(a) Three particles are located in a coordinate system shown in Figure (a). Find

Question

(a) Three particles are located in a coordinate system shown in Figure (a). Find the center of gravity.
m

(b) 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))? xcg = m ycg = m
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. xcg = m ycg = m

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. xcg = m ycg = m
xcg = m ycg = m Consider the figure below, wherem1 = 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. 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

b)

The X-axis of centre of gravity does not change,

Xcg =0.7137 m

Ycg=(m1*(1)+m2*0+m3*(-0.5))/(m1+m2+m3)=(5.39*1+2.32*0+4.5*(-0.5))/(5.39+2.32+1.98)

Ycg=0.324 m

c)

Xcg= (0.7137*(m1+m2+m3)+ 1.52*0)/(m1+m2+m3+m4)

=(0.7137*(5.39+2.32+1.98)+ 1.52*0)/(5.39+2.32+1.98+1.52)

Xcg=0.617 m

Ycg= (0.324*(m1+m2+m3)+ 1.52*0.282)/(m1+m2+m3+m4)

=(0.324*(5.39+2.32+1.98)+ 1.52*0.282)/(5.39+2.32+1.98+1.52)

Ycg=0.318 m

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