Question
xcg = m
ycg = m Consider the figure below, where m1 = 5.39 kg, m2 = 2.32 kg, m3 = 4.50 kg, d1 = 0.370 m, and d2 = 1.98 m. Treat the objects as point particles. Three particles are located in a coordinate system shown in Figure (a). Find the center of gravity. m How does the answer change if the object on the left is displaced upward by 1.00 m and the object on the right is displaced downward by 0.500 m (Figure (b))? Use the values from PRACTICE IT to help you work this exercise. If a fourth particle of mass 1.52 kg is placed at (0 m, 0.282 m) in Figure (a), find the x- and y-coordinates of the center of gravity for this system of four particles.
Explanation / Answer
a)y=0
x=(-m1d1+m2*0+m3d3)/(m1+m2+m3)=(-5.39*0.37+4.5*1.98)/(5.39+2.32+4.5) =0.5664m
b)x' =0.5664 (remains same)
y'=(m1*1-m3*0.5)/(m1+m2+m3)=(5.39-4.5*0.5)/(5.39+2.32+4.5)=0.257166m
c)X=((m1+m2+m3)*x+(1.52*0))/(m1+m2+m3+1.52)=0.5037m
Y=((m1+m2+m3)*0+(1.52*0.282))/(m1+m2+m3+1.52)=0.03122m