In this explainer, we will learn how to find the position of the center of gravity of a lamina composed of standard shapes in 2D.
The weight of a body is a force, and a force acts at a point. For a body that is a particle, we can say that the weight of the body acts at the position of the body.
Consider however a rigid, horizontal rod. The rod can be modeled as a system of particles, each separated by a distance , as shown by the following figure.
We can imagine comparing a single particle, P, to the system of particles. The mass of P is equal to the sum of the masses of the particles. Because P has no length, it cannot rotate, so any force that acts on it can only produce translational acceleration.
If a net force acts on the system of particles at a particular point, , that is unique to the system, each particle in the system is accelerated due to equivalently to the acceleration of P due to .
If acts at , the system is given only translational acceleration due to . The system does not rotate due to .
When the acceleration of P and that of the system are equal, point is that of the center of gravity of the system.
Let us define the center of gravity of a one-dimensional system of particles.
Definition: The Center of Gravity of a One-Dimensional System of Particles
The position of , the center of gravity of a one-dimensional system of particles, is given by where is the mass of the particle of index and is the distance from the origin of a coordinate system of the particle of index .
The approximation of a system of particles can be made more realistic by allowing to tend to zero and, hence, to tend to infinity. This requires replacing the summation with the following integral:
In the case of a uniform rod, is given by where is the length of the rod and is, therefore, at the midpoint of the rod.
A force acting on the center of gravity of a rod acts equivalently to a set of forces acting in the same direction at every point along the rod. The same substitution of one force for a set of forces can be considered for the force of the weight of the rod.
A rod resting on a surface exerts forces at every point of contact between the rod and the surface, but the weight of a rod can be modeled as a single force that acts at the rodβs center of gravity.
If a rod is in contact with a surface, the surface will produce a reaction force, , in the opposite direction to the weight of the rod, , where will act along a line passing through the center of gravity. In order for and to act to make the body be in equilibrium, the center of gravity of the rod must be vertically above a part of the surface that provides , as shown in the following figure.
The position of the center of gravity of the rod relative to the surface determines whether or not the weight will produce a rotational effect.
In this explainer, only bodies of negligible thickness are considered. These bodies are called laminas. A lamina is two-dimensional.
For a lamina, the center of gravity can be defined by a 2D coordinate in a given coordinate system. The origin of the coordinate system used to define the center of gravity of a lamina can, for convenience, be one of the vertices of the lamina, assuming the lamina has any vertices.
Let us look at an example of determining the position of the center of gravity of a lamina.
Example 1: Finding the Center of Gravity of a Lamina in the Shape of a Right Triangle
In the figure shown, find the position of the center of gravity of the uniform triangular lamina , considering to be the origin point.
Answer
For uniform laminas, the center of gravity of the body is the geometric center of the body. The lamina is regular, so its center of gravity is its geometric center.
The geometric center of a right triangle is the point of intersection of two lines where for each line one end intersects a vertex of the triangle and the other end intersects the midpoint of the line opposite that vertex. These lines are the medians of the triangle.
The following figure shows two lines that intersect at the point , the geometric center of , called its centroid. The coordinates of the intercepts with the - and -axes of these lines are included in the figure.
The values of and can be determined algebraically by finding the values of and where the equation of line equals the equation of line .
The equation of is
The equation of is
These equations are equal, where and are the coordinates of the centroid of :
There is a common factor of that can be eliminated to give
This can be rearranged as follows to determine :
Substituting this value of into the equation of gives
To obtain the -coordinate of the point , the value of must be multiplied by , so
Therefore, the coordinates of the center of gravity of are given by
The method of finding the centroid by intersection of medians could have been simplified greatly as is a triangle, and we know that centroid divides the median in the ratio . This means that the position of the centroid is necessarily of the length of its median, and so the -coordinate value is necessarily . As the length of the side of in the -direction is the length of the side in the -direction, the -coordinate value is necessarily .
When we are locating the center of gravity of a uniform lamina, we need to identify the geometric center of the lamina. This is because when we are considering a uniform lamina, the two are equal. Let us consider why this is the case.
Consider the following uniform elliptical lamina with focus points at and .
The horizontal line of symmetry has been added, which passes through both and . Now, if we were to split the ellipse along this line of symmetry, we would end up with two congruent shapes. Since this is a uniform lamina, the two shapes would also have equal mass. Let us now consider the centers of gravity of these two new laminas. Since the laminas are congruent and uniform, the centers of gravity will be in the same place on both of them.
Note that since the ellipse is a regular shape and this is a uniform lamina, it is logical that the centers of gravity of the two shapes will lie on the vertical line of symmetry of the ellipse. In the diagram, we can see the centers of gravity, and , and that the line segment is perpendicular to .
We know that the mass of each of these laminas is the same, so the center of gravity of the ellipse will lie at the midpoint of , which will be at the point where it intersects . Hence, the center of gravity of the ellipse is at the intersection of the lines of symmetry.
By looking at this ellipse, we have seen a property of uniform laminas.
Property: The Center of Gravity of a Symmetrical Lamina
If a geometrical axis of symmetry of a uniform lamina exists, then the center of gravity lies along that line of symmetry.
Similarly, if a uniform lamina has more than one geometrical line of symmetry, then the center of gravity will lie at the intersection of these lines of symmetry.
The point at which the lines of symmetry of the lamina intersect can also be called the geometric center. Hence, the center of gravity lies at the same point as the geometric center of a uniform lamina.
We will only be focusing on laminas in this lesson, but the same concept can be extended to 3D shapes with uniform density. If they have any planes of symmetry, then the center of gravity will lie somewhere in that plane. If they have two planes of symmetry, we can narrow it down to the line where the two planes intersect, and if it has three or more planes of symmetry, then the center of gravity will be at the point where all three planes intersect.
In the next example, we will see how we can find the center of gravity of a system involving a lamina and some masses.
Example 2: Finding the Center of Gravity of a Lamina with Additional Masses Added
A uniform lamina in the form of a square of side length 28 cm has a mass of 54 grams. Masses of 10, 8, 4, and 8 grams are fixed at , , , and respectively. Find the coordinates of the center of gravity of the system.
Answer
What has been presented in the question can be considered as a system consisting of five particles. The four suspended masses are effectively particles, and the center of gravity of the lamina is also effectively a particle. Because the lamina is a uniform square, the center of gravity of the lamina is at the center of the square.
The system of particles is a two-dimensional system rather than a one-dimensional system. It is, however, possible to model the two-dimensional system as two one-dimensional systems, in the - and -directions respectively.
The one-dimensional system for the -direction can be represented by the following table:
| Total | ||||||
|---|---|---|---|---|---|---|
| Mass | 10 | 8 | 54 | 8 | 4 | 84 |
| Distance from | 0 | 0 | 14 | 28 | 28 | |
| Mass Distance | 0 | 0 | 756 | 224 | 112 | 1βββ092 |
The center of gravity of the system in the -direction is given by
Finding the position of the center of gravity of the system in the -direction is equivalent to using the following formula: where the values of , , and are found by summing the values of that are found at a given -value. Therefore,
The square has sides of length 28 cm, and the mass of the square acts at the midpoint of these lines. The value of , taking as being at , is then given by
The same method can be applied in the -direction to give the position of the center of gravity in the -direction, and it so happens that the distribution of masses around the system of particles gives the same values for , , and and values for , , and equal to , , and , respectively, as shown in the following figure.
Therefore, the value of is also 13 cm.
The position of the center of gravity of the system is
The center of gravity of the system is very close to the center of gravity of the lamina. This is to be expected, firstly, as most of the mass of the system is due to the lamina, secondly, because the other masses are geometrically distributed symmetrically around the center of the lamina, and, thirdly, because the range of the values of the other masses is small compared to the mass of the lamina.
Let us look at an example of a compound lamina.
Example 3: Finding the Center of Gravity of a Lamina Composed of a Square and a Triangle
A uniform square lamina has a side length . Another uniform lamina of the same density, shaped as an isosceles triangle, is attached to the square such that lies outside the square and . Given that the squareβs side length is times the triangleβs height, find the center of gravity of the system.
Answer
The centers of gravity of the square and the triangle that the compound lamina consists of are both readily determinable.
The center of gravity of the square is the center of the square, so it has the coordinates taking as the origin of a two-dimensional coordinate system.
The -position of the center of gravity of the triangle is at a point of the length of the median of the triangle from its base; the coordinates of this point are
The lamina is uniform, so the masses of the square and the triangle are proportional to their relative areas.
The area of the square is and the area of the triangle is and so the mass of the triangle is the mass of the square.
The centers of gravity of the square and the triangle form a one-dimensional system for the -direction that can be represented by the following table:
| Total | |||
|---|---|---|---|
| Mass | 1 | ||
| Distance from | |||
| Mass Distance |
The center of gravity of the system in the -direction is given by
This gives the coordinates of the center of gravity of the two-dimensional system as .
Now, let us look at an example where a uniform lamina is folded.
Example 4: Finding the Coordinates of the Center of Gravity of a Nonuniform Rectangular Lamina
A uniform rectangular lamina has a length of 63 cm and a width of 59 cm. It is divided into three equal-sized rectangles along its length; the last of these rectangles has been folded over so that it lies flat on the middle rectangle as show in the figure. Find the coordinates of the center of gravity of the lamina in this form.
Answer
The lamina is divided into three equal-sized rectangles. The center of gravity of each rectangle is at the geometric center of each of the rectangles. Each rectangle has a mass .
After the lamina is folded, the area bounded by the dashed line has the same coordinates as the shaded area in the figure. As the center of gravity of each rectangle is at the geometric center of each of the rectangles, the positions of the centers of gravity of the two rectangles are shown in the following figure, including their distances from in the -direction.
The shaded area consists of two stacked rectangles, so the mass of this area is double the mass of the unshaded area.
The one-dimensional system for the -direction can be represented by the following table:
| Total | |||
|---|---|---|---|
| Mass | 1 | 2 | 3 |
| Distance from | |||
| Mass Distance |
The -coordinate of the center of gravity of the system in the -direction is given by
The result can be confirmed using the following formula:
Substituting the values for and gives
The factor of appears in both the numerator and the denominator, so it can be eliminated to give
Folding the lamina does not change the distribution of its mass in the -direction, so the center of gravity of the lamina in the -direction is simply the midpoint of the 59 cm length of the lamina in the -direction.
The coordinates of the center of gravity of the lamina are, therefore, .
The weight of a body acts at the center of gravity; therefore, a body that is acted on by its weight and by an applied force can only be in equilibrium if the applied force and the weight have the same line of action.
Consider a uniform square lamina that is suspended at equilibrium from a light string at a point , as shown in the following figure.
The vertically upward force on the body due to the string tension has the same line of action as the weight of the lamina.
If the lamina was nonuniform, suspending it as shown would only establish that the line of action of the applied force passed through the center of gravity of the lamina, not the position of the center of gravity. The position of the center can be determined by suspension only by suspending the lamina from two points and finding the intersections of the lines of action of the applied forces that act at the suspension points, as shown in the following figure.
Let us look now at an example where a compound lamina is suspended from a point.
Example 5: Finding the Angle between a Straight Line and the Vertical Line in a Uniform Lamina Given That the Lamina Is Hung Freely from a Point
Find, to the nearest degree, the angle that the line makes with the vertical if the given uniform lamina is hung freely from .
Answer
Rather than suspending the lamina from a point to determine the position of its center of gravity, the question involves determining the position of the center of gravity of the lamina in order to determine the angle a side of the lamina will make from the vertical when it is suspended from a point.
The position of the center of gravity of the lamina can be determined by first determining the positions of the centers of gravity of the rectangular parts of the lamina. The center of gravity of each rectangular part of the lamina is the geometric center of that rectangle.
Before making any calculations, it is possible to approximately locate the center of gravity of the lamina by a scale drawing method, as shown in the following figure.
The scale drawing indicates that the center of gravity of the lamina has an -direction position very near to that of the point from which the lamina is to be suspended.
Modeling the lamina as two rectangles, the positions of the centers of gravity of the rectangles, and , are shown by the following figure.
The coordinates of are and the coordinates of are
As the rectangles are uniform, their relative masses are proportional to their relative areas; therefore,
From this, it can also be seen that and so
The one-dimensional system for the -direction can be represented by the following table:
| Total | |||
|---|---|---|---|
| Mass | 1 | ||
| Distance from | 4 | 13 | |
| Mass Distance | 13 |
The -position of the center of gravity of the lamina is given by
The result can be confirmed using the following formula:
Substituting known values gives
The factor of appears in both the numerator and the denominator and can be eliminated to give
This can be rearranged as follows:
Following an equivalent procedure to find the -position of the center of gravity of the lamina gives
The angle between the center of gravity and the line of action of the force that acts to suspend the lamina is shown in the following figure. The angle is very small.
When the lamina is suspended, it will rotate counterclockwise through this angle about its suspension point.
The vertical distance from suspension point to center of gravity is given by and the horizontal distance from suspension point to center of gravity is given by
Therefore, angle is given by giving a value for of
The line will rotate through the same angle. As is initially horizontal, its initial angle to the vertical is . Decreasing this angle by results in a final angle, rounded to the nearest degree, of .
Key Points
- If a net force acts on a system of particles at the center of gravity of the system, each particle in the system is accelerated due to equivalently to the acceleration by of a single particle that has a mass equal to the mass of the system.
- The weight of a body acts at its center of gravity.
- For a uniform lamina, the position of the center of gravity is the geometric center of the lamina.
- The position of the center of gravity of a one-dimensional system of particles is given by where is the mass of the particle of index and is the distance from the origin of a coordinate system of the particle of index .
- A two-dimensional system of particles can be treated as a pair of one-dimensional systems.
- The center of gravity of a lamina is the point that intersects the lines of action of multiple forces that each individually act to suspend the lamina at equilibrium.