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,
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平面汇交力系,
各力的作用线都在同一平面内且
汇交于一点的力系。
引 言
① 平面汇交力系
平面力系 ②平面平行力系 (平面力偶系是其中的特殊情况 )
③ 平面一般力系 (平面任意力系 )
研究方法：几何法，解析法。
例：起重机的挂钩。
力系分为：平面力系、空间力系
平面特殊力系：指的是平面汇交力系、平面力偶系和平面平
行力系。
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§ 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
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§ 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
第二章 平面特殊力系
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§ 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
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§ 2-1 平面汇交力系合成与平衡的几何法
一、合成的几何法
?c o s2 212221 FFFFR ???
)180s i n (s i n
1
?? ??
RF
2,任意个共点力的合成
为力多边形
1.两个共点力的合成
合力方向由正弦定理,
由余弦定理,
?? c o s)1 8 0c o s ( ????由力的平行四边形法则作，
也可用力的三角形来作。
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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,
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结论,即,
即：平面汇交力系的合力等于各分力的矢量和，合力的作用
线通过各力的汇交点。
二、平面汇交力系平衡的几何条件
?? FR
4321 FFFFR ????
在上面几何法求力系的合力中，合力为
零意味着力多边形自行封闭。所以平面
汇交力系平衡的必要与充分的几何条件
是,
平面汇交力系平衡的充要条件是,
? ?? 0FR
力多边形自行封闭
或
力系中各力的矢量和等于零
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[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 ?,
,
,
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[例 ] 已知压路机碾子重 P=20kN,r=60cm,欲拉过 h=8cm的障碍
物。求：在中心作用的水平力 F的大小和碾子对障碍物的压力。
577.0)(tg
22
?? ??? hr hrr?
又由几何关系,
① 选碾子为研究对象
② 取分离体画受力图
解,
∵ 当碾子刚离地面时 NA=0,拉力 F最大,这时
拉力 F和自重及支反力 NB构成一平衡力系。
由平衡的几何条件，力多边形封闭，故
?tg?? PF ?co sPN B ?
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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
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由作用力和反作用力的关系,碾子对障碍物的压力等于 23.1kN。
此题也可用力多边形方法用比例尺去量。
F=11.5kN,NB=23.1kN 所以
几何法解题步骤：①选研究对象；②作出受力图；
③作力多边形，选择适当的比例尺；
④求出未知数
几何法解题不足,①精度不够，误差大 ②作图要求精度高；
③不能表达各个量之间的函数关系。
下面我们研究平面汇交力系合成与平衡的另一种方法,
解析法 。
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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?
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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
一、力在坐标轴上的投影
X=Fx=F·cos?,
Y=Fy=F·sin?=F ·cos?
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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,
即,
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二、合力投影定理 由图可看出，各分力在 x轴和在 y
轴投影的和分别为,
????? XXXXR x 421
??????? YYYYYR y 4321
?? YR y?? XR x
合力投影定理：合力在任一轴上的投影，等于各分力在同一
轴上投影的代数和。
即,
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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
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合力的大小,
方向,
作用点,
x
y
R
R
??tg ?
?
?? ??
X
Y
R
R
x
y 11 tgtg?
∴
为该力系的汇交点
三、平面汇交力系合成与平衡的解析法
从前述可知：平面汇交力系平衡的必要与充分条件是该力系
的合力为零。 即,
?
?
??
??
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
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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；
?
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解,①研究 AB杆
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
由 EB=BC=0.4m,
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；
?
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[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 ?
①
②
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[例 ] 已知如图 P,Q,求平衡时 =？ 地面的反力 ND=？ ?
解,研究球受力如图,
选投影轴列方程为
PP-TN D 3Q60s i n2Qs i n-Q 02 ????? ?
由②得
060???
212c o s 21 ??? PPT
T?
由①得
0?? X
0?? Y
0co s 12 ??? TT ?
0Qs i n2 ???? DNT ?
①
②
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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,
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又,
?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
解,研究块,受力如图,
解力三角形,
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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,
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再研究球，受力如图,
作力三角形
解力三角形,
?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(
时球方能离开地面当 hR hRhPF ? ??? )2(
NB=0时为球
离开地面
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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,
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1、一般地，对于只受三个力作用的物体，且角度
特殊时用 几 何法（解力三角形）比较简便。
解题技巧及说明,
3、投影轴常选择与未知力垂直，最好使每个方程中
只有一个未知数。
2、一般对于受多个力作用的物体，且角度不特殊或
特殊，都用解析法。
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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,
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5、解析法解题时，力的方向可以任意设，如果求出
负值，说明力方向与假设相反。对于二力构件,
一般先设为拉力，如果求出负值，说明物体受压
力。
4、对力的方向判定不准的，一般用解析法。
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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
about a point,Introduction,
② F↑,d↑rotation is apparent,
④ Unit,N?m or kgf?m,
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① 是代数量。 )( FM
O
当 F=0或 d=0时,=0。 )( FM
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
力对物体可以产生 移动效应 --取决于力的大小、方向
转动效应 --取决于力矩的大小、方向
§ 2-3 力矩、力偶的概念及其性质
- +
dFFM O ???)(
一、力对点的矩
说明,
② F↑,d↑ 转动效应明显。
④ 单位 N?m，工程单位 kgf?m。
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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
odoAo ADRM o ??? ?2)(
again∵
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定理,平面汇交力系的合力对平面内任一点的矩，等于所
有各分力对同一点的矩的代数和
即,
二、合力矩定理
由合力投影定理有：
证毕现 )()()( 21 FmFmRm ooo ??
[证 ]
?
?
?
n
i
iOO FmRm
1
)()(
od=ob+oc
oboAo ABFM o ???? 2)( 1
ocoAo ACFM o ???? 2)( 2
odoAo ADRM o ??? ?2)(
又 ∵
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[Example] Determine and,if F,Q and l are
known,see diagram below,
Solution,① Use the moment of a
force about a point,
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 ???)(
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[例 ] 已知：如图 F,Q,l,求,和
解,①用力对点的矩法
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 ???)(
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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
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① 两个同向平行力的合力 大小,R=Q+P
方向：平行于 Q,P且指向一致
作用点,C处
确定 C点，由合力距定理
)()( QmRm BB ? QPR ???又
ABQCBR ????
代入CBACAB ?? Q
P
CB
AC ?整理得
三、平面力偶及其性质
力偶,两力大小相等，作用线不重合的反向平行力叫力偶。
性质 1：力偶既没有合力，本身又不平衡，是一个基本力学量。
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
方向：平行于 Q,P且与较大的相同
作用点,C处 （推导同上）
P
Q
CA
CB ?
性质 2：力偶对其所在平面内任一点的矩恒等于力偶矩，而
与矩心的位置无关，因此力偶对刚体的效应用力偶矩度量。
力偶 无合力 R=F'-F=0
1' ??
F
F
CA
CB? CACB ??
???? CBdCBCB 必有成立若,
处合力的作用点在无限远????? d
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
为有限量证明 ??? 0)( Rm O
xFdxFFmFm OO ?????? ')()'()(?
)( RmdF O????
说明：① m是代数量，有 +,-；
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 ???
由于 O点是任取的
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
of balanced forces(Q,Q'),adding Q
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
性质 3：平面力偶等效定理
作用在同一平面内的两个力偶，只要它的力偶矩的大小相等，
转向相同，则该两个力偶彼此等效。
[证 ] 设物体的某一平面
上作用一力偶 (F,F')
现沿力偶臂 AB方向
加一对平衡力 (Q,Q'),
Q',F'合成 R',
再将 Q,F合成 R,
得到新力偶 (R,R'),
将 R,R'移到 A',B'点，则 (R,R')，取 代了原力偶 (F,F' )
并与原力偶等效。
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
不变，可以任意改变力偶中力
的大小和相应力偶臂的长短，
而不改变它对刚体的作用效应。
由上述证明可得下列 两个推论,
比较 (F,F')和 (R,R')可得
m(F,F')=2△ ABD=m(R,R')
=2 △ ABC
即△ ABD= △ ABC,
且它们转向相同。
① 力偶可以在其作用面内任
意移动，而不影响它对刚体
的作用效应。
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
平面力偶系,作用在物体同一平面的许多力偶叫平面力偶系
设有两个力偶
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
平面力偶系平衡的充要条件是,所有各力偶矩的代数和
等于零。
?
?
?????
n
i
in mmmmM
1
21 ?
即 0
1
??
?
n
i
im
结论,
平面力偶系合成结果还是一个力偶,其力偶矩为各力偶矩
的代数和 。
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
的孔,每个钻头的力偶矩为
求工件的总切削力偶矩和 A, B端水平反力?
mN154321 ????? mmmm
mN60)15(4
4321
??????
???? mmmmM
02.0 4321 ?????? mmmmN B
N3 0 02.060 ??? BN N 3 0 0??? BA NN
解, 各力偶的合力偶距为
根据平面力偶系平衡方程有,
由力偶只能与力偶平衡的性质，
力 NA与力 NB组成一力偶。
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
平面平行力系,各力的作用线在同一平面内且相互平行的力系叫 ～
一、平面平行力系的合成
设在刚体上作用一平面平行力系,现求其
合成结果。
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 ??
根据两个平行力合成理论可知，力 与 合成一个合力
5 1 FF 1R
iFFFFFFRR ???????? )()( 4325121
所以
,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
当力系平行于 y轴时,
21 RR ??
iFRRR ???? 21
ii YFR ?? ??
合力作用线的位置, 由
)()()()( 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
合力偶
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
由平面平行力系合成分析过程可知，平面平行力系总可以
与两个平行力 和 等效，由公理 1，二力 和 平衡的充要
条件是：等值、反向、共线，即 ( ) 和
平面平行力系平衡的充要条件为,
力系中各力的代数和等于零，同时，各力对平面内任一点
的矩的代数和也等于零 。即,
1R 2R 2R1R
21 RR ?? 0?? iF
)()(0)( 21 RmRmFm ooio ????
二、平面平行力系的平衡条件
同时满足。因此,
? ?
? ? ??
0)(
0
io
ii
Fm
YF
平面平行力系的平衡方程
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
P=700kN,the full load is W=200kN,
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
full load,Determine the reaction
forces the railway apply on the wheels
of the crane at the point A and B,
64
平面平行力系的平衡方程也可用两矩式表示，即
? ? 0)( iA Fm
? ? 0)( iB Fm
其中,A,B两点的连线
必须不与各力线平行
[例 ] 已知：塔式起重机 P=700kN,
W=200kN (最大起重量 )，尺寸如
图。求：①保证满载和空载时不
致翻倒，平衡块 Q=？ ②当
Q=180kN时，求满载时轨道 A,B
给起重机轮子的反力？
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
限制条件,
解得
解, ⑴ ①首先考虑满载时，起
重机不向右翻倒的最小 Q为,
② 空载时,W=0 由 ? ? 0)( Fm
A 0)22(2)26( ?????? BNPQ
限制条件 为,0?BN 解得 kN 350?Q
因此保证空、满载均不倒 Q应满足如下关系,
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
⑵ Q=180kN,full load W=200kN,determine NA,NB,
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
由平面平行力系的平衡方程可得,
解得,
69
70
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
平面汇交力系,
各力的作用线都在同一平面内且
汇交于一点的力系。
引 言
① 平面汇交力系
平面力系 ②平面平行力系 (平面力偶系是其中的特殊情况 )
③ 平面一般力系 (平面任意力系 )
研究方法：几何法，解析法。
例：起重机的挂钩。
力系分为：平面力系、空间力系
平面特殊力系：指的是平面汇交力系、平面力偶系和平面平
行力系。
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
第二章 平面特殊力系
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
一、合成的几何法
?c o s2 212221 FFFFR ???
)180s i n (s i n
1
?? ??
RF
2,任意个共点力的合成
为力多边形
1.两个共点力的合成
合力方向由正弦定理,
由余弦定理,
?? c o s)1 8 0c o s ( ????由力的平行四边形法则作，
也可用力的三角形来作。
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
结论,即,
即：平面汇交力系的合力等于各分力的矢量和，合力的作用
线通过各力的汇交点。
二、平面汇交力系平衡的几何条件
?? FR
4321 FFFFR ????
在上面几何法求力系的合力中，合力为
零意味着力多边形自行封闭。所以平面
汇交力系平衡的必要与充分的几何条件
是,
平面汇交力系平衡的充要条件是,
? ?? 0FR
力多边形自行封闭
或
力系中各力的矢量和等于零
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
物。求：在中心作用的水平力 F的大小和碾子对障碍物的压力。
577.0)(tg
22
?? ??? hr hrr?
又由几何关系,
① 选碾子为研究对象
② 取分离体画受力图
解,
∵ 当碾子刚离地面时 NA=0,拉力 F最大,这时
拉力 F和自重及支反力 NB构成一平衡力系。
由平衡的几何条件，力多边形封闭，故
?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
由作用力和反作用力的关系,碾子对障碍物的压力等于 23.1kN。
此题也可用力多边形方法用比例尺去量。
F=11.5kN,NB=23.1kN 所以
几何法解题步骤：①选研究对象；②作出受力图；
③作力多边形，选择适当的比例尺；
④求出未知数
几何法解题不足,①精度不够，误差大 ②作图要求精度高；
③不能表达各个量之间的函数关系。
下面我们研究平面汇交力系合成与平衡的另一种方法,
解析法 。
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
一、力在坐标轴上的投影
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,
即,
18
二、合力投影定理 由图可看出，各分力在 x轴和在 y
轴投影的和分别为,
????? XXXXR x 421
??????? YYYYYR y 4321
?? YR y?? XR x
合力投影定理：合力在任一轴上的投影，等于各分力在同一
轴上投影的代数和。
即,
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
合力的大小,
方向,
作用点,
x
y
R
R
??tg ?
?
?? ??
X
Y
R
R
x
y 11 tgtg?
∴
为该力系的汇交点
三、平面汇交力系合成与平衡的解析法
从前述可知：平面汇交力系平衡的必要与充分条件是该力系
的合力为零。 即,
?
?
??
??
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
解,①研究 AB杆
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
由 EB=BC=0.4m,
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 ?
①
②
24
[例 ] 已知如图 P,Q,求平衡时 =？ 地面的反力 ND=？ ?
解,研究球受力如图,
选投影轴列方程为
PP-TN D 3Q60s i n2Qs i n-Q 02 ????? ?
由②得
060???
212c o s 21 ??? PPT
T?
由①得
0?? X
0?? Y
0co s 12 ??? TT ?
0Qs i n2 ???? DNT ?
①
②
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
又,
?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
解,研究块,受力如图,
解力三角形,
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
再研究球，受力如图,
作力三角形
解力三角形,
?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(
时球方能离开地面当 hR hRhPF ? ??? )2(
NB=0时为球
离开地面
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
特殊时用 几 何法（解力三角形）比较简便。
解题技巧及说明,
3、投影轴常选择与未知力垂直，最好使每个方程中
只有一个未知数。
2、一般对于受多个力作用的物体，且角度不特殊或
特殊，都用解析法。
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
负值，说明力方向与假设相反。对于二力构件,
一般先设为拉力，如果求出负值，说明物体受压
力。
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
about a point,Introduction,
② F↑,d↑rotation is apparent,
④ Unit,N?m or kgf?m,
34
① 是代数量。 )( FM
O
当 F=0或 d=0时,=0。 )( FM
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
力对物体可以产生 移动效应 --取决于力的大小、方向
转动效应 --取决于力矩的大小、方向
§ 2-3 力矩、力偶的概念及其性质
- +
dFFM O ???)(
一、力对点的矩
说明,
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
odoAo ADRM o ??? ?2)(
again∵
36
定理,平面汇交力系的合力对平面内任一点的矩，等于所
有各分力对同一点的矩的代数和
即,
二、合力矩定理
由合力投影定理有：
证毕现 )()()( 21 FmFmRm ooo ??
[证 ]
?
?
?
n
i
iOO FmRm
1
)()(
od=ob+oc
oboAo ABFM o ???? 2)( 1
ocoAo ACFM o ???? 2)( 2
odoAo ADRM o ??? ?2)(
又 ∵
37
[Example] Determine and,if F,Q and l are
known,see diagram below,
Solution,① Use the moment of a
force about a point,
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
解,①用力对点的矩法
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
方向：平行于 Q,P且指向一致
作用点,C处
确定 C点，由合力距定理
)()( QmRm BB ? QPR ???又
ABQCBR ????
代入CBACAB ?? Q
P
CB
AC ?整理得
三、平面力偶及其性质
力偶,两力大小相等，作用线不重合的反向平行力叫力偶。
性质 1：力偶既没有合力，本身又不平衡，是一个基本力学量。
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
方向：平行于 Q,P且与较大的相同
作用点,C处 （推导同上）
P
Q
CA
CB ?
性质 2：力偶对其所在平面内任一点的矩恒等于力偶矩，而
与矩心的位置无关，因此力偶对刚体的效应用力偶矩度量。
力偶 无合力 R=F'-F=0
1' ??
F
F
CA
CB? CACB ??
???? CBdCBCB 必有成立若,
处合力的作用点在无限远????? d
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
为有限量证明 ??? 0)( Rm O
xFdxFFmFm OO ?????? ')()'()(?
)( RmdF O????
说明：① m是代数量，有 +,-；
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 ???
由于 O点是任取的
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
of balanced forces(Q,Q'),adding Q
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
性质 3：平面力偶等效定理
作用在同一平面内的两个力偶，只要它的力偶矩的大小相等，
转向相同，则该两个力偶彼此等效。
[证 ] 设物体的某一平面
上作用一力偶 (F,F')
现沿力偶臂 AB方向
加一对平衡力 (Q,Q'),
Q',F'合成 R',
再将 Q,F合成 R,
得到新力偶 (R,R'),
将 R,R'移到 A',B'点，则 (R,R')，取 代了原力偶 (F,F' )
并与原力偶等效。
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
不变，可以任意改变力偶中力
的大小和相应力偶臂的长短，
而不改变它对刚体的作用效应。
由上述证明可得下列 两个推论,
比较 (F,F')和 (R,R')可得
m(F,F')=2△ ABD=m(R,R')
=2 △ ABC
即△ ABD= △ ABC,
且它们转向相同。
① 力偶可以在其作用面内任
意移动，而不影响它对刚体
的作用效应。
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
平面力偶系,作用在物体同一平面的许多力偶叫平面力偶系
设有两个力偶
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
平面力偶系平衡的充要条件是,所有各力偶矩的代数和
等于零。
?
?
?????
n
i
in mmmmM
1
21 ?
即 0
1
??
?
n
i
im
结论,
平面力偶系合成结果还是一个力偶,其力偶矩为各力偶矩
的代数和 。
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
的孔,每个钻头的力偶矩为
求工件的总切削力偶矩和 A, B端水平反力?
mN154321 ????? mmmm
mN60)15(4
4321
??????
???? mmmmM
02.0 4321 ?????? mmmmN B
N3 0 02.060 ??? BN N 3 0 0??? BA NN
解, 各力偶的合力偶距为
根据平面力偶系平衡方程有,
由力偶只能与力偶平衡的性质，
力 NA与力 NB组成一力偶。
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
平面平行力系,各力的作用线在同一平面内且相互平行的力系叫 ～
一、平面平行力系的合成
设在刚体上作用一平面平行力系,现求其
合成结果。
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 ??
根据两个平行力合成理论可知，力 与 合成一个合力
5 1 FF 1R
iFFFFFFRR ???????? )()( 4325121
所以
,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
当力系平行于 y轴时,
21 RR ??
iFRRR ???? 21
ii YFR ?? ??
合力作用线的位置, 由
)()()()( 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
合力偶
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
由平面平行力系合成分析过程可知，平面平行力系总可以
与两个平行力 和 等效，由公理 1，二力 和 平衡的充要
条件是：等值、反向、共线，即 ( ) 和
平面平行力系平衡的充要条件为,
力系中各力的代数和等于零，同时，各力对平面内任一点
的矩的代数和也等于零 。即,
1R 2R 2R1R
21 RR ?? 0?? iF
)()(0)( 21 RmRmFm ooio ????
二、平面平行力系的平衡条件
同时满足。因此,
? ?
? ? ??
0)(
0
io
ii
Fm
YF
平面平行力系的平衡方程
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
P=700kN,the full load is W=200kN,
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
full load,Determine the reaction
forces the railway apply on the wheels
of the crane at the point A and B,
64
平面平行力系的平衡方程也可用两矩式表示，即
? ? 0)( iA Fm
? ? 0)( iB Fm
其中,A,B两点的连线
必须不与各力线平行
[例 ] 已知：塔式起重机 P=700kN,
W=200kN (最大起重量 )，尺寸如
图。求：①保证满载和空载时不
致翻倒，平衡块 Q=？ ②当
Q=180kN时，求满载时轨道 A,B
给起重机轮子的反力？
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
限制条件,
解得
解, ⑴ ①首先考虑满载时，起
重机不向右翻倒的最小 Q为,
② 空载时,W=0 由 ? ? 0)( Fm
A 0)22(2)26( ?????? BNPQ
限制条件 为,0?BN 解得 kN 350?Q
因此保证空、满载均不倒 Q应满足如下关系,
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
⑵ Q=180kN,full load W=200kN,determine NA,NB,
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
由平面平行力系的平衡方程可得,
解得,
69
70
