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Centres of Bodies: Center of gravity

Center of gravity, center of mass, and centroid are three similar concepts, which can be closely related under certain circumstances. This chapter will start with discussing the center of gravity, which is affected by the distribution of weight (or gravitational forces). Gravitational acceleration can be considered constant at any part of a body within the scope of this textbook. In this case, the center of mass coincides with the center of gravity of the same body. In some engineering analysis, we only need to consider the mass distribution of a body, not the weight. The center of mass will be discussed in the second section. Further with constant density all over a body, the center of gravity and the center of mass coincide with the centroid.  In the last section, we introduce the centroid and prove that all the three concepts become equivalent under some conditions.

In many problems involving force analysis, the weight distribution or the total weight (resultant force) of a body may not be ignored and therefore should be involved in calculations . In those problems, the total weight and its location should be determined.

Assume a discrete system consisting of n particles that each has a mass of m_i (Fig. 9.1a). Located in the Earth’s gravitational field, the weight of each particle is a force with a magnitude of W_i=m_ig (where g\approx 9.81\rm m/s^2) and a direction toward the Earth. The weights of all particles form a system of parallel forces as shown in Fig. 9.1a.  The weights can be represented by an equivalent total (resultant) weight,

(9.1a)   \[\bold W=\sum_{i=1}^n \bold W_i \]

or in a scalar form,

(9.1b)   \[W=\sum_{i=1}^n W_i \]

which acts at a point called the center of gravity of the system and denoted by \rm G.

Fig. 9.1 Particles and their weights produced by the gravitational field of Earth parallel to the z axis.

The location of G, of a system is determined by the requirement that the moment of the original force system about any point must equal the moment caused by the single resultant force about the same point (see Section 3.4). A Cartesian coordinate system at some point O in the space, as shown in Fig. 9.1b, can be set up to find the coordinates (\bar x, \bar y, \bar z) of G. The coordinates of the particles and the G of a system can inform us about the location of G relative to the particles. The moment of \bold W about each Cartesian axis should equal to the sum of the moments of \bold W_i about the same axis.

As indicated by Fig. 9.1b, \bold W and \bold W_i are parallel to the z axis and therefore they produce moments only about the x and y axes. Equating the moments of the individual \bold W_i and \bold W about each of the x and y axes in scalar formulation,

About the x axis: W\bar y=\sum_{i=1}^n W_iy_i
About the y axis: W\bar x=\sum_{i=1}^n W_ix_i

leads to (the limits of the sum are not shown for the sake of conciseness),

(9.2)   \[\begin{split}\bar x&=\frac{\sum x_iW_i}{W}\\\bar y&=\frac{\sum y_iW_i}{W}\end{split}\]

where W and W_i are the magnitudes of the weights, and the coordinates (\bar x, \bar y) and (x_i,y_i,z_i) are signed distances from the origin. The sign of a coordinate determines the sign of the moment.

Equations 9.2 sufficiently determine the points through which the lines of action of \bold W_i passes. These points have the common x and y coordinates of (\bar x, \bar y) and an arbitrary z coordinate (Fig. 9.1b). To determine the z coordinate (\bar z) of G, the body and the coordinate system are rotated about the x axis while preserving the location of the origin as demonstrated in Fig. 9.2 (the body and the coordinate system can be rotated arbitrarily). The weights are now parallel to the y axis. Equating the components of the moments of the two systems about the x axis indicates that,

(9.3)   \[\bar z=\frac{\sum z_iW_i}{W} \]

Fig. 9.2 Rotation of the body and the coordinate system about the x axis while preserving the location of the origin.

Equations 9.2 and 9.3 imply that the coordinates of G are unique and determined using,

(9.4)   \[\begin{split}\bar x&=\frac{\sum x_iW_i}{W}\\\bar y&=\frac{\sum y_iW_i}{W}\\\bar z&=\frac{\sum z_iW_i}{W}\end{split}\]

Remark: the location of the center of gravity does not depend on a coordinate system chosen for the calculations; it has an absolute location with respect to any point of the body.

Remark: center of gravity of a body is not necessarily a point of the body, but rather, it can be a point of the geometric space surrounding the body.

Example 9.1.1

Determine the center of gravity, G, of the set of particles shown.

SOLUTION

The total weight W is W=15 + 20 +10=45\text{ N}. By Eq. 9.5, we can write,

    \[\begin{split}\bar x&=\frac{\sum x_iW_i}{W}= \frac{(4)(15)+(-2)(20)+(0)(10)}{45}=0.44 \text{ m}\\\bar y&=\frac{\sum y_iW_i}{W}= \frac{(6)(15)+(1)(20)+(-3)(10)}{45}=1.78\text{ m}\\\bar z&=\frac{\sum z_iW_i}{W}= \frac{(-1)(15)+(3)(20)+(1)(10)}{45}=1.22\text{ m}\end{split}\]

The following interactive example demonstrates the center of gravity of a set of particles. The particles are all located in a plane, say z=0. Add particles (up to 10 particles) with different weights and arrange their locations by dragging them around to observe the variations in the location of the center of gravity.

Equations 9.6 define the center of gravity for a discrete system which consists of a set of (a finite number of) particles or parts. A continuous physical body, however, consists of an infinite number of infinitely small particles. Determining the center of gravity of a continuous body is explained below.

Consider a continuous body in three dimensions (Fig. 9.3). The body can be divided into n small elements with size of \Delta V_i, as shown in Fig 9.3. Each element can be regarded as a particle with weight of \Delta W_i. If the coordinate of the center of gravity of an element is denoted as (\tilde x_i, \tilde y_i, \tilde z_i), The coordinates of the center of gravity of the body can be approximately calculated as,

(9.5)   \[\begin{split}\bar x&=\frac{\sum_i^n \tilde x_i\Delta W_i}{\sum_i^n\Delta W_i}\\\bar y&=\frac{\sum_i^n \tilde y_i\Delta W_i}{\sum_i^n\Delta W_i}\\\bar z&=\frac{\sum_i^n \tilde z_i\Delta W_i}{\sum_i^n\Delta W_i}\end{split}\]

Fig. 9.3 Center of gravity of a continuous body.

If the number of elements increases in the limit as n\to \infty (the sizes of the elements become infinitely small), then the above sums attain the following integral forms in the limit and uniquely determine the coordinates of the center of gravity of the body as,

(9.6)   \[\begin{split}\bar x&=\frac{\int_V \tilde x dW}{\int_V dW}\\\bar y&=\frac{\int_V  \tilde y dW}{\int_V dW}\\\bar z&=\frac{\int_V \tilde z dW}{\int_V dW}\end{split}\]

where \int_V denotes integration over the volume (V) of the body. The term \int_V dW equals W being the total weight of the body (Fig 9.3).

Since the integrals should be evaluated over all points of the the volume of the body, the concept of specific weight is used to write the integrals in the form of volume integrals. The Specific weight, \gamma of a material is defined as its weight per unit volume, i.e. \gamma=W/V. Letting dW=\gamma dV the integrals in Eq. 9.6 can be written as,

(9.7)   \[\begin{split}\bar x&=\frac{\int_V \tilde x \gamma dV}{\int_V \gamma dV}\\\bar y&=\frac{\int_V  \tilde y \gamma dV}{\int_V \gamma dV}\\\bar z&=\frac{\int_V  \tilde z \gamma dV}{\int_V \gamma dV}\end{split}\]

The specific weight \gamma can also be a function of location, i.e. (x,y,z),  meaning the specific weight of a body can be different at different places.

To implement the integrals in Eq. 9.7, we need to know the location of the centers of gravity of any differential element. If the elements are infinitely small (in all directions), we can assume the specific weight is constant for each element and the center of gravity can be considered at the geometric center (see Section 9.3) of the element. If the differential elements are cubes or in a cuboid shape, the size of each element can be written as dV=dxdydz, and since this element is infinitely small in all directions \tilde x=x, \tilde y=y and \tilde z=z. Thereby, Eqs. 9.7 become,

(9.8a)   \[\begin{split}\bar x&=\frac{\int_V x \gamma dV}{\int_V \gamma dV}\\\bar y&=\frac{\int_V y \gamma dV}{\int_V \gamma dV}\\\bar z&=\frac{\int_V z \gamma dV}{\int_V \gamma dV}\end{split}\]

and can take the form of triple integrals as,

(9.8b)   \[\begin{split}\bar x&=\frac{\iiint_V x \gamma dxdydz}{\iiint_V \gamma dxdydz}\\\bar y&=\frac{\iiint_V y \gamma dxdydz}{\iiint_V \gamma dxdydz}\\\bar z&=\frac{\iiint_V z \gamma dxdydz}{\iiint_V \gamma dxdydz}\end{split}\]

EXAMPLE 9.1.2

Find the total weight and the center of gravity of a right rectangular prism made of a material with a constant specific weight \gamma. The dimensions of the sides of the body are a, b, and c as shown.

SOLUTION

Setting a Cartesian coordinate system as shown and using a cuboidal (cubic) differential element dV=dxdydz (see figure below), calculate the weight using the integral W=\int_V \gamma dV. As \gamma is constant, we can write,

    \[W=\int_V \gamma dV=\gamma\int_V  dV=\gamma V= \gamma (abc)\]

Note that the term \int_V dV equals the volume of the body which is, here, a rectangular cube with a volume of V=abc.

Considering a cuboidal differential element dV=dxdydz as shown, calculate the triple integrals in Eqs 9.8 for determining the center of gravity.

    \[\begin{split}\bar x&=\frac{\int_V x \gamma dV}{\int_V \gamma dV}=\frac{\int_V x dV}{\int_V dV}= \frac{1}{abc}\int_0^a\int_0^b\int_0^c x dxdydz\\&=\frac{1}{abc}\int_0^c\left(\int_0^b\left(\int_0^axdx\right)dy\right)dz=\frac{1}{abc}(bc\frac{a^2}{2})\\&=\frac{a}{2}\end{split}\]

Similarly,

    \[\begin{split}\bar y&=\frac{b}{2}\\\bar z&=\frac{c}{2}\end{split}\]

briefly shown as,

Using the volumetric differential element dV=dxdydz leads to triple integrals in a Cartesian frame. However, depending on the geometry of a body, it is possible to use other shapes for differential elements and write the integrals in Eqs. 9.7 as single-variable (one-dimensional) integrals. One approach is to choose prismatic elements with differential length along only one axis (as apposed to dV=dxdydz being differential along three axes). To demonstrates this approach, we use differential elements having the shape of a right rectangular prism or a cuboid and solve the previous example again as follows.

EXAMPLE 9.1.3

Find the center of gravity and total weight of a right rectangular prism made of a material with a constant specific weight \gamma and dimensions as shown.

SOLUTION

Each equation in Eqs. 9.7 can be calculated independently, therefore, we can choose different volumetric elements for each equation.

To find \bar x, choose a differential element (with a volume of) dV=(bc)dx as shown in the following figure. The differential element is a right rectangular prism or a cuboid whose length in the x direction is dx. Because dx is small enough (approaching to zero), the center of gravity of the differential element, \tilde x, in the x direction is equal to x indicating the location of the differential element along the x axis.

Consequently,

    \[\begin{split}\bar x&=\frac{\int_V \tilde x \gamma dV}{\int_V \gamma dV}=\frac{\int_V x \gamma (bc)dx}{\int_V \gamma (bc)dx}=\frac{\int_0^a x dx}{\int_0^a dx}= \frac{a^2}{2a}\\&=\frac{a}{2}\end{split}\]

To calculate \bar y, consider a volumetric differential element dV=(ac)dy as shown in the figure. This is a prismatic element whose length dy in the y direction is infinitesimal; therefore, \tilde y = y for the element.

Similar to the previous calculations, we can write,

    \[\begin{split}\bar y&=\frac{\int_V \tilde y \gamma dV}{\int_V \gamma dV}=\frac{\int_V y \gamma (ac)dy}{\int_V \gamma (ac)dx}=\frac{\int_0^b y dy}{\int_0^b dx}= \frac{b^2}{2b}\\&=\frac{b}{2}\end{split}\]

To calculate \bar z, consider a volumetric differential element dV=(ab)dz shown in the figure. This implies that \tilde z = z for the element.

Therefore,

    \[\begin{split}\bar z&=\frac{\int_V \tilde z \gamma dV}{\int_V \gamma dV}=\frac{\int_V z \gamma (ab)dz}{\int_V \gamma (ab)dz}=\frac{\int_0^c z dz}{\int_0^c dz}= \frac{c^2}{2c}\\&=\frac{c}{2}\end{split}\]

As a result, the center of gravity of the rectangular prism is located at C(\frac{a}{2}, \frac{b}{2},\frac{c}{2}) as also seen in Example 9.1.2.

Calculate the weight using W=\int_V \gamma dV. As \gamma is constant, any of the differential element can be used to write the integral as a single-variable integral.

    \[\begin{split}W=&\int_V \gamma dV=\int_0^a \gamma (bc)dx =\int_0^b \gamma (ac)dy=\int_0^c \gamma (ab)dz\\=&\gamma (abc) = \gamma V\text{ N}\end{split}\]


Remark: avoiding triple integrals by using prismatic elements makes the calculations easier in many cases.

In the previous example, the center of gravity of the rectangular prism was located at its geometric center C(\frac{a}{2}, \frac{b}{2},\frac{c}{2}) naturally understood for the shape of a rectangular prism. The geometric center will be precisely defined in Section 9.3. Nevertheless, the center of gravity of a rectangular prism does not necessarily coincide with its geometric center if the body is made of a material with a varying specific weight. The following example clarifies this point.

EXAMPLE 9.1.4

The container shown in the figure contains a material (e.g fluid) with a specific weight linearly varying from the bottom to the top of the container, i.e. \gamma (z)=\gamma_0(\frac{-1}{c}z + 2). Find the center of gravity of the fluid and its total weight.

SOLUTION

Each equation in Eqs. 9.7 can be calculated independently, therefore, we can choose different volumetric elements for each equation.

To find \bar x, choose a differential element dV=(bc)dx as shown in the following figure.

Consequently,

    \[\begin{split}\bar x&=\frac{\int_V \tilde x \gamma (z) dV}{\int_V \gamma (z) dV}=\frac{\int_V x \gamma (z) (bc)dx}{\int_V \gamma (z) (bc)dx}=\frac{\int_0^a x \gamma (z) dx}{\int_0^a \gamma (z) dx}= \frac{\int_0^a x \gamma (z) dx}{\int_0^a \gamma (z) dx}=\frac{\int_0^a x dx}{\int_0^a dx}\\&=\frac{a}{2}\end{split}\]

Note that \gamma is a function of z and the integration is with respect to x.

To calculate \bar y, consider the volumetric differential element dV=(ac)dy shown in the figure. This implies that \tilde y = y for the element.

Similar to the previous calculations, we can write,

    \[\begin{split}\bar y&=\frac{\int_V \tilde y \gamma (z) dV}{\int_V \gamma (z) dV}=\frac{\int_V y \gamma (z) (ac)dy}{\int_V \gamma (z) (ac)dy}=\frac{\int_0^b y \gamma (z) dy}{\int_0^b \gamma (z) dy}= \frac{\int_0^b y \gamma (z) dy}{\int_0^b \gamma (z) dy}=\frac{\int_0^b y dy}{\int_0^b dy}\\&=\frac{b}{2}\end{split}\]

We can also obtain \bar x = a/2 and \bar y = b/2  by noticing that the weight of each element is constant, or the specific weight is uniform in the x and y directions.

To find \bar z, consider the following differential element and write the integration for the z direction. Note that \gamma is a function of z and the integration is also with respect to z.

    \[\begin{split}\bar z&=\frac{\int_V \tilde z \gamma (z) dV}{\int_V \gamma (z) dV}=\frac{\int_V z \gamma_0(\frac{-1}{c}z + 2) (ab)dz}{\int_V \gamma_0(\frac{-1}{c}z + 2) (ab)dz}=\frac{\int_0^c z (\frac{-1}{c}z + 2) dz}{\int_0^c (\frac{-1}{c}z + 2) dz}\\ &=\frac{4}{9}c\end{split}\]

Briefly, \bar x = \frac{a}{2},\ \bar y = \frac{b}{2},\ \bar z = \frac{4}{9}c

Note that, in general, if \gamma is not constant in least one direction, the center of gravity does not locate at the (geometric) center for a rectangular prism.

To calculate the total weight, we should use a prismatic differential element (right rectangular prism) inside which \gamma (z) is almost constant. Because \gamma (z) is a function of z, its value within the element dV=(ab)dz is constant, or approaching a constant value, as dz is small enough. Therefore,

    \[W = \int_V \gamma dV= \int_0^c \gamma_0(\frac{-1}{c}z + 2) (ab)dz=\frac{3}{2}\gamma_0 (abc)= \frac{3}{2}\gamma_0 V\text { N}\]

Composite bodies

A composite body is a (continuous) body that is composed of several sub-bodies possibly  with different or same specific weights. To calculate the center of gravity of a composite body composed of several sub-bodies with volumes V_1, V_2, \dots, V_n and specific weights \gamma_1, \gamma_2, \dots, \gamma_n, each integration in Eq. 9.7 (or Eqs. 9.6) becomes the sum of the integrations over all the composing bodies,

    \[\begin{split}\bar x&= \frac{\sum_{i=1}^n\int_{V_i}  \tilde x \gamma_i dV}{\sum_1^n\int_{V_i} \gamma_i dV}\\\bar y&= \frac{\sum_{i=1}^n\int_{V_i}  \tilde y \gamma_i dV}{\sum_1^n\int_{V_i} \gamma_i dV}\\\bar z&= \frac{\sum_{i=1}^n\int_{V_i} \tilde z \gamma_i dV}{\sum_1^n\int_{V_i} \gamma_i dV}\end{split}\]

With W_i = \int_{V_i} \gamma_i dV, the above equations can be written as,

(9.9)   \[\begin{split}\bar x&= \frac{\sum \tilde x_i W_i}{\sum W_i}\\\bar y&= \frac{\sum \tilde y_i W_i}{\sum W_i}\\\bar z&= \frac{\sum \tilde z_i W_i}{\sum W_i}\end{split}\]

where \tilde x_i, \tilde y_i and \tilde z_i denote the coordinates of the center of gravity of each sub- body. Equation 9.9 indicates that each composing body with weight W_i and its center of gravity located at (\tilde x_i, \tilde y_i, \tilde z_i) is replaced by its own total weight W_i and located at (\tilde x_i, \tilde y_i, \tilde z_i),  as what we have seen in Eqs. 9.4.

EXAMPLE 9.1.5

Find the center of gravity of the composite body shown. Each body has a constant \gamma.

SOLUTION

To determine the center of gravity of the composite body, we need to determine the weight of each sub-body and express the centers of gravity with respect to the origin of one fixed coordinate system.

Since \gamma is constant for each body, determine its center of gravity according to the formulation derived in Example 9.1.2. The following figure shows the calculated center of gravity of each body. The locations of the centers of gravity are independently determined for each body.

The weights of the bodies are W_1=(2)(5)(1)(0.7) =7 \text{ KN}, W_2= W_3 = 10\text{ KN}. Therefore,

    \[\begin{split}\bar x &= \frac{(0.5+0.5)(7)+(1)(10)+(1)(10)}{7+10+10}= 1\text{ m}\\\bar y &= \frac{(2.5)(7)+(0.75)(10)+(5-0.75)(10)}{7+10+10}=2.5\text{ m}\\\bar z &= \frac{(4.5+0.35)(7)+(2.25)(10)+(2.25)(10)}{7+10+10}=2.92\text{ m}\end{split}\]