Calculate the x y and zcoordinates of the mass center of the

Calculate the x-, y-, and z-coordinates of the mass center of the fixture formed from aluminum plate of uniform thickness.

Solution

>> Considering Quadrant of a Circle.

It\'s Cross Section is in X-Z Plane

As, Center of Mss of Such Shape = (4r/3, L/2, 4r/3)

r = radius of Circle = 170 mm = 0.17 mm

L = Length upto which it is extended along Y Axis = 230 mm = 0.23 m

=> COM1 = (0.072, 0.115, 0.072) m

and, Area, A1= r²/4 = *0.19²/4 = 0.0284 m²

>> Considering Right Triangle

It lies in X-Y Plane

As, Center of Mass of Such Shape lies at (b/3, h/3, 0)

b = width of Triangle = 170 mm = 0.17 mm

h = height of Triangle = L = 230 mm = 0.23 m

=> COM2 = (0.0567, 0.0767, 0)

and, Area, A2 = (1/2)*b*h = 0.5*0.17*0.23 = 0.0196 m²

>> Considering Flat Plate

It Lies in X-Z Plane

As, Center os Mass of Such Shape Lies at (b/2, 0, - w/2)

b = Side of Plate = 170 mm = 0.17 m

w = widthof plate = 90 mm = 0.09 m

=> COM3 = (0.085, 0 , - 0.045 ) m

and, Area, A3 = b*w = 0.17*0.09 = 0.0153 m²

>> Now, Let\'s Assume, COM of whole system lies at (x,y,z)

>> Considerng X - Direction

A*x = A1*x1 + A2*x2 + A3*x3

As, A = A1 + A2 + A3 = 0.0284 + 0.0196 + 0.0153 = 0.0633 m²

=> 0.0633*x = 0.0284*0.072 + 0.0196*0.0567 + 0.0153*0.085

=> x = 0.0704 m = 70.4 mm

>> Considerng Y - Direction

A*y = A1*y1 + A2*y2 + A3*y3

As, A = A1 + A2 + A3 = 0.0284 + 0.0196 + 0.0153 = 0.0633 m²

=> 0.0633*y = 0.0284*0.115 + 0.0196*0.0767 + 0.0153*0

=> y = 0.07534 m = 75.34 mm

>> Considerng Z - Direction

A*z = A1*z1 + A2*z2 + A3*z3

As, A = A1 + A2 + A3 = 0.0284 + 0.0196 + 0.0153 = 0.0633 m²

=> 0.0633*x = 0.0284*0.072 + 0.0196*0 + 0.0153*- 0.045

=> z = 0.02143 m = 21.43 mm