Center of gravity

In most equilibrium problems, one of the forces acting on the body is its weight. 
We need to be able to calculate the torque of this force. The weight doesn’t 
act at a single point; it is distributed over the entire body. But we can always 
calculate the torque due to the body’s weight by assuming that the entire force 
of gravity (weight) is concentrated at a point called the center of gravity 
(abbreviated “cg”). The acceleration due to gravity decreases with altitude; but 
if we can ignore this variation over the vertical dimension of the body, then the 
body’s center of gravity is identical to its center of mass (abbreviated “cm”), 
which we defined in Section 8.5. We stated this result without proof in Section 10.2, 
and now we’ll prove it.

First let’s review the definition of the center of mass. For a collection of par-
ticles with masses m1, m2, c and coordinates 1x1, y1, z12, 1x2, y2, z22, c,

Also, xcm, ycm, and zcm are the components of the position vector r

Scm of the center 

of mass, so Eqs. (11.3) are equivalent to the vector equation

Now consider the gravitational torque on a body of arbitrary shape (Fig. 11.2). 
We assume that the acceleration due to gravity g
S
 is the same at every point in 
the body. Every particle in the body experiences a gravitational force, and the 
total weight of the body is the vector sum of a large number of parallel forces. A 
typical particle has mass mi and weight w
Si = mi gS. If rSi is the position vector of 
this particle with respect to an arbitrary origin O, then the torque vector TSi of the weight wSi with respect to O is, from Eq. (10.3),

The total torque due to the gravitational forces on all the particles is

When we multiply and divide this by the total mass of the body,

The fraction in this equation is just the position vector rScm of the center of mass, with components xcm , ycm , and zcm , as given by Eq. (11.4), and MgS is equal to the total weight wS of the body. Thus

The total gravitational torque, given by Eq. (11.5), is the same as though the total weight wS were acting at the position rScm of the center of mass, which we also call the center of gravity. If gS has the same value at all points on a body, its center of gravity is identical to its center of mass. Note, however, that the center of mass is defined independently of any gravitational effect.

While the value of gS varies somewhat with elevation, the variation is ex-tremely slight (Fig. 11.3). We’ll assume throughout this chapter that the center of gravity and center of mass are identical unless explicitly stated otherwise.

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