1. Definisi bagi Pusat graviti
Jasad tegar terdiri daripada banyak zarah; oleh itu, daya graviti bertindak ke atas setiap zarah ini. Dalam erti kata lain, setiap zarah mempunyai beratnya sendiri. Pusat graviti sesuatu objek ialah titik pada objek di mana berat semua bahagian objek dianggap berpusat pada titik tersebut.
If an object is homogeneous (the density of each part of the object is the same or the object is composed of similar material) and the shape of the object is symmetrical (for example square, rectangular, circle)
then the weight of the object coincides with the center of mass of the object, located at the center of the object. For triangles, the center of mass is located at 1/3 h (h = height of the triangle).
2. Equation of the center of gravity
If the shape of the object is symmetrical and the object is homogeneous, then the center of gravity of the object coincides with the center of the mass of the object,
where the center of gravity and center of mass is located in the center of the object. Conversely, if the object is homogeneous but not symmetrical, then the position of the object’s weight can be determined using the following formula.
The coordinates of the object’s weight on the x-axis:
The coordinates of the object’s weight on the y-axis:
x = the midpoint of the object on the x-axis, y = the midpoint of the object on the y-axis, A = area. If the object is in three dimensions, besides determining the coordinates of the object’s weight on the x and y axes, it also determines the coordinates of the object’s weight on the z-axis. Area (A) is replaced by volume (V).
Contoh masalah 1.
Determine the coordinates of the center of gravity of the homogeneous object in the figure on the side!
penyelesaian:
Divide object into three parts.
A1 = (10-0)(10-0) = (10)(10) = 100
A2 = (20-10)(30-0) = (10)(30) = 300
A3 = (30-20)(10-0) = (10)(10) = 100
x1 = 1⁄2 (10-0) = 1⁄2 (10) = 5
x2 = 1⁄2 (20-10) + 10 = 1⁄2 (10) + 10 = 5 + 10 = 15
x3 = 1⁄2 (30-20) + 20 = 1⁄2 (10) + 20 = 5 + 20 = 25
y1 = 1⁄2 (10-0) = 1⁄2 (10) = 5
y2 = 1⁄2 (30-0) = 1⁄2 (30) = 15
y3 = 1⁄2 (10-0) = 1⁄2 (10) = 5
The coordinates of the object’s weight on the x-axis:
The coordinates of the object’s weight on the y-axis:
The coordinates of the object’s weight are (15; 11)
Contoh masalah 2.
Determine the coordinate of the object’s weight in the figure!
penyelesaian:
Divide object into two parts, area 1 = square, area 2 = triangle.
A1 = (5-1)(3-0) = (4)(3) = 12
A2 = 1⁄2 (6-0)(9-3) = 1⁄2 (6)(6) = (3)(6) = 18
x1 = 1⁄2 (5-1) + 1 = 1⁄2 (4) + 1 = 2 + 1 = 3
x2 = 1⁄2 (6-0) = 1⁄2 (6) = 3
y1 = 1⁄2 (3-0) = 1⁄2 (3) = 1.5
y2 = 1/3(9-3) + 3 = 1/3(6) + 3 = 2 + 3 = 5
The coordinates of the object’s weight on the x-axis:
The coordinates of the object’s weight on the y-axis:
The coordinate of the object’s weight is (3; 3.6)
