Oh, dude, the intersection of the three lines must be the center of gravity of the irregularly shaped lamina because that's just how gravity works. Like, gravity pulls everything towards the center of mass, so if you want to find where all the forces balance out, you gotta look at where those lines meet. It's like the universe's way of saying, "Hey, this is where things chill out."
The center of gravity of a triangular lamina lies at the point of intersection of the medians of the triangle, which is also known as the centroid. It is located one-third of the distance from each vertex to the midpoint of the opposite side along the median.
The centre of mass of a rectangular lamina lies at the point of intersection of its diagonals.
If the lamina is in two dimensions (i.e. not curled round into a third dimension) then the centre of gravity will be somewhere within the flat shape. The position of the centre of gravity will depend on the distribution of mass across the lamina. If the lamina is curled round into a third dimension then the centre of gravity will be somewhere within the volume enclosed, fully or partially, by the lamina; this may or may not be on the lamina.
The center of mass of a uniform triangular lamina lies at the intersection of the medians of the triangle, which is also known as the centroid. It is located one-third of the distance from each vertex along each median.
If you mean that to be a question, then I must assume that you're asking "Where is it ?" Assuming also that the thickness of the plate is the same everywhere, the location of that point completely depends on the shape of the outer edge of the plate.
The center of gravity of a triangular lamina lies at the point of intersection of the medians of the triangle, which is also known as the centroid. It is located one-third of the distance from each vertex to the midpoint of the opposite side along the median.
The centre of mass of a rectangular lamina lies at the point of intersection of its diagonals.
If the lamina is in two dimensions (i.e. not curled round into a third dimension) then the centre of gravity will be somewhere within the flat shape. The position of the centre of gravity will depend on the distribution of mass across the lamina. If the lamina is curled round into a third dimension then the centre of gravity will be somewhere within the volume enclosed, fully or partially, by the lamina; this may or may not be on the lamina.
The center of gravity of an irregular lamina is the point at which the entire weight of the lamina can be considered to act. It can be determined by balancing the lamina on a point and finding the point of equilibrium. Mathematically, it can be calculated by finding the weighted average of the x and y coordinates of all the points on the lamina.
Only if the lamina is the same shape as the rectangle!
The center of mass of a uniform triangular lamina lies at the intersection of the medians of the triangle, which is also known as the centroid. It is located one-third of the distance from each vertex along each median.
If the object is a thin lamina with uniform thickness (e.g. a piece of paper), the the centre of gravity of the object is at its geometrical centre. It can be determined by suspending a load (e.g. pendulum) on an edge of the lamina twice and the point where the plumb lines intersect is the centre of gravity.
The centroid of a lamina is the point at which it could be balanced if it was suspended. It represents the center of mass of the lamina and is the point where all the mass could be concentrated to achieve balance. The centroid is an important concept in engineering and physics for determining the equilibrium and stability of objects.
If you mean that to be a question, then I must assume that you're asking "Where is it ?" Assuming also that the thickness of the plate is the same everywhere, the location of that point completely depends on the shape of the outer edge of the plate.
The plural form of lamina is laminae.
Elachista lamina was created in 1948.
Lamina Sankoh died in 1964.