Question

Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: (1) 0.300kg , (0.300m , 0.400m ); (2) 0.400kg , (0.200m , -0.400m ); (3) 0.200kg , (-0.300m , 0.700m ).

Find the x-coordinate of the center of mass of the system of three chocolate blocks.

Find the y-coordinate of the center of mass of the system of three chocolate blocks.

Answer 1

use the center of mass formula to solve for the x and y coordinates separately:

x = (m1x1 + m2x2 + m3x3)/(m1 + m2 +m3) = [(.300)(.300)+(.4)(.2)+(.2)(-.3)]/(.0.3 + 0.4+0.2)

=0.122

y = (m1y1 + m2y2 + m3y3)/(m1 + m2 +m3) = [(.3)(.4)+(.4)(-.4)+(.2)(.7)]/(.0.3 + 0.4+0.2)

= .111

so the coordinate is: (.122, .111)

Answer 2

X-coordinate m231 tm2x2 + m3'3 m1 + m2 ms -0.12 m 0.3+0.4+0.2 Y-coordinate m1 + m2 +m 0.11 m 0.3+0.4+0.2

Answer 3

ANSWER :

Let the x-coordinate of the mass centre be x.

Total mass at mass centre = 0.400 + 0.300 + 0.200 = 0.900 kg

So, 

0.900 * x = 0.300 * 0.300 + 0.400 * 0.200 - 0.200 * 0.300

=> x = (0.300 * 0.300 + 0.400 * 0.200 - 0.200 * 0.300) / 0.900 = 0.122 m 

So, x-coordinate of mass centre is 0.122 m (ANSWER).

Let the y-coordinate of the mass centre be y.

Total mass at mass centre = 0.400 + 0.300 + 0.200 = 0.900 kg

So, 

0.900 * y = 0.300 * 0.400 - 0.400 * 0.400 + 0.200 * 0.700

=> y = (0.300 * 0.400 - 0.400 * 0.400 + 0.200 * 0.700) / 0.900 = 0.111 m 

So, y-coordinate of mass centre is 0.111 m (ANSWER).