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. Answers: (X, Y, Z) = (,, ) mm

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 = 150 mm = 0.15 mm

L = Length upto which it is extended along Y Axis = 315 mm = 0.315 m

=> COM1 = (0.0637, 0.1575, 0.0637) m

and, Area, A1= r²/4 = *0.15²/4 = 0.0177 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 = 150 mm = 0.15 mm

h = height of Triangle = L = 315 mm = 0.315 m

=> COM2 = (0.05, 0.105, 0)

and, Area, A2 = (1/2)*b*h = 0.5*0.15*0.315 = 0.0236 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 = 150 mm = 0.15 m

w = widthof plate = 85 mm = 0.085 m

=> COM3 = (0.075, 0 , - 0.0425) m

and, Area, A3 = b*w = 0.15*0.085 = 0.0128 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 = 0.0177 + 0.0236 + 0.0128 = 0.0541 m²

=> 0.0541*x = 0.0177*0.0637 + 0.0236*0.05 + 0.0128*0.075

=> x = 0.0604 m = 60.4 mm

>> Considerng Y - Direction

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

As, A = A1 + A2 + A3 = 0.0177 + 0.0236 + 0.0128 = 0.0541 m²

=> 0.0541*y = 0.0177*0.1575 + 0.0236*0.105 + 0.0128*0

=> y = 0.09733 m = 97.33 mm

>> Considerng Z - Direction

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

As, A = A1 + A2 + A3 = 0.0177 + 0.0236 + 0.0128 = 0.0541 m²

=> 0.0541*x = 0.0177*0.0637 + 0.0236*0 + 0.0128*- 0.0425

=> z = 0.01078 m = 10.78 mm