1. A 3.02 kg particle has the xy coordinates (-1.31 m, 0.182 m), and a 5.94 kg p
ID: 1449287 • Letter: 1
Question
1. A 3.02 kg particle has the xy coordinates (-1.31 m, 0.182 m), and a 5.94 kg particle has the xy coordinates (0.209 m, -0.251 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 3.59 kg particle such that the center of mass of the three-particle system has the coordinates (-0.439 m, -0.983 m)? 2. In the figure, three uniform thin rods, each of length L = 49 cm, form an inverted U. The vertical rods each have a mass of 11 g; the horizontal rod has a mass of 33 g. What are (a) the x coordinate and (b) the y coordinate of the system's center of mass?Explanation / Answer
m1 = 3.02 kg, x1 = -1.31, y1 = 0.182
m2 = 5.94 kg, x2 = 0.209, y2 = -0.251
m3 = 2.59 kg, y3 = ?, x3= ?
now, m1x1 + m2x2 + m3x3 = (m1+m2+m3)X
X = (3.02*(-1.31) + 5.94*0.209 + 2.59*x1)/(3.01+5.94+2.59) = -0.439
x1 = -2.35571 m
and m1y1 + m2y2 + m3y3 = (m1+m2+m3)Y
3.02*0.182 + 5.94*(-0.251) + 2.59Y3 = (3.02+5.94+2.59)(-0.983)
y3 = -4.02 m
(x3,y3) = (-2.255,-4.02)m
2. m1 = m2 = m3 = 0.011 = m
(X,Y) = ((mx1 +mx2+mx3)/3m , (my1+my2+my3)/3m) = ((0+24.5+49)/3,(-24.5+0+-24.5)/3) = (24.5,-16.33) with origin at the top left corner
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