1
Theoretical Mechanics
2
3
Coplanar system of concurrent forces,
Forces whose lines of action intersect at one
point are called concurrent,if all the concurrent
forces acting on a body lie in one plane,they
form a coplanar system of concurrent forces,
Introduction
① coplanar system of concurrent forces
② coplanar system of parallel forces (coplanar
system of force couples is a special cast)
③ general case of a force system in a plane
The methods of study,the graphical method
and the analytical method,
Example,the
hook of a crane
Force system,coplanar force system or three-dimensional
force system
Coplanar
force
system,
Special cast of force system in a plane,coplanar system of
concurrent forces,coplanar system of force couples and coplanar
system of parallel forces,
4

① 平面汇交力系

③ 平面一般力系 (平面任意力系 )

5
§ 2–1 The graphical method of composition
and the equilibrium of a coplanar system of
concurrent forces
§ 2–2 The analytical method of composition
and the equilibrium of a coplanar system of
concurrent forces
§ 2–3 The concepts and the character of a moment
and a force couple
§ 2–4 The composition and the equilibrium of a
coplanar system of force couples
§ 2–5 The composition and the equilibrium of a
coplanar system of parallel forces
Chapter 2,Special cases of force systems in a plane
6
§ 2–1 平面汇交力系合成和平衡的几何法
§ 2–2 平面汇交力系合成和平衡的解析法
§ 2–3 力矩,力偶的概念及其性质
§ 2–4 平面力偶系的合成与平衡
§ 2–5 平面平行力系的合成与平衡

7
§ 2–1 The graphical method of composition and the
equilibrium of a coplanar system of concurrent forces
Ⅰ The graphical method of the composition of forces
?c o s2 212221 FFFFR ???
2,The composition of any
coplanar concurrent forces
The force
polygon,
1.The composition of two concurrent forces
According to the law of cosines,
the direction of the resultant force is,
According to the law of cosines,
?? c o s)1 8 0c o s ( ????By the parallelogram rule or
constructing a force triangle,
)180s i n (s i n
1
?? ??
RF
8
§ 2-1 平面汇交力系合成与平衡的几何法

?c o s2 212221 FFFFR ???
)180s i n (s i n
1
?? ??
RF
2,任意个共点力的合成

1.两个共点力的合成

?? c o s)1 8 0c o s ( ????由力的平行四边形法则作，

9
Conclusion,In general,
The resultant of a coplanar system of concurrent forces is
equal to the geometrical sum of these forces and it applies
to the point of intersection of these forces,
Ⅱ The graphical condition of equilibrium
?? FR
4321 FFFFR ????
If the resultant force is zero,the force
polygon draw with these forces is closed,
So the graphical condition of equilibrium of
a coplanar system of concurrent forces is,
The necessary and sufficient condition is,? ?? 0FR
The force polygon is closed or
The geometrical sum of all forces is zero,
10

?? FR
4321 FFFFR ????

? ?? 0FR

11
[Example] A road-roller runs over a barrier,the weight of the roller is
P=20kN,its radius is r=60cm and the height of the barrier h=8cm,
Determine the magnitude of the horizontal force F acting on the center
of the roller and the force with which the roller acts on the barrier,
577.0)(tg
22
?? ??? hr hrr?
By the geometrical relation
① choose the roller as the body to be studied;
② isolate the body,draw the force diagram,
Solution,
∵ When the roller just left the ground NA=0,F is
maximum,F,gravitation and reaction
force NB form of a balanced force system,
According to the graphical condition of equilibrium
?tg?? PF ?co sPN B ?,
,
,
12
[例 ] 已知压路机碾子重 P=20kN,r=60cm,欲拉过 h=8cm的障碍

577.0)(tg
22
?? ??? hr hrr?

① 选碾子为研究对象
② 取分离体画受力图

∵ 当碾子刚离地面时 NA=0,拉力 F最大,这时

?tg?? PF ?co sPN B ?
13
By the relation between action force and reaction force,the force
with which the roller acts on the barrier is 23.1kN,
This problem can also be solved by the method of the force
polygon,measure the magnitude of the force by a scale,
F=11.5kN,NB=23.1kN,Therefore
The steps of the graphical method are,
① choose a body to study; ② draw the force diagram;
③ draw the force polygon with a proper scale;
④ determine the unknown quantities,
The shortages of the graphical method are,
① the precision is not enough,the error is large;
② drawing the diagram need higher precision;
③ it can not show the functional relations among the forces,
We shall study the other method of the composition and equilibrium
of coplanar systems of concurrent forces,
The analytical method
14

F=11.5kN,NB=23.1kN 所以

③作力多边形，选择适当的比例尺；
④求出未知数

③不能表达各个量之间的函数关系。

15
F
F
F
X x???c o s
F
F
F
Y y???c o s
22
yx FFF ??
§ 2-2 The analytical method of composition and the
equilibrium of a coplanar system of concurrent forces
Ⅰ Projection of a force on an axis
X=Fx=F·cos?,
Y=Fy=F·sin?=F ·cos?
16
F
F
F
X x???c o s
F
F
F
Y y???c o s
22
yx FFF ??
§ 2-2 平面汇交力系合成与平衡的解析法

X=Fx=F·cos?,
Y=Fy=F·sin?=F ·cos?
17
Ⅱ The law of projection
of a resultant force From the diagram the projections of
all forces on the x and y Axes are,
????? XXXXR x 421
??????? YYYYYR y 4321
?? YR y?? XR x
The law of projection of a resultant force,
The projection of the resultant force on an axis is equal to the
algebraic sum of the projections of all forces on the same axis,

18

????? XXXXR x 421
??????? YYYYYR y 4321
?? YR y?? XR x

19
The magnitude of
the resultant force,
Its direction,
Its point of application,
x
y
R
R
??tg ?
?
?? ??
X
Y
R
R
x
y 11 tgtg?∴
The point of intersection of all forces,
Ⅲ The analytical method of composition and the equilibrium
See above,for a coplanar system of concurrent forces to be in
equilibrium it is necessary and sufficient that the resultant of these
forces is zero,
?
?
??
??
0
0
YR
XR
y
x
This is the necessary and sufficient condition
of equilibrium,called balanced equations,
00 22 ???? yx RRR
?? ???? 2222 yxRRR yx
20

x
y
R
R
??tg ?
?
?? ??
X
Y
R
R
x
y 11 tgtg?

?
?
??
??
0
0
YR
XR
y
x 为平衡的充要条件，也叫平衡方程
00 22 ???? yx RRR
?? ???? 2222 yxRRR yx
21
Solution,
① study the rod AB;
② draw its force diagram;
③ write down the balanced equations;
④ solve the equations,
0?? X
0??Y
045c o sc o s 0 ???? CDA SR ?
045s i ns i n 0 ?????? CDA SRP ?
[Example] Determine SCD and RA,if P=2kN,
EB=BC=0.4m,
3
1
2.1
4.0tg ???
AB
EB?
kN 24.4tg45c o s45s in 00 ???? ?PS CD kN 16.3
c o s
45c o s 0 ???
?CDA SR；
?
22

②画出受力图
③列平衡方程
④解平衡方程
0?? X
0??Y
045c o sc o s 0 ???? CDA SR ?
045s i ns i n 0 ?????? CDA SRP ?
[例 ] 已知 P=2kN 求 SCD,RA

3
1
2.1
4.0tg ???
AB
EB?解得,
kN 24.4tg45c o s45s in 00 ???? ?PS CD kN 16.3
c o s
45c o s 0 ???
?CDA SR；
?
23
[Example] P,Q are known,Determine and the reaction force
ND in equilibrium,
?
Solution,study the ball,choose an axis of
projection and write down the balanced
equations,
PP-TN D 3Q60s i n2Qs i n-Q 02 ????? ?
from② we find,
060???
212c o s 21 ??? PPT
T?
from① we find,
0?? X
0?? Y
0co s 12 ??? TT ?
0Qs i n2 ???? DNT ?

24
[例 ] 已知如图 P,Q,求平衡时 =？ 地面的反力 ND=？ ?

PP-TN D 3Q60s i n2Qs i n-Q 02 ????? ?

060???
212c o s 21 ??? PPT
T?

0?? X
0?? Y
0co s 12 ??? TT ?
0Qs i n2 ???? DNT ?

25
Again,
?co s
FN ?
)2(1)(c o s
22
hRhRR hRR ??????
)2( hRh
RFN
??
???
[Example] Determine the magnitude of F when the ball just left the
ground if we know P,R,h,
Solution,study the little block,the forces on it are shown below,
Solve the force triangle,
26

?co s
FN ?
)2(1)(c o s
22
hRhRR hRR ??????
)2( hRh
RFN
??
???
[例 ] 求当 F力达到多大时，球离开地面？已知 P,R,h

27
Then study the ball,the force
diagram is given below
Draw the force triangle,
Solve the force triangle,
?s i n??? NP?
R
hRA g a i n ???s i n? NN ??
R
hR
hRh
RFNP ??
??
?????
)2(
s i n ?
)2(
)(
hRh
hRFP
?
???
hR
hRhPF
?
??? )2(
When the ball left
the ground,NB=0
hR
hRhPF
?
??? )2(When,the ball can leave the ground,
28

?s i n??? NP?
R
hR ???s i n?又 NN ??
R
hR
hRh
RFNP ??
??
?????
)2(
s i n ?
)2(
)(
hRh
hRFP
?
???
hR
hRhPF
?
??? )2(

NB=0时为球

29
1,Generally,when a body is subjected to the action
of three forces and the angles are special,it is
convenient to use the graphical method (solve
the force triangle),
The skills and introductions of problem solving,
3,Usually,it is best to choose the coordinate axes
perpendicular to the unknown forces,try to
simplify every equation with only one unknown
quantity,
2,In general,when a body is subjected to the action
of many forces and angles are special or not,it can
be solved by the analytical method,
30
1、一般地，对于只受三个力作用的物体，且角度

3、投影轴常选择与未知力垂直，最好使每个方程中

2、一般对于受多个力作用的物体，且角度不特殊或

31
5,Solving a problem by the analytical method the
direction of the forces can be assumed to be
arbitrarily,If the result is negative,it denotes that
the direction of the force is opposite to the
assumed one,If a component subjected to the
action of two forces,we commonly first assume
that the forces is a tension,If the result is negative
it denotes that the body is under compression,
4,If you can not judge the direction of the forces
correctly,we used to employ the analytical method,
32
5、解析法解题时，力的方向可以任意设，如果求出

4、对力的方向判定不准的，一般用解析法。
33
① is a scalar,)( FM
O
When F=0 or d=0,=0,)( FM
O
③ is the exclusive factor causing a
rotation
)( FM O
⑤ =2⊿ AOB=F?d,twice the area
of the ⊿ AOB,
)( FM O
A force acting on displace it--the displacement depends on the
a body tends to, magnitude and direction of the forces
rotate it-----the rotation depends on the magnitude
and direction of the moments
§ 2-3 The concepts and character
of a moment and a force couple
- +
dFFM O ???)(
Ⅰ Moment of force
② F↑,d↑rotation is apparent,
④ Unit,N?m or kgf?m,
34
① 是代数量。 )( FM
O

O
③ 是影响转动的独立因素。 )( FM O
⑤ =2⊿ AOB=F?d,2倍 ⊿ 形面积。 )( FM
O

§ 2-3 力矩、力偶的概念及其性质
- +
dFFM O ???)(

② F↑,d↑ 转动效应明显。
④ 单位 N?m，工程单位 kgf?m。
35
law,the moment of the resultant of a coplanar system of concurrent
forces about any center is equal to the algebraic sum of the
moments of the component forces about that center,
Ⅱ the law for the moment of a resultant force
[Proof] By the law of
projection of the resultant force,
)()()( 21 FmFmRm ooo ???
?
?
?
n
i
iOO FmRm
1
)()(
od=ob+oc
oboAo ABFM o ???? 2)( 1
ocoAo ACFM o ???? 2)( 2
again∵
36

[证 ]
?
?
?
n
i
iOO FmRm
1
)()(
od=ob+oc
oboAo ABFM o ???? 2)( 1
ocoAo ACFM o ???? 2)( 2

37
[Example] Determine and,if F,Q and l are
known,see diagram below,
Solution,① Use the moment of a
② By the law of the
moment of a resultant you get,
)( Fm O )(Qm o
? s i n)(
lFdFFm
O ????
lQQm o ???)(
?c t g)( ????? lFlFFm yxO
lQQm o ???)(
38
[例 ] 已知：如图 F,Q,l,求,和

②应用合力矩定理
)( Fm O )(Qm o
? s i n)(
lFdFFm
O ????
lQQm o ???)(
?c t g)( ????? lFlFFm yxO
lQQm o ???)(
39
① Composition of two parallel forces of the same sense
magnitude,R=Q+P
direction,parallel to Q,P,
and is of same direction
the point of application,point C
determine point C,by the law of the
moment of a resultant,
)()( QmRm BB ? QPRa g a i n ???
ABQCBR ???? CBACAB ?? Q
P
CB
AC ??
Ⅲ Coplanar force couple and their character
Force couple,A system of two parallel forces of the same magnitude
and opposite direction acting on a rigid body,
Character 1,The resultant of a force couple does not exist;
Moreover,a force couple can not be in equilibrium
too,It is a basic mechanical quantity,
Take into account
40
① 两个同向平行力的合力 大小,R=Q+P

)()( QmRm BB ? QPR ???又
ABQCBR ????

P
CB
AC ?整理得

41
② Composition of two parallel forces of opposite sense
magnitude,R=Q-P
direction,parallel to Q and P,and has
the same sense as the greater force,
the point of application,
point C P
Q
CA
CB ?
Character 2,The algebraic sum of the moments of the
forces of a couple about any point in its plane of action is
independent of the location of that point and is equal to the
moment of the couple,
Force couple has zero resultant R=F'-F=0
1' ?? FFCACB? CACB ??
????? CBdCBCBif,
????? d The point of application of the resultant is at infinity,
42
② 两个反向平行力的合力 大小,R=Q-P

P
Q
CA
CB ?

1' ??
F
F
CA
CB? CACB ??
???? CBdCBCB 必有成立若,

43
??? 0)( Rm O ????? 0)'()( FmFm OO
xFdxFFmFm OO ?????? ')()'()(?
)( RmdF O????
Introduction,① m is a scalar,but has +,-;
② F,d are not independent on each other,
only the moment of the couple
is independent;
③ the magnitude,m=± 2⊿ ABC;
④ unit,N? m,
dFm ???
Because point O is arbitrary
dFm ???? — +
d
??? 0)( Rm O
Proof that
is a finite quantity
44
??? 0)( Rm O ????? 0)'()( FmFm OO

xFdxFFmFm OO ?????? ')()'()(?
)( RmdF O????

② F,d 都不独立，只有力偶矩 是独立 量；
③ m的值 m=± 2⊿ ABC ；
④单位,N? m
dFm ???

dFm ???? — +
d
45
Character 3 (the equivalent law of coplanar force couples),
A couple acting on a rigid body can be replaced by any other
couple of the same moment lying in the same plane without
altering the external effect on that body,
[Proof]
Assume a force couple (F,F') acting
on a body is in a plane; along the arm
of the force couple AB add a couple
and F we get R; adding Q' and F',
we get R‘,then,we get a new force
couple (R,R'),Move R and R' to the
point A‘ and B‘,we get a new force couple (R,R'),this is equivalent
to the old force couple(F,F' ),
46

[证 ] 设物体的某一平面

Q',F'合成 R',

47
② It is possible to change the
magnitudes of the forces of a
couple or the perpendicular
distance between them arbitrarily
without changing its moment,
From the proof above we get two corollaries,
By the comparison between
(F,F') and (R,R') we get,
m(F,F')=2△ ABD=m(R,R')
=2 △ ABC
Hence,△ ABD= △ ABC,
Moreover,they cause rotations
in the same direction,
① A force couple can be
transferred anywhere in its
plane of action freely,
48
② 只要保持力偶矩大小和转向

m(F,F')=2△ ABD=m(R,R')
=2 △ ABC

① 力偶可以在其作用面内任

49;111 dFm ??
222 dFm ??
dPma g a i n 11 ?
dPm 22 ???
'21 PPR A ??
2'1 PPR B ??
§ 2-4 The composition and the
equilibrium of a coplanar system of force couples
Coplanar system of force couples,many force couples acting on a
rigid body lie in the same plane
Assume there are two force couples in a plane,
d d
21
'
21
'
21 )( mmdPdPdPPdRM A ????????
Therefore the total moment of the
resultant force couples is
50;111 dFm ??
222 dFm ??
dPm 11 ?又
dPm 22 ???
'21 PPR A ??
2'1 PPR B ??
21'21'21 )( mmdPdPdPPdRM A ????????? 合力矩
§ 2-4 平面力偶系的合成与平衡

d d
51
Necessary and sufficient conditions for the equilibrium of a
coplanar system of force couples,the algebraic sum of their
moments has to be zero,
?
?
?????
n
i
in mmmmM
1
21 ?
H e n c e,0
1
??
?
n
i
im
Conclusion,
A coplanar system of force couples is equivalent to a single
couple lying in the same plane,the moment of which equals the
algebraic sum of the moments of the component couples,
52

?
?
?????
n
i
in mmmmM
1
21 ?

1
??
?
n
i
im

53
[Example] A workpiece lies on a drill press,four equidimensional
holes are drilled in the workpiece,the force couple of every drill
is,
Determine the total moment of the resultant of the force couples
and the horizontal reaction forces at point A and B,
mN154321 ????? mmmm
mN60)15(4
4321
??????
???? mmmmM
02.0 4321 ?????? mmmmN B
N3 0 02.060 ??? BN N 3 0 0??? BA NN
Solution,The resultant force couple is
By the equilibrium equations
of a coplanar system of force
couples,
According to the character of two force
couples in equilibrium,NA and NB make
up a new force couple,
54
[例 ] 在一钻床上水平放置工件,在工件上同时钻四个等直径

mN154321 ????? mmmm
mN60)15(4
4321
??????
???? mmmmM
02.0 4321 ?????? mmmmN B
N3 0 02.060 ??? BN N 3 0 0??? BA NN

55
§ 2-5 The composition and the
equilibrium of a coplanar system of parallel forces
Coplanar system of parallel forces,a force system in which the
forces are coplanar and the
lines of action of these forces
are parallel,Ⅰ The composition of a coplanar system of parallel forces,
Assuming a coplanar system of parallel forces
acting on a rigid body being given determine the resultant of these
forces,
,,321 ?FFF
56
§ 2-5 平面平行力系的合成和平衡

321 ?FFF,、
57
)()()()()()()( 4325121 FmFmFmFmFmRmRm ooooooo ??????
)(?? io Fm
,511 FFR ?? )()()( 511 FmFmRm ooo ??
By the law of the composition of a coplanar system of parallel
forces the resultant of and is,
5 1 FF 1R
iFFFFFFRR ???????? )()( 4325121
Consequently,4322 FFFR ??? )()()()( 4322 FmFmFmRm oooo ???
In a similar way we get
58
)()()()()()()( 4325121 FmFmFmFmFmRmRm ooooooo ??????
)(?? io Fm
,511 FFR ?? )()()( 511 FmFmRm ooo ??

5 1 FF 1R
iFFFFFFRR ???????? )()( 4325121

,4322 FFFR ??? )()()()( 4322 FmFmFmRm oooo ???同理
59
⑴ When,the resultant force of the original force
system is
When the force system is parallel to the y axis,
21 RR ??
iFRRR ???? 21
ii YFR ?? ??The position of The lines of action of the resultant force,
From,
)()()()( 21 ioooo FmRmRmRm ????
?
?? ?????
i
iiii
R F
xF
R
xFx
⑵ When ( ),the result of the composition of
the original force system is the moment of the resultant force
couple,
21 RR ?? 0?? iF
iiio xFFmm ??? ?? )(
??????? iiR xFxFxFxFxR 552211 ?
Hence,
60
⑴, 当 时，原力系的合成结果是一个合力

21 RR ??
iFRRR ???? 21
ii YFR ?? ??

)()()()( 21 ioooo FmRmRmRm ????
??????? iiR xFxFxFxFxR 552211, ?即
?
?? ?????
i
iiii
R F
xF
R
xFx
⑵ 当 时 (即 时 )，原力系合成结果是一

21 RR ?? 0?? iF
iiio xFFmm ??? ?? )(
61
the necessary and sufficient conditions are,
The algebraic sum of the forces is equal to zero,In addition,the
algebraic sum of the moments of the forces about an arbitrary
point is equal to zero,
Namely,
From the analytical method for a coplanar system of parallel forces,
we can know that a coplanar system of parallel forces is equivalent to
two parallel forces and, By the law 1,the necessary and
sufficient conditions for the equilibrium of two forces and are,
they have to be collinear,equal in magnitude and of opposite direction,
( ) and must be
satisfied together,Consequently
1R 2R
2R1R
21 RR ?? 0?? iF )()(0)( 21 RmRmFm ooio ????
Ⅱ The necessary and sufficient conditions for the equilibrium
of a coplanar system of parallel forces,
? ?
? ? ??
0)(
0
io
ii
Fm
YF
These are the equations of equilibrium
of a coplanar system of parallel force,
62

1R 2R 2R1R
21 RR ?? 0?? iF
)()(0)( 21 RmRmFm ooio ????

? ?
? ? ??
0)(
0
io
ii
Fm
YF

63
The equilibrium equations of a coplanar system of parallel
forces can also be expressed by two moment equations,
hence,
? ? 0)( iA Fm
? ? 0)( iB Fm
The line between the point A
and B cannot be parallel to the
lines of action of the forces
[Example] The weight of a crane is
dimensions are given in the fig,
Determine:① the range of the
magnitude of Q to insure the crane
never turns,② let Q=180kN,W is the
forces the railway apply on the wheels
of the crane at the point A and B,
64

? ? 0)( iA Fm
? ? 0)( iB Fm

[例 ] 已知：塔式起重机 P=700kN,
W=200kN (最大起重量 )，尺寸如

Q=180kN时，求满载时轨道 A,B

65
? ? 0)( Fm B
0)22()212(2)26( ???????? ANWPQ
0?AN
kN 75?Q
Limit condition,
Result,
Solution,⑴ ① When the crane is full
load,the minimum of Q is determined
by the condition that the crane does
not turn right,
② When W=0,? ? 0)( Fm
A 0)22(2)26( ?????? BNPQ
Limit condition,0?BN Result,kN 350?Q
Consequently,the range of magnitude of Q to insure the crane
does not turn is,
kN 3 5 0kN 75 ?? Q
66
? ? 0)( Fm B
0)22()212(2)26( ???????? ANWPQ
0?AN
kN 75?Q

② 空载时,W=0 由 ? ? 0)( Fm
A 0)22(2)26( ?????? BNPQ

kN 3 5 0kN 75 ?? Q
67
04)212(2)26( ???????? BNWPQ? ? 0)( Fm A
,0? ?iF 0?????? BA NNWPQ
kN 8 7 0
,kN 2 1 0
?
?
B
A
N
N
Solution,According to the equilibrium equations of a coplanar
system of parallel forces,
Result,
68
04)212(2)26( ???????? BNWPQ? ? 0)( Fm A
,0? ?iF 0?????? BA NNWPQ
kN 8 7 0
,kN 2 1 0
?
?
B
A
N
N
⑵ 求当 Q=180kN，满载 W=200kN时,NA,NB为多少

69
70