Because g varies so little over the extent of most structures, any structure\'s
ID: 1370126 • Letter: B
Question
Because g varies so little over the extent of most structures, any structure's center of gravity effectively coincides with its center of mass. Here is a fictitious example where g varies more significantly. The figure shows an array of six particles, each with mass m, fixed to the edge of a rigid structure of negligible mass. The distance between adjacent particles along the edge is 2.70 m. Following are the values of g at each particle's location:
Particle 1: g = 7.95 m/s2
Particle 2: g = 7.74 m/s2
Particle 3: g = 7.66 m/s2
Particle 4: g = 7.50 m/s2
Particle 5: g = 7.66 m/s2
Particle 6: g = 7.74 m/s2
Using the coordinate system shown, find (a) the x coordinate xcom and (b) the y coordinate ycom of the center of mass of the six-particle system. Then find (c) the x coordinate xcogand (d) the y coordinate ycog of the center of gravity of the six-particle system
Explanation / Answer
Let us find the x coordinate of center of mass. To do this first of all we need to sum the product of mass and x coordinate of all six particle and then devide . so the x-coordinate comes out to be,
x = (m*0+m*0+m*0+m*2.7+m*2.7+m*2.7)/(6m) = 2.7/2 = 1.35 m
Similarly we can find the y coordinate of center of mass,
y = (m*0+m*0+m*2.7+m*2.7+m*5.4+m*5.4)/6m = 2.7 m
So the coordinate of center of mass will be (1.35 m,2.7 m).
Now let us find the center of gravity for the particles. First we will focus on finding the x coordinate,
x = (7.95*0+7.74*0+7.66*0+7.5*2.7+7.66*2.7+7.74*2.7)/(7.95+7.74+7.66+7.50+7.66+7.74) = 1.34 m
and y cordinate,
y = (7.95*0+7.74*0+7.66*2.7+7.5*2.7+7.66*5.4+7.74*5.4)/(7.95+7.74+7.66+7.50+7.66+7.74) = 2.68 m
So the center of gravity will be (1.34 m,2.68 m).