George Minchin Minchin.

A treatise on statics, with applications to physics online

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new circumstances it is clear that if we prevent the motion of
any one point, we prevent the motion of the system. Suppose
the motion of the point 4 to be stopped by the application of a
force, jF, in the direction A^A, A' being the point to which A
moves. Now, equilibrium exists under the action of (a) the
given external forces, (^) the newly-introduced geometrical con-
nections, and (y) the force F\ hence the sum of the virtual works
of these forces = for every displacement. Choose that dis-
placement which the system is supposed actually to undergo
when the force 2^ is not applied at A. Now, by the last Article,
since none of the geometrical conditions (^) are violated by this
displacement, the forces proceeding from them will do no work.
Hence the equation of work is

where 2P8/? denotes the virtual work of the given acting forces.
But, by hypothesis, ^Php = for every displacement, and there-
fore for this one; hence F.AA^=:0, i.e. either AA' = 0, or
F=z 0, either of which signifies that no motion of the system
takes place. Hence the system is in equilibrium.

In Fig. 135, «!, flgj ^3> ••• supposed to be virtual positions
of the points of application of the forces Pj, Pg, P3, . . .•

109.] Remarks on the Equation of Virtual Work. Equation
(1) of last Article, though strictly true in the case of forces
acting on a particle, is not so when these forces are applied at
points in a body of finite extension, or to a system of particles
connected in any manner. In fact, the internal forces of the
system have been eliminated from equations (1), (2), and (3) of
Art. 106, by assuming that ^^g— «»2^3 = ^' Now, we know
that this quantity is not strictly equal to zero, but equal to an
infinitesimal of the second order, if the angular displacement of
the line %^2 ^s regarded as an infinitesimal of the first order.
It is more correct, therefore, to say that for the equilibrium of

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a body the virtual work of the ap2)lied forces is an infinitesimal of
the second order^ if the greatest displacement in the system is re-
garded as an infinitesimal of the first order.

lie] General Uniplanar Displacement of a Bigid Body.
Since the general condition of equilibrium of a rigid body re-
quires the vanishing of the virtual work of the acting forces for
every virtual displacement which could be an actual one, it is
evidently necessary to investigate all the kinds of displacement
which such a body could undergo. Now, evidently, the position
of a right line is known, if the positions of any two of its points
are known ; and also the position of any body is known, if the
positions of any three* of its points which are not in directum
are known. Hence, to investigate the displacements to which
a rigid body may be subject, it is sufficient to determine the
general displacements of a system formed of three points.

In Fig. 134 let such a system be ;»j«»2^3> ^^^ '^^ ^^^ ^® ^^7
displacement whatever of this system in its own plane. Then
it is clear that if we moved % into the position a, and then got
^2 into the position b, the remaining point, m^ would take up
the position c. This follows from Prop. VII of the first book of
Euclid. Now what is necessary to move the line mi w^g ^^^ ^^^
position ai ? Two things —

(a) The point % must be moved up to a, by a simple motion
of translation ; and

(j3) When this is done, the line ^Wg must be rotated about
a so as to bring ^Zg i^^ ^^^ position b. This second motion is
called a motion of rotation.

If we suppose that in the first motion (a) the line m^m^ is
moved parallel to itself, while m^ is moved to a, the subsequent
motion of rotation which brings m^ into the position b will be
a small one, the position abc being only slightly different from
m-jin^ m^ ,

Hence — If a rigid body receives any displacement parallel to a
fixed plane, it may be brought from its old into its new position
by (a) a motion of translation which has the same magnitude and
direction for all its points, and (j3) a motion of rotation which has
also the same angular magnitude and sense for all its points,

* If, as in the present chapter, the displacement is made parallel to one plane,
the positions of two points wiU suffice. We use the term uniplanar to signif;^
' confined to one plane.*

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Thus, in Fig. 136, by the motion of translation common to
all the points, w^ is carried to a, while m^ is carried to b\ and

^3 to (f^ the lines m-i^m^^
m^m^, and niim^ being car-
ried parallel to themselves
to ab% iV, and ad', respee-*
tively. Then, by the motion
of rotation ab' is tamed round
to ab^ and c' is made to co-
incide with c.

111.] Beduotion of Die-
Fig. 136. " placement to notation.

Every uniplanar displace-
ment of a rigid body can be produced by rotation simply. For let
^i%^8 ^ ^^^ position, and abc any other position, of the body,
a^b^c being the displaced positions of ^x* ^2> ^3> respectively.
Draw a perpendicular to the line m^a at its middle point, and a
perpendicular to ^2^ ^^ ^^^ middle point, and let these two
perpendiculars intersect in /. Then m^m^m^ can be brought into
the position abc by a pure rotation round /. For, comparing
the triangles m-^Im>^ and a/i, we see that, since the three sides
of the one are equal to the three sides of the other, the angles % -^^
and m^Ib are equal. Hence, if the body is rotated round /
through the angle w^ / a, so as to bring % by a circular arc to
a, this rotation will bring m^ to i, and therefore every other
point of the body to its proper displaced position. If the dis-
placed position is very close to the original position, instead of
bisecting m^a and m,^b and erecting perpendiculars, we may
erect these perpendiculars at m^ and m^ to the directions, m^a, m^b,
of the displacements of % and m^. In this case the point I is
called in Kinematics the Instantaneous Centre of rotation of the

A ilisnlfl/»Ampnt of translation is one such that the centre of

Work corresponding to a Virtual Motion of
t a rigid body (Fig. 137) be in equilibrium under
' forces in one plane, Pj, Pg* Aj - > ^^^ le* ^^
>d to receive a motion of translation parallel
ne. Ox, whereby the points, A^, A^, A^^..., of
B diflferent forces receive virtual displacements,

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AiC^, A^a^, ^aflfg,..., all parallel to Oce, and equal to a. Then
(Art. 56, p. 79), the virtual work of the force Pi is a x projection
of Pi along Ox. Let the
projection of P^ along
Ox be Xi : then the vir-
tual work of Pj is aX^.
Similarly, if Xg, ^,...,
be the components of Pg,
P3,... along Ox, the vir-
tual works of these forces
will be oXg, 0X3,.., Hence


the equation of virtual
work is a(Xi + X, + Z3+...) = 0,

or a2X=0,

Consequently, since a is arbitrary, we have

2X=0. (2)

Hence — For the equilibrium of a rigid body it is necessary that
the sum of the components of the acting forces along every arbitrary
right line shall be zero.

This condition is not sufficient, since every virtual displace-
ment of a body is not one of translation alone.

113.] Virtual Work corresponding to a Motion of Bota-
tion. Let several forces, P^, Pg, P3,... (Pig. 138), acton a body
at points ^1, A^, ^3,..., and suppose that the body is rotated
through a small angle = «, round an axis perpendicular to the
plane of the forces through an ar-
bitrary point, 0. Then the points
A^, A^, A^y,,, will describe small
circular arcs, A^a^y A^a^, A^a^,...
having as their common centre,
and subtending the same angle a>,
at 0. Let $1 be the angle between
OAi and the direction of Pj. Then,
evidently, the projection of Aia^ on
the direction of Pj is Aia^. sin $1.
But Aiai = (ji).OAi; therefore the
virtual work of P^ is ©Pi.O^i sin O^.

Ifpi = the perpendicular, Oqi, from on the line of action of
Pi, this is evidently (aP^^Pi.

Fig. 138.

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Similarly, the virtual work of Pg is oaP^.p^i and that of P3 is
— WP3.JO3. Hence the equation of virtual work is

«(A/?i + -P2/>2-^3^3+-) = 0, (1)

or 2Pit? = 0. (2)

But the product of a force, P, and the perpendicular, jo, let
fall upon it from the point 0, is the moment of the force with
respect to the point 0, or rather with respect to an axis through
perpendicular to the plane of the figure.

Hence, equation (2) asserts that for equiliirium the mm {with
their prosier signs) of the moments of the forces with respect to any
point in their plane is zero.

As regards the signs to be given to the moments. Pi /? 1, P2 JP2 j • • •
of the forces, we see that —

Those forces which tend to rotate the hody in the same sense
round the point give virtual worh of the same sign, and thenfore
have moments of the same sign with respect to 0.

Thus, in Fig. 138, the forces P^ and P^ tend to turn the body
round 0, in a sense opposite to that of watch-hand rotation,
while P3 tends to turn it in the opposite sense. Hence, in
the Equation of Moments, as the equation


is called, PiPi and P^P^ ^^^ve the same sign, and P^p^ has an
opposite sign.

Since (Art. Ill) every uniplanar displacement of a rigid body can
be produced by a rotation, and since a rotation gives an equation
of virtual work which is simply one of moments round the
corresponding centre of rotation, it is clear that the necessary and
suflScient conditions of equilibrium of a system of coplanar forces
acting on a rigid body are exhausted in the statement — the sum
of the moments of the forces round every point in their plane
is zero.

Also since all possible displacements of a deformable system
are by no means exhausted in motions of translation and rotation
common to all its parts, the equation of virtual work for such a

d to the above conditions as suflScient.

1 Expression for the Displacement of a

shall now investigate the changes produced
of any point in a rigid body by given small

ion and rotation. Let the motion of trans-

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Fig. 139.

lation first take place. Then draw any two rectangular axes, Ox
and Oy, through 0.(Pig. 139) the new position of a point Oj.
Let the motion of translation Ofi^ common to all parts of the
body, be resolved in two compo-
nents, a and d, parallel to Ox and

Then, if a? and^ denote the co-
ordinates of a point Q^ in the body
with reference to fixed axes drawn
through Pi parallel to Ox and 0^,
these quantities will be increased
by a and i, respectively, by the
motion of translation. To find how much they will be subse-
quently altered by an angular rotation = w round 0, let Q
describe a small arc of a circle, Q^, round 0.

Let fall the perpendiculars QM and qm on Ox^ and Qp on qm.
It is evident that OMz=i x and Q Jf = y. Then the increase oiy pro-
duced by the rotation = qp^ and the increase in a? = — Qp. Now

Qp = Qj'.sin QO«=a).OQ.sin QOa? = a).Qif=cay;

and g^= Qq, cosQOa? = a).OQ.cos Q,Ox = (a. OJf=a)a?.

Hence, if hx and hy denote the changes produced in x and y by
the two motions combined,

hx = a-^tay^ (l)

5^ = i-l-wa?. (2)

These are the general analytical expressions for the displace-
ments of a particle in the body, (They can obviously be obtained
by differentiating the equations a? = r cos ^, y = r sin ^, on the
supposition that 6 alone varies by a quantity 8^ = 0), and then
adding a and b to the results.)

115.] Analytical Conditions of Equilibrium. If any forces,
P^ P29 -Paj'-j act on a rigid body in one plane, the condition
necessary and sufficient for equilibrium is (Art. 108)

P,hp^+P^bp2-{-P^bp,-\-... = 0. (1)

Let X^ and Y^ be components of P^ along two rectangular axes,*
Ox and Oy, and let Xi and y^ be the co-ordinates of the point at
which Pi acts. Then (Art. 57^ p. 80)

P,bp, = X^bx^+r^by,. (2)

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Making similar substitutions for P2^P29 ■Pa^/'sj*-., equation (1)

X^bx^-i-T^by^ + X^ba^^ F^by^ + ... = 0, (3)

or 2(XbX'hrby)=zO. (4)

Substituting in (4) the values of bx and by given in the last
Article, we have

2 {X(fl— a:y)+ r(J4-«a?)} = 0,
or a.2X+b. 2r+».2(a?r-yZ) = 0,... (5)

since a, d, and a> are common to all points of the body, and may
be taken outside the sign of summation.

Now the displacements a, by and o) are completely independent
of each other, and therefore equation (5) requires that

2X=0, 27=0) .V

2(a?r-yX) = or ^ '

For, choose another virtual displacement in which a and 6 are
the same as before and ta different. Then we have

a2X+ i2r+©'2(a?r-yX) = 0. (7)

Subtracting (7) from (6),

(a)-a>')2(a?r-yX) = 0.

But since co — © is not = 0, this equation requires that

2(a?r-yX) = 0.

Similarly, by making a alone variable, we prove that 2X = 0,
and by making b alone variable, 2 J = 0.

The three equations (6) constitute the analytical conditions of
equilibrium of the body.

116.] Varignon's Theorem of Moments. TAe moment of the
resultant of two forces with respect to any point in their plane is
equal to the sum of the moments of the forces with respect to this

point. The following is the
proof of this proposition by the
principle of Virtual Work.
Let B (Fig. 140) be the re-

^^ /^^-^-""^ ^ sultant of two forces, P and Q,

^ applied at a point -4, and let

^' '^^* be any point in their plane.

Then the virtual work of B for any displacement of A = the

virtual work of Ph- the virtual work of Q. Let the virtual dis-

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placement of A be one of rotation round 0, through a small
angle = co. Then, as in Art. 113, the virtual work of i2 is a> . i2 . OA.
sin OAR ; but this = « . 22 x the perpendicular from on i2 = <» x
the moment of H with respect to 0. Similarly, the virtual work
of P = 0) X moment of P with respect to ; and virtual work
of Q =5 0) X moment of Q with respect to 0. Therefore, &c. —


In precisely the same way, the moment of the resultant of any
number of forces is proved to be equal to the sum of the moments
of the forces separately.

117.] Partioalar case in which the Besultant of Transla-
tion vanishes. When forces applied to a particle have no
resultant of translation, their whole effect is null. It is not
necessarily so, however, if they are applied to a body of finite
dimensions. For emmple —

If the forces acting upon a rigid body form by their magnitudes
and lines of action the sides of a closed polygon taken in order ^ their
resultant of translation vanishes^ and they have a constant moment
with respect to all points in their plane.

Let forces Pj, Pg, P3, ... (Fig. 141) act at points, -4^,^, -ig,...
in a body in one plane, and let
these forces be represented in mag-
nitudes and lines of action by the
sides of the polygon formed by their
points of application.

Now since (Art. 55) the sum of
the projections of the sides of this
polygon on any arbitrary line = 0,
the condition of Art. 112 is fulfilled,
and the forces have no resultant of

Let be any point inside the polygon, and take the sum of
the moments of the forces round it. If the perpendiculars from
on the sides A^A^, A^^,..., be ^1, i?2> - > ^^^ sum of the
moments will be

Aa + Ai»2 + ^8A + - = ^» suppose.
And since P^, Pg,... are equal to the sides of the polygon, is
evidently = 2 . area of polygon. This is constant for all points
inside the polygon.

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Now if we take the sum of the moments round any external
point, (7, we shall have

since P^ turns the body round C in a sense opposite to that
in which the other forces turn it. But this sum is equal to

and this is again equal to 2 . area of polygon.

Hence for all points in the plane, the sum of the moments, (?,
is constant.

118.] Theorem, ^a mimber of forces acting upon a rigid body
in one plane have a constant moment with respect to all points in the
plane, they can have no resultant force, and must be reducible to
a couple.

For, suppose that they have a resultant = JB, then if p is the
perpendicular let fall on R from any point, 0, the sum of the
moments of the forces = B.p (Art. 116). Hence by varying the
position of 0, the sum of the moments varies, which is contrary
to hypothesis. They are reducible to two equal, parallel, and
opposite forces. For their resultant is zero ; hence, compound-
ing them in pairs, they must reduce to two parallel, equal, and
opposite forces forming a couple, or to two such forces directly
opposite to each other in a right line. But in the latter case the
sum of their moments about any point would be zero ; therefore
if this moment is not zero, the forces must reduce to a couple.

119.] Two Parallel Forces. To fnd the resultant of two
parallel forces, P and Q, acting in the same sense.

Let AB (Fig. 142) be the shortest distance between P and Q,
and let the forces be supposed to act at A and JB. Also let the

reversed resultant, R, act at
any point, 0, in AB. Since
the forces are in equilibrium,
their virtual work = for every
virtual displacement (Art. 109).
Choose first a virtual displace-
ment of translation along AB,
J. For this displacement the vir-

tual work of the forces P and
the virtual work of -B = 0, therefore R is
Q. Again^ choose a virtual displacement of

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rotation about through an angle = co. The virtual work of P
is then OA^ and that of Q is — Q.o^OB, while that of R is
zero. Hence

P.OA^Q.O£=zO, (1)

'• OB" P'
Finally, to find the magnitude of 5, take a virtual displacement
of translation parallel to the forces. This evidently gives

It=P+Q. (2)

Therefore tke resultant cf two parallel forces acting in the same
sense is a force parallel to them in the same sense, equal to
their sum, and dividing the line joining their points of application
in the inverse ratio of the forces.

Equation (l) asserts that the moments of two parallel forces
with respect to any point on their resultant are equal and

If P and Q act in opposite senses
(Fig. 143), the resultant is obtained in
magnitude and direction by simply chang-
ing the sign of Q.

Thus (1) becomes q^*-


^ - ^ , /ox Fig. 143.

OB" P' ^^

which shows that is on the production of AB at the side of
the greater force; and (2) gives

Ii = P-Q. (4)


1. To solve example 12, p. 138, by the principle of Virtual Work.

Imagine a displacement in which the ends A and B remain in con-
tact with the planes. Then the virtual works of B and S are zero,
and if y is the height of G above the horizontal plane, the equation
of virtual work is

-Tray-P.(£(il(7) = 0. (1)

Now y = a sin ^, AC = (a + b) cos 6 ; .'. c?y = a cos ^ d0, and

d(AC)=: '-{a+h)Bmddei

.'. (1) gives Facosi? = P(a + 6)sin^, or tan^= p . . »

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2. To solve example 13, p. 138, by the principle of Virtual Work.
Choosing a virtual displacement which keeps A and B in contact

with the planes, the equation of work is

'-'Wdy'-T.d{AC)z=iO. (!)

Now jPC"= BP^cobV + AP^ an% and this equation also holds in the
displaced position. Hence we may differentiate it, and we then obtain
FC.d (PC) = -{PB^-PA^ sin cos OdO

=r ^{a'\'b){PB'-PA)tanecoBede

,. ,i(P(7)=-(a + 6)(^-?i5S|)8indcosd(W

x= —(a + 6)sin(d— <^)ci^.

Also y = a sin d, .*. c2^ =5 a cos 6d0 ; and substituting these values
of d{PC) and dy in (1), we obtain the value of T,

3. Four rigid bars, freely jointed together at their extremities,
form a quadrilateral, ABCD ; the opposite vertices are connected by

strings, AG and BD^ in a state of
tension; compare the tensions of
these strings.

Let the bar AB be considered as
fixed, and let the quadrilateral
undergo any slight deformation.
Then the bars AD and BG will
turn round the points A and B,
that is, the points D and G will
describe small paths, Dd and (7c,
perpendicular to iLZ> and ^(7. Hence
(Art. Ill) the point, 7, of inter-
section oi AD and BG is the in-
stantaneou^ centre for the bar GD,
and the angles Did and GIc are
equal. Denote their common value by b$. Then Dd = ID . dd, and

Now, since in the displacement of the system none of the geo-
metrical conditions — ^namely, the constancy of the lengths of the bars
— are violated, the reactions of the bars will not enter into the equa-
tion of virtual work. Hence if T and T^ denote the tensions of the
strings AG and BD, this equation will be (see p. 90),

T.bAG-^r.bBD:=:0. (1)

But bAG = projection of Gc on AG

=: Gc . Bin AGB =i IG .mi AGB . b e ;

and similarly bBD =^ —ID , Bin BD A . b6. Hence (1) becomes

T.IG . siuilC^ = r.ID . sin^2>il. (2)

Again, _______


Fig. 144.

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[19.] EXAMPLBa 175

Substituting in (2), we obtain


Another solution of this problem (quoted from Euler) will be found
in Walton's Mechanical Problema, p. 101.

4. Four rigid bars, freely jointed at their extremities, form a
quadrilateral, A BCD ; the bars AB and AD are connected by a string,
aa in a state of tension, a being a given point in AB, and a a given
point in AD ; in the same way, BA and BG are connected by a string
6)3 ; CB and CD are connected by a string cy; and DC and DA by
a string db ; find the relation between the tensions of these strings.

K the lengths of the strings aa, 5)3, cy and db are denoted by x, y, z,
and w, and the tensions in them by X, Y, Z^ TT, the equation of virtual
work for a slight deformation will be

X&»+ Yby-hZbz-^ Wbw = 0. (1)

Now a?=:zAa*+Aa^'^2Aa.AacoBA ^Aa^-{'Aa*

therefore xbx = 2-r^^-j^-BD .iBD.


Substituting this value of b«, and similar values of 5y, bx, iw, in
(1), we have

But from the last Example, we have


, ^ ,. fX Aa.Aa Z Cc.Cy \ BD^

hence, finally, (- ' J^TID -^ T ' CBTCd) OBTO


_(Y^ Bh.Bp W Dd.Db N AC^
"Vv 'bA.BC'^ to ^ DC.Da)\

For a different solution, see Walton, i&t d.

5. Six equal heavy beams are freely jointed at their extremities ;
one is fixed on a horizontal plane, and the system lies in a vertical
plane ; the middle points of the two upper non-horizontal beams are
connected by a rope in a state of tension. Show that the tension
oftiiisropeis BTTcotd,

W being the weight of each beam, and B the inclination of the non-
horizontal beams to the horizon.

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Let X be the length of the rope, y the height of the centre of
gravity of the system, 2 a the length of each beam, and T the tension
of the rope. Then the virtual work of the tension is—Tbx (see p. 90),
and the virtual work of the weight of the system is— 6 Wby, Hence

But a? = 2a (1 + cos 0\ and y = 2a sin 0, and the deformation imagined
is one in which the upper horizontal beam moves vertically through
a small space. Hence the values of y and x will be of the same forms
as before, and

bx= —2aBmdbO, by = 2acoB6b$.

Substituting these values of bx and 5y, we have


6. A body receives a small general displacement parallel to one
plane ; find the co-ordinates of the instantaneous centre.

J£ the components of the motion of translation parallel to the axes
of X and y are ba and bb, and the rotation is 5a>, the equations
of Art. 114 give for the displacement of any point whose co-ordinates
are jb, y,

djB = ba^ybci)

by = bb + xboi.

Now, the displacement of the instantaneous centre is zero ; hence,
if (ou, y) be its co-ordinates, we have

b<a "^ "" do)

A particular case may be noticed. K any body in contact with a
suiface receives any small displacement parallel to one plane, the body
still remaining in contact with the surface, the instantaneous centre
lies on the normal to the surface of contact. In the rolling of one
figure on another the point of contact is the instantaneous centre.

7. A uniform beam, AB (Fig. 127, p. 148), rests as a tangent at a
point F against a smooth curve in a vertical plane, one extremity, A,
resting against a smooth vertical plane; find the position of equi-
librium, and the nature of the curve so that the beam may rest in
all positions.

Let the weight of the beam through G, and the normal reactions

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