Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are 100 g, 150 g, and 200 g respectively? Each side of the equilateral triangle is 0.5 m long.

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Answer 1 · 2017-12-30T12:34:38

The centre of mass is #=(0.28,0.12)#

Let the corner of the triangle #{(A,100)}# be at the origin #O(0,0)#

Then,

the corner #{(B, 150)}# wiil be at #(0.5,0)#

and the corner #{(C,200)}# will be at #(0.25, 0.25sqrt3)#

Let #G(x_G,y_G)# be the center of mass of the particles

Then,

#x_G=(ax_a+bx_b+cx_c)/(a+b+c)=(100*0+150*0.5+200*0.25)/(100+150+200)#

#=(125)/(450)#

#=0.28#

#y_G=(ay_a+by_b+cy_c)/(a+b+c)=(100*0+150*0+200*0.25sqrt3)/(100+150+200)#

#=(51.7)/(450)#

#=0.12#