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.
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
1 answer