The mass of an object is a measure of how much stuff there is in the object. An object has
mass equal to one kilogram, a unit of mass in the metric system, if it would exactly balance a
known one kilogram object when placed on a balance. The known object is one kilogram by
definition. The mass of an object does not depend on where the balance is used. It would be one
kilogram on the moon as well as on the earth. The weight of an object is something else. It is the
force exerted on the object by gravity and has magnitude gm where g is a constant called the
acceleration of gravity. Thus the weight of a one kilogram object would be different on the
moon which has much less gravity, smaller g, than on the earth. An important idea is that
of the center of mass. This is the point at which an object will balance no matter how it is
turned.

Definition 3.5.4Let an object consist of p point masses m_{1},

⋅⋅⋅

,m_{p}with the position of the k^{th}of theseat R_{k}. The center ofmass of this object R_{0}is the point satisfying

∑p
(Rk − R0 )× gmku = 0
k=1

for all unit vectors u.

The above definition indicates that no matter how the object is suspended, the total torque on it due to
gravity is such that no rotation occurs. Using the properties of the cross product

( ∑p ∑p )
Rkgmk − R0 gmk × u = 0 (3.30)
k=1 k=1

(3.30)

for any choice of unit vector u. You should verify that if a × u = 0 for all u, then it must be the case that
a = 0. Then the above formula requires that

∑p ∑p
Rkgmk − R0 gmk= 0.
k=1 k=1

dividing by g, and then by ∑_{k=1}^{p}m_{k},

∑p
R0 = -∑kp=1Rkmk--. (3.31)
k=1 mk

(3.31)

This is the formula for the center of mass of a collection of point masses. To consider the center of mass of
a solid consisting of continuously distributed masses, you need the methods of calculus.

Example 3.5.5Let m_{1} = 5,m_{2} = 6, and m_{3} = 3 where the masses are in kilograms. Suppose m_{1}is located at 2i + 3j + k,m_{2}is located at i− 3j + 2k and m_{3}is located at 2i−j + 3k.Find the centerof mass of these three masses.