(H4 
MECHANICS, 
lei directions, and, as they are proportional to the weights 
of" the particles, they will be in equilibrio about the centre 
of gravity. 
Prop. XXXV. To find the centre of gravity of any number of 
bodies placed in a firaig fit line. —Let A, B, C, D, E, (fig. 64..) be 
any number of bodies whofe common centre of gravity is <p. 
In the ftraight line AE take any point X. Then, fince all the 
bodies are in equilibrio about their common centre of gra¬ 
vity <j>, we have, by the property of the lever, AX A$+ B X 
B 4 >=Cx C<5>-f-DX D<j>4-E x E<*> ; but, fince X<p —XA = A<f, 
and X$— XB=B<|>, and fo on with the reft, we have by 
fubftitution A X X<j>—X A-J-Bx X<j>—XB=iC x Xt—XC + 
DxX<|)—XD-j-ExX<f—XE. Hence, by multiplying and 
tra'nfpofing, we obtain A X X<j>-j-BxX<f>-|-C>< Xif-f^XXli 
+ ExX<f=AxXAJ r BxXB+CxXC + DxXD-l-Ex 
XE; then dividing by A+B+C-f-D-J-K, we have 
X'P— 
A X XA-f BX XB+GXXC+DXXD+EX XE 
A-f-B-J-C+D-j-E 
Now AxXA, BxXB, &c. are evidently the momenta of 
the bodies A, B, &c. and the divifor A-J-B-j-C+D-J-E is 
the fum of the weights of ail the bodies; therefore the 
diftance of the point X from the centre of gravity <p is 
equal to the fum of the momenta of all the weights di¬ 
vided by the fum of the weights. 
Cor. 1. If the point X had been taken between A and 
E, at x for example, then the quantity AxXA would 
have been reckoned negative, as lying on a different fide 
of the point X. 
Cor. 2. From this Propofition we may deduce a general 
rule for finding the centre of gravity in' any body or fyf- 
tem of bodies. Let any point be affumed at the extre¬ 
mity of the fyfiem; then the product of the momenta of 
all the bodies (or the product arifing from the continual 
multiplication of each body by its diftance from the 
point), divided by the fum of the weights of all the bo¬ 
dies, will be a quotient which exprefles the diftance of 
the centre of gravity from the point affumed. 
Prop. XXXVI. If, in a fyfiem of bodies , a perpendicular 
be let fall from each upon a given plane, the fum of the pro- 
duEls of each body multiplied by its perpendicular difiance from 
the plane is equal to the fum of all the bodies multiplied by the 
perpendicular dijlance of their common centre of gravity from the 
given plane .—Let A, B, C, (fig. 65.) be the bodies which 
compofe the fyitem, and MN the given plane; by Prop. 
XXXIV. find F the centre of gravity of A and B, and 
G the centre of gravity of the three bodies; and from 
A,F, B,G,C, draw A a, Ff, B b, Gg, C c, perpendicular 
to the plane MN. Through F draw xFy, meeting A a 
produced in x, and B b in y; then, in the fimilar triangles 
AkF, Bjy F, we have A x : By= AF : BF, that is, (Prop. 
XXXI.) as B : A ; hence, AxAairrBxB^; that Is, 
A y,x a — An = BxBi — y b, or, on account of the equa¬ 
lity of the lines x a,Ff,B b -, A X Ff —A c~BxB b — Ff-, 
therefore, by multiplying and tranfpofing, we have A-^-B 
X Ff— A X A a Bx B b. In the very fame way, by 
drawing zv G z parallel to the plane, it may be fhown, that 
A + B + CXG£=Ax A« + BxB i + CxCc. 
Cor. By dividing by A+B-j-C, we have G= 
A'X A «-{- B X B £ + (J X C c 
A-fB+C 
Prop. XXXVII. To find the centre of gravity of a Jlraight 
fine, compofedof material particles .—If w>e confider the Itraight 
line as compofed of a number of material particles of the 
fame fize and denfity, it is evident that its centre of gra¬ 
vity will be a point in the line equidiftant from its extre¬ 
mities. For, if we regard the line as a lever fupported 
upon its middle point as a fulcrum, it will evidently be in 
equilibrio in every pofition, as the number of particles or 
weights on each fide of the fulcrum is equal. 
Prop. XXXVIII. To find the centre of gravity ofa parallelo¬ 
gram. —Let ABCD (fig. 66.) be a parallelogram of uniform 
denfity ; bifeCt AE inF, and, having drawn F f parallel to 
AC or BD, bifeCt it in ; the point <j> will be the centre of 
gravity of the parallelogram. The parallelogram may be 
regarded as compofed of lines A B, ab, parallel to one 
another, and confiding of material particles of the fame 
lize and denfity. Now, by Prop. XXXIV. the centre of 
gravity of AB is F, and the centre of gravity of a b is c; 
and in the fame way it may be fhown that the centre of 
gravity of every line of which the furface is compofed 
lies in the line F f. But Ff may be confidered as com¬ 
pofed of a number of material particles of uniform den- 
lity, each being equal in weight to the particles in the 
line A B, therefore, its centre of gravity w ill be in <?, its 
middle point. 
Prop. XXXIX. To find the centre of gravity of a triangle. 
—Let ABC (fig. 67.) be a triJngle of uniform denfity, and 
let AB, BC, be bifected in the points E and D. Join 
CE, AD, and the point of interfeCtion F will be the cen¬ 
tre of gravity of the triangle ABC. The triangle may¬ 
be confidered as compofed of a numberof parallel lines of 
material particles, B C, be, 13 x ; but in the fimilar tri¬ 
angles ADC, Arc; AD : DC—A e : e c ; and in the tri¬ 
angles ADC, ADB, A e b ; BD : DA=zb e : e A ; hence, 
by compofition, BD : TtCezzb e : e c ; but BD and DC are 
equal; therefore, be=zec, and the line be, fuppofed to 
eonfift of material particles, will be in equilibrio about e. 
In the fame way it may be fhown that every other line (3 k 
will be in equilibrio about a point iituated in the line AD ; 
confequently, the centre of gravity is in that line. For the 
fame reafon it follows, that the centre of gravity is in the 
lineCE; that is, it will be in F, the point of interfeCtion of 
thefe two lines. In order to determine the relation be¬ 
tween FA and FD, join ED; then, fince BE=EA, and 
BD=DC, BE : EA:=ED : DC ; and, confequently, ED 
is parallel to AC, and the triangles BED, BAG, fimilar. 
We have, therefore, CA : CB —DE : DB, and by alter¬ 
nation CA : DE=CB : DB ; that is, CA : DE=2 : 1. 
In the fimilar triangles CFA, DFE, AF : AC=DF : DE, 
and by alternation AF : DF=AC : DE ; that is, AF : 
DFzr2 : 1, or AFzrifAD. 
Cor. 1. By a well-known theorem in geometry, we have 
AB 2 + AC 2 r=aBD 2 + 2 AB 2 =j-BC 2 -j-|-AF 2 
AB 2 4-BC 2 2=2CC 2 +aBG 2 =iAC 2 +|CfH' 
AC 2 + B C 2 =:2 AE 2 -)-2 E C 2 =f AB 2 -}-|bF" 2 . 
By adding thefe three equations, and removing the frac¬ 
tions, we have AB 2 -f-BC 2 -j-AC 2 :=3AF 2 -f3CF 2 +3BF 2 ; 
or, in any plane triangle, the fum of the fquares of the 
three fides is equal to thrice the fum of the fquares of the 
diftances of the centre of gravity, from each of the angu¬ 
lar points. 
Cor. 2. By refolving the three quadratic equations 
in the preceding corollary, we obtain A F = 
if 2 AB 2 +2 A C 2 ; CF-j- F2BA 2 +2BC 2 — AC*j 
and BF—ifz BC 2 -|-2 AC 2 —AB 2 ; formulae which ex- 
prefs the affiances of the Centre of gravity from each of 
the angular points. 
Prop. XL. To find the centre of gravity of a trapezium, or 
any redilinealfigure. —Let A B C D E, (fig. 68.) be the tra¬ 
pezium, and let it be divided into the triangles A B C, 
ACE, E C D, by the lines AC, E C. By the laft Propo¬ 
fition, find m, n, 0, the centres of gravity of the triangles, 
and take the point F in the line m n, fo that F» : F m=z 
triangle ABC : triangle ACE; then F will be the centre 
of gravity of thefe triangles. Join Fe>; and find a point f, 
fo that fo : Ff= triangle ABC -}- triangle ACE : trian¬ 
gle CED ; then all the triangles will he in equilibrio about 
f \ that is, / is the centre of gravity of the rectilineal 
figure A B C D E. The fame method may be employed in 
finding the centre of gravity of a trapezium, whatever be 
the number of its fides. 
Prop. XLI. If any momenta be communicated to the parts 
of a fyfiem,, its centre of gravity will move in the fame manner 
that a body, equal to the fum of the bodies in the Jyfiem, would, 
move, were it placed in that centre, and the fame momenta, in 
the fame dir dims, communicated to it, —Let A, B, C, fig. 6 9. 
be 
