Question

An object is made from a uniform piece of sheet metal. the object has dimensions of a = 1.25 ft , where a is the diameter of the semi-circle,b = 3.17 ft , and c = 1.95 ft . a hole with diameter d = 0.750 ft is centered at (1.05,0.625). find x¯,y¯, the coordinates of the body's centroid.

Answer 1

From the given figure, we can divide it into three separate shapes: a semicircle, a rectangle and a right triangle.

The centroid of the body is the arithmetic mean of the centroid of the individual shapes.

Centre of the diameter of the semicircle is located at
`\left( (1.25)/(2) ,\ (1.25)/(2)\right)=(0.625,\ 0.625)` and the centroid of the semicircle is located at
`\left( (0.625)/(2) ,\ 0.625\right)=(0.3125,\ 0.625)`.

The length of the base of the right triangle is obtained as follows:


`√((1.95)^2-(1.25)^2) = √(3.8025-1.5625) = √(2.24) =1.4967ft`

The length of the base of the rectangle is given by


`3.17-1.4967-0.625=1.0483ft`

The centroid of the rectangle is given by


`\left( (0.625+1.0483)/(2) ,\ 0.625\right)=\left( (1.6733)/(2) ,\ 0.625\right)=(0.8367,\ 0.625)`

The centroid of the rectangle with the hole is given by


`\left( (0.8367+1.05)/(2) ,\ 0.625\right)=\left( (1.8867)/(2) ,\ 0.625\right)=(0.9434,\ 0.625)`

The midpoint of side c of the triangle is given by


`\left( (0.625+1.0483+3.17)/(2),\ 0.625\right)=\left( (4.8433)/(2),\ 0.625\right)=(2.4217,\ 0.625)`

The midpoint of the altitude of the right triangle is given by (0.625 + 1.0483, 0.625) = (1.6733, 0.625)

The equation of the line joining points (1.6733, 0) and (2.4217, 0.625) is given by y = 0.8351x - 1.3974

The equation of the line joining points (3.17, 0) and (1.6733, 0.625) is given by y = -0.4176x + 1.3238

The x-value of the point of intersection of line y = 0.8351x - 1.3974 and y = -0.4176x + 1.3238 is given by

0.8351x - 1.3974 = -0.4176x + 1.3238
1.2527x = 2.7212
x = 2.1723

The y-value of the point of intersection of line y = 0.8351x - 1.3974 and y = -0.4176x + 1.3238 is given by

y = 0.8351(2.1723) - 1.3974 = 1.8141 - 1.3974 = 0.4167

Thus, the centroid of the right triangle is given by (2.1723, 0.4167).

Therefore, the centroid of the object is given by:


`\left( (0.3125+0.8367+2.1723)/(3) ,\ (0.625+0.625+0.4167)/(3) \right) \\ \\ = \left((3.3215)/(3) ,\ (1.6667)/(3)\right)=(1.1072,\ 0.5556)`