Answer:
([tex]\frac{37a}{34}[/tex],[tex]\frac{25a}{34}[/tex])
Step-by-step explanation:
Given data,
Side of square = a
center of gravity of x,y
[tex]x_{cg}[/tex] = [tex]\frac{w_{1}x_{1} +w_{2}x_{2}+......+w_{n}x_{n} }{w_{1}+w_{2}+......+w_{n} }[/tex]
[tex]y_{cg}[/tex] = [tex]\frac{w_{1}y_{1} +w_{2}y_{2}+......+w_{n}y_{n} }{w_{1}+w_{2}+......+w_{n} }[/tex]
now, find the center of mass of each plate
For plate 1
W = 10N
x- coordinate = a/2
y- coordinate = a+a/2 = 3a/2
For plate 2
W = 30N
x- coordinate = a+a/2=3a/2
y- coordinate = a+a/2 = 3a/2
For plate 3
W = 70N
x- coordinate = a+a/2=3a/2
y- coordinate = a/2
For plate 4
W = 60N
x- coordinate = a/2
y- coordinate = a/2
[tex]x_{cg}[/tex] = [tex]\frac{10\times\frac{a}{2} +30\times\frac{3a}{2}+70\times\frac{3a}{2}+60\times\frac{a}{2}}{10+30+70+60 }[/tex] =[tex]\frac{185a}{170}[/tex] =[tex]\frac{37a}{34}[/tex]
[tex]y_{cg}[/tex] = [tex]\frac{10\times\frac{3a}{2} +30\times\frac{3a}{2}+70\times\frac{a}{2}+60\times\frac{a}{2}}{10+30+70+60 }[/tex] = [tex]\frac{125a}{170}[/tex] = [tex]\frac{25a}{34}[/tex]
Therefor ([tex]\frac{37a}{34}[/tex],[tex]\frac{25a}{34}[/tex]) will the center of gravity
