≈ 6.062 m.

Hence, the y-coordinate of the center of gravity (with respect to the origin at the 14 kg particle) is 6.062 m.

To summarize:
A) The x-coordinate of the center of gravity is 3.5 m.
B) The y-coordinate of the center of gravity is 6.062 m.