1
Theoretical Mechanics
2
3
Practically,there exists many force systems in which the action
lines of the forces are not coplanar,these are force systems in space,
They are the most general force systems,
Diagram (a) is a concurrent forces system in space,
Diagram (b) is an arbitrary force system in space,if neglecting the
forces of the wind from the side it is a parallel force system in space,
The force of
wind from
the front
The force of
wind from
the side b
4
工程中常常存在着很多各力的作用线不在同一平面内的力
系，即空间力系，空间力系是最一般的力系。
(a)图为空间汇交力系； (b)图为空间任意力系；
(b)图中去了风力为空间平行力系。
迎 面
风 力
侧 面
风 力
b
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Chapter 5,Force system in space
§ 5–1 Concurrent force system in space
§ 5–2 System of force couples in space
§ 5–3 The moment of a force about a point or an axis
§ 5–4 Reduction of a force system in space to a given
point
§ 5–5 Discussion of the result of the reduction of a
force system in space
§ 5–6 Equilibrium equations of a force system in
space and their applications
§ 5–7 The center of a system of parallel forces and the
center of gravity of an object
Class of exercises
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第五章 空间力系
§ 5–1 空间汇交力系
§ 5–2 空间力偶系
§ 5–3 力对点的矩与力对轴的矩
§ 5–4 空间一般力系向一点的简化
§ 5–5 空间一般力系简化结果的讨论
§ 5–6 空间一般力系的平衡方程及应用
§ 5–7 平行力系的中心与物体的重心
习题课
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§ 5-1 Concurrent force system in space
1,The projection and decomposition of a force on the axes in space,
(1)The representation of force in space,
a force is given by its
magnitude,its direction and its
point (line) of application
magnitude,
point of application,the point at
which the force acts onto the
object,
direction,it can be determined by
the three angles ?,?,g or by the angle of inclination ? and the angle
of depression ?,
?
g
?
Fxy
O
FF ?
8
一、力在空间轴上的投影与分解,
1.力在空间的表示,
力的三要素,
大小、方向、作用点 (线 )
大小,
作用点,在物体的哪点就是哪点
方向,
由 ?,?,g三个方向角确定
由仰角 ? 与俯角 ? 来确定。
?
g
?
Fxy
O
FF ?
§ 5-1 空间汇交力系
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(2) The method of direct projection
From the fig we see
g
?
?
c o s
,c o s
,c o s
??
??
??
FZ
FY
FX
????g c o sc o sc o sc o ss i n ???????? FFFX xy
????g s i nc o ss i ns i ns i n ???????? FFFY xy
?g s i nco s ???? FFZ
(3) The method of indirect projection
When the angles ?,? and g are not
easy to determine we can project F
onto the plane xy firstly and then
project the result Fxy onto the axes x
and y,
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2、一次投影法（直接投影法）
由图可知,
g
?
?
c o s
,c o s
,c o s
??
??
??
FZ
FY
FX
????g c o sc o sc o sc o ss i n ???????? FFFX xy
????g s i nc o ss i ns i ns i n ???????? FFFY xy
?g s i nco s ???? FFZ
3、二次投影法（间接投影法）
当力与各轴正向夹角不易
确定时，可先将 F 投影到 xy
面上，然后再投影到 x,y轴上,
即
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(4) Decomposition of a force along the coordinate axes,
If the three perpendicular components
of a force along the coordinate axes
are
then,zyx
FFF a n d,
zyx FFFF ???
222 ZYXF ????
F
Z
F
Y
F
X ??? g?? c o s,c o s,c o s
kZFjYFiXF zyx ???,,
due to
kZjYiXF ???
We get
Fx
Fy
Fz
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4、力沿坐标轴分解,
若以 表示力沿直角
坐标轴的正交分量，则,
zyx FFF,,
zyx FFFF ???
222 ZYXF ????
F
Z
F
Y
F
X ??? g?? c o s,c o s,c o s
kZFjYFiXF zyx ???,,
而,
kZjYiXF ???
所以,
Fx
Fy
Fz
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(1) The graphical method,
It is same as in the case of the composition of a coplanar system of
concurrent forces,It can be solved by the force polygon,
The resultant force is equal to the geometrical sum of the
components,
in FFFFFR ??????? ?321
(2)The analytical method,
Because,substitute it into the equation above
We get
is the projection of the resultant force onto the
axis x,hence
kZjYiXF iiii ???
kZjYiXR iii ??? ???
? iX
?? ix XR
?? iy YR
?? iz ZR
2,The composition of a concurrent force system in space,
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1、几何法,与平面汇交力系的合成方法相同，也可用力多
边形方法求合力。
即：合力等于各分力的矢量和
in FFFFFR ??????? ?321
2、解析法,
由于 代入上式
合力
由 为合力在 x轴的投影,∴
kZjYiXF iiii ???
kZjYiXR iii ??? ???
? iX
?? ix XR
?? iy YR
?? iz ZR
二、空间汇交力系的合成,
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(3) The law of projection of the resultant force,
The projection of the resultant force of a force system in space onto
an axis is equal to the algebraic sum of the projections of the
component forces onto the same axis,
The resultant force is given by
??? ?????? 222222 )()()( ZYXRRRR zyx
R
R
R
R
R
R zyx ???? g?? c o s,c o s,c o s
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3、合力投影定理,
空间力系的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。
??? ?????? 222222 )()()(,ZYXRRRR zyx合力
R
R
R
R
R
R zyx ???? g?? c o s,c o s,c o s
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3,The equilibrium of a concurrent force system in space:
? ?0X
? ?0Y
? ?0Z
They are called equilibrium equations
of a concurrent force system in space,
Hence the necessary and sufficient condition of equilibrium in
the analytical method is given by the equations
Hence the necessary and sufficient condition of equilibrium in
the graphical method is that the forces form a closed polygon,
0?? ? iFR
The necessary and sufficient condition of equilibrium is that
the resultant force is zero,
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三、空间汇交力系的平衡,
? ?0X
? ?0Y
? ?0Z
称为平衡方程
空间汇交力系的平衡方程
∴ 解析法 平衡充要条件为,
∴ 几何法 平衡充要条件为该力系的 力多边形封闭 。
0?? ? iFR
空间汇交力系平衡的充要条件是,力系的合力为零,即,
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§ 5–2 System of force couples in space
Beside the magnitude and the turning direction,the action plane of a
force couple in space has to be determined,therefore a force couple
is represented by a vector,
1,A force couple can be represented by a vector
The turning direction of the force
couple is determined by the right-
hand rule,Looking from the bottom
of the vector,the clockwise rotation is
positive,
Force couple in space is a free vector,
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§ 5-2 空间力偶系
由于空间力偶除大小、转向外，还必须确定力偶的作用面，
所以空间力偶矩必须用矢量表示。
一、力偶矩用矢量表示,
力偶的转向为右手螺旋定则。
从力偶矢末端看去，顺时针转动
为正。
空间力偶是一个自由矢量。
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Proof,① Draw II//Ⅰ,cd // ab
② Draw a couple of equilibrium
forces R and R' at point E,
(-R=R' should be satisfied)
③ Composing F1 and R we get
F2 at the point d composing
F1'and R' we get F2' at point c,R-F1=F2,R'- F1'= F2 ',
④ The force couple (F1,F1') in plane I is replaced by (F2,F2') in
plane II,
2,The law of equivalent law of a force couple in space
A force couple acting on a rigid body can be replaced by any other
force couple of the same moment and turning direction lying in a
parallel plane without altering the external effect on that body,
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[证 ] ① 作 II//Ⅰ, cd // ab
② 作一对平衡力 R,R' (在 E点,且
使 -R=R')
③ 由反向平行力合成得,
F1与 R合成得 F2，作用在 d点
F1'与 R'合成得 F2'，作用在 c点
且 R-F1=F2, R'- F1'= F2'
④ 在 I内的力偶（ F1,F1'）等效变成 II内的（ F2,F2' ）
二、空间力偶的等效定理
作用在同一刚体的两平行平面的两个力偶，若它们的转向相
同，力偶矩的大小相等，则两个力偶等效。
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Because the force couple in space is a free vector it has three elements
① the magnitude of the moment of the force couple
② the direction of the the moment of the force couple is the
normal of the action plane of the force couple,
③ the turning direction,given by the right-hand rule,
m
3,Composition and equilibrium of a system of force couple in space
Because the moment of a force couple in space is a free vector,it
can be translated to any point without changing its direction,
Especially,it can be translated to the concurrent point,Because the
moment are vectors they can be composed according to usual
vector rules,The resultant moment of force couple is the sum of the
moments of the component force couples,
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由此可得出,空间力偶矩是自由矢量,它有三个要素,
①力偶矩的大小 =
② 力偶矩的方向 ——与力偶作用面法线方向相同
③转向 ——遵循右手螺旋规则。
m
三、空间力偶系的合成与平衡
由于空间力偶系是自由矢量，只要方向不变，可移至任意
一点，故可使其滑至汇交于某点，由于是矢量，它的合成符合
矢量运算法则。 合力偶矩 = 分力偶矩的矢量和
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?
?
??????
n
i
in mmmmmm
1
321 ?
The projections are
0?? xm
0?? ym
0?? zm
m
m
m
m
m
mmmmm zyx
zyx ?????? g?? c o s,c o s,c o s;
222
0?? ? imm
Obviously the equilibrium condition of the force couple system in
space is
26
?
?
??????
n
i
in mmmmmm
1
321 ?
投影式 为,
0?? xm
0?? ym
0?? zm
m
m
m
m
m
mmmmm zyx
zyx ?????? g?? c o s,c o s,c o s;
222
0?? ? imm
显然空间力偶系的平衡条件是,
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In the plane it is an algebraic quantity,
In space,it is a vector,
[Example] the spherical hinges on the
reflective mirror of a car
§ 5–3 The moment of a force about a point or an axis
1,The moment of a force about a point
)a r e a (2)( A O BdFFm O ?????
If r is the radius vector at point A then
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在平面中：力对点的矩是代数量。
在空间中：力对点的矩是矢量。
[例 ] 汽车反镜的球铰链
§ 5-3 力对点的矩与力对轴的矩
一、力对点的矩的矢量表示
面积A O BdFFm O ???? 2)(
如果 r 表示 A点的矢径，则,
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The moment of the force about the
point o is equal to the vector
product of the radius vector of the
force at the point and the force,
dFFrFrFmFrFm OO ??????? ),s i n ()(,)(
21 ???
a n d B e c a u s e kZjYiXF ??? kzjyixr ???
ZYX
zyx
kji
FrFm O ???? )(
kFmjFmiFm
kyXxYjxZzXizYyZ
zOyOxO )]([)]([)]([
)()()(
???
??????
The angle between
the two vectors is
O
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即,力对点的矩等于矩心到该力
作用点的矢径与该力的矢量积。
dFFrFrFmFrFm OO ??????? ),s i n ()(,)(
21 ???
kZjYiXF ???由于 kzjyixr ???
ZYX
zyx
kji
FrFm O ???? )(
kFmjFmiFm
kyXxYjxZzXizYyZ
zOyOxO )]([)]([)]([
)()()(
???
??????
两矢量夹角为
O
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Definition,
It is an algebraic quantity,Direction is + –
)''a r e a (2)()( BOAdFFmFm xyxyOz ??????
2,The moment of a force about an axis
Conclusion,the moment of a
force about an axis which is
parallel to the force is zero,
In other words,if the force F
and the axis are coplanar the
moment is zero,
)()()()( xyOxyzzzz FmFmFmFm ???
[Proof]
32
定义,
它是代数量，方向规定 + –
的面积''2)()( BOAdFFmFm xyxyOz ??????
二、力对轴的矩
结论, 力对 //它的轴的
矩为零。即力 F与轴共
面时，力对轴之矩为零。
)()()()( xyOxyzzzz FmFmFmFm ???
[证 ]
33
The moment of a force about an axis which is parallel to the force is
zero,In other words,if the force F and the axis are coplanar the
moment is zero,
34
力对 //它的轴的矩为零。即力 F与轴共面时，力对轴之矩为零。
35
)(c o s)(:H e n c e FmFm zO ?? g )()]([ FmFm zzO ?
3,The relationship between the moment of the force about a point
and the moment of the force about an axis through the point
)a r e a (2)( A O BFm O ??
Proof,
''2)()( BOAFmFm xyzz ???
Draw an arbitrary axis Z through the
point O then
''co s BOAO A B ?g? ??
From the geometrical relationship,
''2c o s2 BOAO A B ?g? ??Hence,
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即,
)(c o s)( FmFm zO ??? g )()]([ FmFm zzO ?
三、力对点的矩与力对通过该点的轴之矩的关系
面积由于 A O BFm O ?2)( ?
[证 ]
''2)()( BOAFmFm xyzz ???
通过 O点作任一轴 Z，则,
''co s BOAO A B ?g? ??
由几何关系,
''2c o s2 BOAO A B ?g? ??所以,
37
Theorem,the projection of the vector of the moment of a force about a
point on an axis through this point is equal to the moment of the force
about the axis,This is the required relationship between the moment of
a force about a point and the moment of the force about an axis
through this point,
)(
)(
c o s,
)(
)(
c o s,
)(
)(
c o s
Fm
Fm
Fm
Fm
Fm
Fm
O
z
O
y
O
x ??? g??
222 ))(())(())(()( FmFmFmFm zyxO ???
kFmjFmiFm zyx )()()( ???
kFmjFmiFmFrFm zOyOxOO )]([)]([)]([)( ?????
Due to
The moment of the force about point O is given by
38
定理, 力对点的矩矢在通过该点的任意轴上的投影等于这力
对于该轴的矩。这就是力对点之矩与对通过该点轴之矩的关系。
)(
)(
c o s,
)(
)(
c o s,
)(
)(
c o s
Fm
Fm
Fm
Fm
Fm
Fm
O
z
O
y
O
x ??? g??
222 ))(())(())(()( FmFmFmFm zyxO ???
kFmjFmiFmFrFm zOyOxOO )]([)]([)]([)( ?????
kFmjFmiFm zyx )()()( ???
又由于
所以力对点 O的矩为,
39
Te method of reduction of a force system in space is the same as
in the planar case,only the planar coordinates have to be extended
to space coordinates,
§ 5–4 Reduction of a force system in space to a given point
nFFFF ?321,,
Suppose there is a force
system in space acting on a
rigid body
Reduce the force
system to an arbitrary
point O,
40
把研究平面一般力系的简化方法拿来研究空间一般力系的
简化问题，但须把平面坐标系扩充为空间坐标系。
§ 5-4 空间一般力系向一点简化
nFFFF ?321,,
设作用在刚体上有
空间一般力系
向 O点简化
（ O点任选）
41
① According to the Theorem of Translation of a Force we
translate every force to the point O,We get a
concurrent force system in space and
an additional force couple system,
[notice] are the moments of the forces about point O,
② Because the force couple in space is a free vector it can be
always translated to the point O,
'',',' 321 nFFFF ?
nmmm ?,,21
nmmm ?,,21
42
① 根据力线平移定理，将各力平行搬到 O点得到一空
间汇交力系,和附加力偶系
[注意 ] 分别是各力对 O点的矩。
② 由于空间力偶是自由矢量，总可汇交于 O点。
'',',' 321 nFFFF ? nmmm ?,,21
nmmm ?,,21
43
③ Composing,we get the principle vector,
In other words,(the principle vector passes
the point O,and it does not depend on the position of the
point O)
Composing,we get the principle moment,
(the principle moment depends on
the position of point O)
'',',' 321 nFFFF ? 'R
?? ?? ii FFR '' 'R
nmmm ?,,21 OM
?? ?? )( iOiO Fmmm OM
44
③ 合成 得主矢
即 （主矢 过简化中心 O,
且与 O点的选择无关）
合成 得主矩
即,（主矩 与简化中心 O有关）
'',',' 321 nFFFF ? 'R
?? ?? ii FFR '' 'R
nmmm ?,,21 OM
?? ?? )( iOiO Fmmm OM
45
If the given center O is the origin of the coordinate system,then
the magnitude of the principle vector is
The direction is
According to the relationship between the moment of the force about
a point and the moment of the force about an axis through the point
The magnitude of the principle moment is
and the direction is,
??? ?????? 222222 )()()('''' ZYXRRRR zyx
'c o s,'c o s,'c o s R
Z
R
Y
R
X ??? ??? g??
????
??
????
??
)(])([ ;)(])([;)(])([
FmFmmFmFmm
mFmFm
zzOOzyyOOy
OxixxiO
222
OzOyOxO MMMM ???
O
Oz
O
Oy
O
Ox
M
M
M
M
M
M ??? 'c o s,'c o s,'c o s g??
46
若取 简化中心 O点为坐标原点，则,
主矢大小
主矢方向
根据力对点之矩与力对轴之矩的关系,
则 主矩大小 为,
主矩方向,
??? ?????? 222222 )()()('''' ZYXRRRR zyx
'c o s,'c o s,'c o s R
Z
R
Y
R
X ??? ??? g??
????
??
????
??
)(])([ ;)(])([;)(])([
FmFmmFmFmm
mFmFm
zzOOzyyOOy
OxixxiO
222
OzOyOxO MMMM ???
O
Oz
O
Oy
O
Ox
M
M
M
M
M
M ??? 'c o s,'c o s,'c o s g??
47
If a force system in space is reduced to a given point,we get a
principle vector and a principle moment,It follows,
§ 5–5 Discussion of the result of
the reduction of a force system in space
1,If the force system is in equilibrium,
0,0' ?? OMR
2,If the force system is equivalent to a force couple,
the moment is equal to the sum of the moments MO of the forces
about the given point,The principle moment does not depend on
the position of the given center,
0,0' ?? OMR
3,If the force system is equivalent to a resultant
force,The principle vector which is equal to the resultant
force vector of the force system,the principle vector passes
the given center O (it depends on the position of the given
center,If the point O is changed,the principle moment is not
zero),
0,0' ?? OMR
'R
R R
48
空间一般力系向一点简化得一主矢和主矩，下面针对主
矢、主矩的不同情况分别加以讨论。
§ 5-5 空间一般力系简化结果的讨论
1,若,则该力系 平衡 （下节专门讨论）。
0,0' ?? OMR
2、若 则力系可合成一个 合力偶,其矩等于原力
系对于简化中心的主矩 MO。 此时主矩与简化中心的位置无关。
0,0' ?? OMR
3,若 则力系可合成为一个 合力,主矢 等
于原力系合力矢,合力 通过简化中心 O点。 （此时
与简化中心有关，换个简化中心，主矩不为零）
0,0' ?? OMR 'R
R R
49
Because
?????? iOOO FRR
M
R
M
ddRM is f o r c er e s u l t a n t t h ea n d
'
g e t w e
4.If,we can discuss from ①
two kinds of situations separately,②
OMR ?'
OMR //'
0,0' ?? OMR
① if
OMR ?'
)( dRM O ??
Turn MO into(R'',R),where R' counteracts R'',so we get R,
50
4,若 此时分两种情况讨论。即,①
②
OMR ?'
OMR //'
0,0' ?? OMR
由于做
??????? iOOO FRR
M
R
M
ddRM 合力,',
① 若 时
OMR ?'
)( dRM O ??
可进一步简化，将 MO变成 ( R'',R)使 R'与 R''抵消只剩下 R。
51
② If,——it is the condition of force screw,
[Example]① Twist the screw ② Shell helix
OMR //'
③ R' is neither parallel with nor perpendicular to M0,there is
an arbitrary angle ? between them,
In this case <1> decompose MO into M// and M?,
<2> deal with M// and M? by ① and ②,respectively,
' '
52
② 若 时,——为力螺旋的情形 （新概念，又移动又转动）
[例 ] ①拧螺丝 ②炮弹出膛时炮弹螺线
OMR //'
③ R′ 不平行也不垂直 M0,最一般的成任意角 ?
在此种情况下,<1>首先把 MO 分解为 M//和 M?
<2>将 M//和 M? 分别按①、②处理。
' '
53
M? make the principle vector
R' translate,the distance is,'
s in
'' R
M
R
MOO O ??? ?
Therefore,there is a force screw
about point O',
Because M// is a free vector it
can be translated to the point O',
M//a is invariant,?
54
M? 使主矢 R'搬家，搬家的矩离,
'
s in
'' R
M
R
MOO O ??? ?
所以在 O'点处形成一个力螺旋 。
因为 M// 是自由矢量,
可将 M//搬到 O'处
M//不变,
?
55
[notice]
The invariant quantities after reduction are R'and M// (they do not
depend on the given center),
If the given center is the point O we get M?,But if the given center
is the point O'it is changed to M?',M// is an invariant quantity (it
does not depend on the given center and the force couple of the
force system)
?
56
[注意 ]
力系简化中的不变量（不随简化中心改变）有,R′, M//
简化中心为 O时：为 M? 当简化中心为 O′ 时，为 M?′
但 M//总是不变的（ 它是
原力系中的力偶与简化
中心无关 ）
?
57
)('' RmROOMM OO ???? ?
?? )(
ism o m e n t p r i n c i p l e t h eB e c a u s e
iOO FmM
??? )()( iOO FmRM
f o r m, in t h e u s e dn o r m a lly is
)()( ?? izz FmRM
After the reduction of the force system in space to the point O,we
get a principle vector R‘ and a principle moment MO, If MO?R' we
can get a new resultant R which acts on the new given center O',
The law of the resultant moment of the force system in space,
58
)('' RmROOMM OO ???? ?
?? )( iOO FmM主矩又 ?
??? )()( iOO FmRM
?? )()( izz FmRM常用投影式
空间力系向 O点简化后得主矢 R'和主矩 MO,若 MO?R'，可
进一步合成为一个 作用在新简化中心 O'点的 合力 R 。
空间力系的合力矩定理,
59
1,The necessary and sufficient conditions for equilibrium of a
force system in space are
0)( ?? ? iOO FmM00' ??? ? FR
??? ??? 222 )()()(':b e c a u s ea g a i n ZYXR
222 ))(())(())(( ??? ??? FmFmFmM zyxO
So the equilibrium equations of a force system in space are
They can also be expressed by 4
moment equations,5 moment
equations or 6 moment equations,
but each case has some limiting conditions,
??
??
??
??
??
??
0)(,0
0)(,0
0)(,0
FmZ
FmY
FmX
z
y
x
§ 5–6 Equilibrium equations of
a force system in space and their applications
60
一、空间任意力系的平衡充要条件是,
0)( ?? ? iOO FmM00' ??? ? FR
??? ??? 222 )()()(' ZYXR?又
222 ))(())(())(( ??? ??? FmFmFmM zyxO
所以 空间任意力系的平衡方程 为,
还有四矩式，五矩式和六矩式,
同时各有一定限制条件。
??
??
??
??
??
??
0)(,0
0)(,0
0)(,0
FmZ
FmY
FmX
z
y
x
§ 5-6 空间一般力系的平衡方程及应用
61
The equilibrium equations of a concurrent force system in space are
Because all action lines of the forces are
concurrent at a point and all axes are
through this point the moment equations are
identities,0
0
0
?
?
?
?
?
?
Z
Y
X
The equilibrium equations of a parallel force system in space,
supposing all the action lines are parallel to the axis z are
they are all identities,?
?
?
?
?
?
0)(
0)(
0
Fm
Fm
Z
y
x ?
?
?
?
?
?
0
0
0)(
Y
X
Fm z
62
空间汇交力系的平衡方程为,
因为各力线都汇交于一点，各轴都通过
该点，故各力矩方程都成为了恒等式。
0
0
0
?
?
?
?
?
?
Z
Y
X
空间平行力系的平衡方程，设各力线都 // z 轴。
因为
均成为了恒等式。 ?
?
?
?
?
?
0)(
0)(
0
Fm
Fm
Z
y
x ?
?
?
?
?
?
0
0
0)(
Y
X
Fm z
63
1,Spherical gemel
2,Constraints in space
There are,in general,six possible motions of a body in space
(translations along three axes and rotations about three axes),
How many kinds of motion are hindered by the constraints in the
following cases,If there are constraints there are reaction forces of
the constraints,too,The impediment of a translation is a reaction
force,the impediment of a rotation is a reaction force couple,
[Examples]
64
1、球形铰链
二、空间约束
观察物体在空间的六种（沿三轴移动和绕三轴转动）可能
的运动中，有哪几种运动被约束所阻碍，有阻碍就有约束反力。
阻碍移动为反力，阻碍转动为反力偶。 [例 ]
65
Spherical gemel
66
球形铰链
67
2,Radial bearing,hinges,ball bearing
68
2、向心轴承，蝶铰链，滚珠 (柱 )轴承
69
3,Sliding bearing
70
3、滑动轴承
71
4,Thrust bearing
72
4、止推轴承
73
5,Splint with a pin
74
5、带有销子的夹板
75
6,Fixed support in space
76
6、空间固定端
77
If a system of parallel forces in
space has a resultant force and the
point of its application is the point
C,point C is called the center of
the system of parallel forces,The
center of gravity of an object is a
special example of the center of a
system of parallel forces,
§ 5–7 The center of a system of parallel
forces and the center of gravity of an object
1,The center of a system of parallel forces and the center of gravity
of object
(1) The center of a system of parallel forces,
From the law of the resultant moment we get,
?? )()( iOO FmRm nnC FrFrFrRr ???????? ?2211
78
空间平行力系，当它有合
力时，合力的作用点 C 就是此
空间平行力系的中心 。而物体
重心问题可以看成是空间平行
力系中心的一个特例。
§ 5-7 平行力系的中心 物体的重心
一、空间平行力系的中心、物体的重心
1、平行力系的中心
由合力矩定理,
?? )()( iOO FmRm nnC FrFrFrRr ???????? ?2211
79
0110,L e t PFFPRR ??
nnC rFrFrFrR ????? ?2211
?
???????
i
iinn
C F
rF
R
rFrFrFr ?2211
R
zF
z
R
yF
y
R
xF
x iiCiiCiiC ??? ???,,a r e e q u a t i o n s p r o j e c t i o n T h e
80
0110,PFFPRR ??令
nnC rFrFrFrR ????? ?2211
?
???????
i
iinn
C F
rF
R
rFrFrFr ?2211
R
zFz
R
yFy
R
xFx ii
C
ii
C
ii
C
??? ???,,:投影式
81
If we treat the gravity on a body
as a system of parallel forces the
problem of determination of the
center of gravity is just the
problem of the determination of
the center of a system of parallel
forces,From the law of the
resultant moment we have
iiC xPxP ??? ?
P
Pz
z
P
Py
y
P
Px
x
i
C
i
C
i
C
?
?
?
?
?
?
?
?
?
,
,
If the body is divided into smaller and smaller
parts the determination of the center of gravity
becomes more accurate,In the limiting case,
(n- ),we used to use an integration to
determine the position of the center of gravity,
?
2,The formula for the center of gravity,
82
如果把物体的重力都看成为
平行力系，则求重心问题就
是求平行力系的中心问题。
由合力矩定理,
iiC xPxP ??? ?
P
Pz
z
P
Py
y
P
Px
x
i
C
i
C
i
C
?
?
?
?
?
?
?
?
?
,
,
物体分割的越多，每一小部分体积越小，
求得的重心位置就越准确。在极限情况下，
(n? ),常用积分法求物体的重心位置。 ?
二、重心坐标公式,
83
?? V dVP g
If gI denote the weight of part No,I and ⊿ Vi is the volume of it,
then, Substitution into the equations above gives in the
limiting case,
Where,this is the accurate formula to determine the center of
gravity,
P
dVz
zP
dVy
yP
dVx
x VCVCVC ??? ???
ggg
,,
iii VP ?g? ?
For a homogeneous body,g =constant,the formula is simplified to,
V
dVz
zV
dVy
yV
dVx
x VCVCVC ???
?
?
?
?
?
?,,
Using the same arguments,we can fine out the formulae for thin plane
and a thin-long rod,
84
设 gi表示第 i个小部分每单位体积的重量, ⊿ Vi第 i个小体积，则
代入上式并取极限，可得,
式中,上式为 重心 C 坐标的精确公式 。
P
dVz
zP
dVy
yP
dVx
x VCVCVC ??? ???
ggg
,,
iii VP ?g? ?
?? V dVP g
对于均质物体,g =恒量，上式成为,
V
dVz
zV
dVy
yV
dVx
x VCVCVC ???
?
?
?
?
?
?,,
同理对于薄平面和细长杆均可写出相应的公式。
85
According to the fact that the center
of the system of parallel forces does
not depend on the direction of the
forces,the action line can be chosen
to be parallel to the axis y,Using the
law of the resultant moment and
determining the moment about the
axis x we find
P
zPzzPPz ii
CiiC
?? ??? ??,
In sum,the equations of the center of gravity are,
P
zP
z
P
yP
y
P
xP
x iiCiiCiiC ???
?
?
?
?
?
?,,
Substituting △ Pi= △ mig and P=Mg into the equations above we
get the equations of center of mass,
M
zmz
M
ymy
M
xmx ii
C
ii
C
ii
C
??? ??????,,
86
根据平行力系中心位置与各平
行力系的方向无关的性质，将力
线转成与 y轴平行，再应用合力矩
定理对 x 轴取矩得,
P
zPzzPPz ii
CiiC
?? ??? ??,
综合上述得 重心坐标公式 为,
P
zPz
P
yPy
P
xPx ii
C
ii
C
ii
C
??? ??????,,
若以△ Pi= △ mig,P=Mg 代入上式可得质心公式
M
zmz
M
ymy
M
xmx ii
C
ii
C
ii
C
??? ??????,,
87
By the same arguments,we can find the equations for a
homogeneous body,a homogeneous board and a homogeneous rod,
They are
V
zV
z
V
yV
y
V
xV
x iiCiiCiiC ???
?
?
?
?
?
?,,:b o d y
A
zA
z
A
yA
y
A
xA
x iiCiiCiiC ???
?
?
?
?
?
?,,:p l a t e
l
zl
z
l
yl
y
l
xl
x iiCiiCiiC ???
?
?
?
?
?
?,,:r o d
88
同理：可写出均质体，均质板，均质杆的形心（几何中心）
坐标分别为,
V
zVz
V
yVy
V
xVx ii
C
ii
C
ii
C
??? ??? ???,,:立体
A
zAz
A
yAy
A
xAx ii
C
ii
C
ii
C
??? ??? ???,,:平板
l
zlz
l
yly
l
xlx ii
C
ii
C
ii
C
??? ??? ???,,:细杆
89
Solution,Due to symmetry the center of this arc must be on the
axis Ox,hence yC=0,Select a small part,
?dRdL ??
? c o s ?? Rx
R
dR
L
dLx
x
L
C
?
??
?
?
2
c o s
2
?
?
?
???
?
?
??
?
?s inRx
C ?
The following is an example of the determination of the center of
gravity of a body by integration,
[Example] Determine the center of gravity of the homogeneous arc
with the radius R and the top angle 2?,
O
90
解,由于对称关系，该圆弧重心必在 Ox轴，即 yC=0。取微段
?dRdL ??
? c o s ?? Rx
R
dR
L
dLx
x
L
C
?
??
?
?
2
c o s
2
?
?
?
???
?
?
??
?
?s inRx
C ?
下面用积分法 求物体的重心实例,
[例 ] 求半径为 R，顶角为 2? 的均质圆弧的重心。
O
91
cm4.6
21
2211
?
?
?
?
?
?
?
SS
ySyS
A
yA
y
ii
C
3,The ways to determine the center of gravity of an object,
① Method of combination,
Solution,
cm248 cm4 21,8 0 cm 212 221 )πR(y,y,π RSS ?????
Determine the center of gravity
of this composed unit,
We know
92
三、重心的求法, ① 组合法
cm4.6
21
2211
?
?
?
?
?
?
?
SS
ySyS
A
yA
y
ii
C
由
解,
cm248 cm4 21,8 0 cm 212 221 )πR(y,y,π RSS ?????
求：该组合体的重心？
已知,
93
0)( F r o m ?? Fm B 01 ????? CW xPlP P
lPx W
C
1???
The area and formulae of the center of gravity of some simple
graphs can be found in the table,
② Experimental method,
<1>hanging method <2> weighting method
Horizontal
line
94
0)( ?? Fm B由 01 ????? CxPlP 称 P
lPx
C
1??? 称
简单图形的面积及重心坐标公式可由表中查出。
② 实验法,
<1>悬挂法 <2>称重法
95
[Example] We know RC=100mm,RD=50mm,Px=466N,Py=352N
and Pz=1400N,When the system is in equilibrium (rotation with
uniform speed) determine the force Q (acting at the nadir point of
roller C) and the reaction forces of bearing A and B,
Solution,① select the object under study ② draw the force diagram
③ select a coordinate system and write out the equations,
If there is only one unknown quantity in every
equation,it is easy to get the solution,
96
[例 ] 已知, RC=100mm,RD=50mm,Px=466N,Py=352N,Pz=1400N
求：平衡时 (匀速转动 )力 Q=？ (Q力作用在 C轮的最低点）
和轴承 A,B的约束反力？
解,①选研究对象 ②作受力图 ③选坐标列方程
最好使每一个方程有一个未知数，方便求解。
97
)N(746,010050;0
)N(352,0;0 F r o m
????????
?????
?
?
QQPm
PYPYY
xzy
yAyA ?
98
)N(746,010050;0
)N(352,0;0
????????
?????
?
?
QQPm
PYPYY
xzy
yAyA ?由
99
)N(385,020s i n ;0
)N(2040,020s i n 50300200 ;0
)N(729,020c o s ;0
)N(437,020c o s 5020050300 ;0
0
0
0
0
???????
???????
???????
???????
?
?
?
?
AzBA
BzBx
AxBA
BByxz
ZQPZZZ
ZQPZm
XQPXXX
XQXPPm
A
A
100
)N(385,020s i n ;0
)N(2040,020s i n 50300200 ;0
)N(729,020c o s ;0
)N(437,020c o s 5020050300 ;0
0
0
0
0
???????
???????
???????
???????
?
?
?
?
AzBA
BzBx
AxBA
BByxz
ZQPZZZ
ZQPZm
XQPXXX
XQXPPm
A
A
101
Method 2,Project the force system in space on the three axes,
change the problem into that of a force system in a plane,and
solve it,Please,practice it by yourselves after the class,
102
方法 (二 ),将空间力系投影到三个坐标平面内，转化为平面力
系平衡问题来求解，请同学们课后自己练习求解。
103
1,Concepts and content,
1)The force couple or the moment of a force about a point is a
vector,
2)The moment of force about an axis,a force couple in plane and
the moment of a force in plane about a point are algebraic quantities,
3)The law of projection of the resultant force of a force system in
space is
4)The law of the resultant moment of the force system in space is
5)The relationship between the moment of the force about a point
and the moment of a force about an axis which passes the point is
)(,FmM o
?????? iziyix ZRYRXR,,
)()( ioo FmRm ??
)()(:e q u a t i o n p r o j e c t i o n izz FmRm ??
? ? )()( FmFm zZo ?
Exercises to chapter 5 the force system in space
104
一、概念及内容,
1、空间力偶及空间力对点之矩是矢量,
2、空间力对轴之矩和平面力偶、平面力对点之矩是代数量。
3、空间力系 合力投影定理,
4、空间力系的 合力矩定理,
5,空间力对点之矩与对轴之矩的关系,
)(,FmM o
?????? iziyix ZRYRXR,,
)()( ioo FmRm ??
)()(,izz FmRm ??投影式
? ? )()( FmFm zZo ?
第五章, 空间力系, 习题课
105
2,Basic Equations
1) The equilibrium equations of a force system in space are
?
?
?
?
?
?
0
0
0
Z
Y
X
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
z
y
x
m
m
m
Z
Y
X
? ?
? ?
? ?
0
0
0
z
y
x
m
m
m
? ?
? ?
? ?
0
0
0
z
y
m
m
X
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
z
y
x
m
m
m
Y
X
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
'x
z
y
x
m
m
m
m
Y
X
The additional conditions of the
equations of four moments,the
equations of five moments and
the equations of six moments
make the equations independent,
A general
force
system in
space
Concur-
-rent
force
system
in space
System
of force
couples
in space
Force
system
in space
on x axis
only
Force
system
in space
on xoy
plane
only
106
二、基本方程
1,空间力系的平衡方程
空
间
一
般
力
系
空
间
汇
交
力
系
?
?
?
?
?
?
0
0
0
Z
Y
X
空
间
力
偶
系
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
z
y
x
m
m
m
Z
Y
X
? ?
? ?
? ?
0
0
0
z
y
x
m
m
m
空
间∥
x
轴
力
系
? ?
? ?
? ?
0
0
0
z
y
m
m
X
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
z
y
x
m
m
m
Y
X
空
间∥
xoy
面
的
力
系
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
'x
z
y
x
m
m
m
m
Y
X
四矩式, 五矩式和六矩式的附加条件均
为使方程式独立。
107
3,Steps,skills and remarkable problems in solution,
1) Steps,① Select the object to study (as in the case of a plane),
② Draw the force diagram,
③ Select a coordinate system,write down the equations,
④ Solve the equations,determine the unknown quantities,
2) Some problems about the force systems in space,
① x,y,z may not be the three axes of the moment,We can select
six axes arbitrarily,
② The equations for the moments can be less than three (because
MO is a vector),
③ There are six independent equations for the force system in space
(because the body in space has six degree of freedom),there are
only three degree of freedom in a plane,
④ The problem of friction is involved,in general,
108
三、解题步骤、技巧与注意问题,
1,解题步骤, ①选研究对象
(与平面的相同 ) ② 画受力图
③选坐标、列方程
④解方程、求出未知数
2、空间力系的几个问题,
① x,y,z (三个取矩轴和三个投影轴可以不重合 )可以任选的
六个轴。
② 取矩方程不能少于三个（ ∵ MO是矢量）
③ 空间力系独立方程六个（ ∵ 空间物体六个自由度）
平面三个自由度
④ 空间力系中也包括摩擦问题。
109
2) Skills,
① It is always convenient to replace the projection axis by the axis
of the moment,
② It’s better,if the coordinate axis ? unknown quantities or if
the axis of the moment is parallel to the unknown quantities
or intersects with the unknown quantities,
③ It is normal to study a system from the whole to the parts,
④ The friction force is F = N f,its direction is opposite to the
tendency of motion,
3) Remarkable problems,
① A force couple does not appear in the projection axes equations,
② A force couple in space is a vector,the force couple in a plane is
an algebraic quantity,
③ The usual way to determine the center of gravity of a body is the
method of combination,The center of gravity,the center of mass
and the centroid of area are the same in the case of a
homogeneous body,
110
2,解题技巧,
①用取矩轴代替投影轴，解题常常方便
②投影轴尽量选在与未知力 ?，力矩轴选在与未知力平行或相交
③一般从整体 —>局部的研究方法。
④摩擦力 F = N f,方向与运动趋势方向相反。
3,注意问题,
① 力偶在投影轴中不出现（即在投影方程中不出现）
② 空间力偶是矢量，平面力偶是代数量。
③ 求物体重心问题常用组合法。
对于均质物体，重心、中心、形心为同一点。
111
[Example 1] Let P=2000N,point C is in the plane Oxy,
Determine the moment of force P about the three axes,
?????
?????
???
???
60c o s45c o s
60s i n45c o s
45c o s
45s i n
PP
PP
PP
PP
y
x
xy
z
Solution,① Select a body to investigate; ② Draw the force diagram,
③ Select a coordinate system and write down the equations,
112
[例 1] 已知,P=2000N,C点在 Oxy平面内
求：力 P对三个坐标轴的矩
?????
?????
???
???
60c o s45c o s
60s i n45c o s
45c o s
45s i n
PP
PP
PP
PP
y
x
xy
z
解,①选研究对象；②画受力图；③选坐标列方程。
113
)mN(2.3860c o s45c o s560s i n45c o s6
0)5(6)()()()(
????????
?????????
PP
PPPmPmPmPm yxzzyzxzz
)mN(8.8445s i n6
600)()()()(
????
??????
P
PPmPmPmPm zzxyxxxx
)mN(7.7045s i n5
500
)()()(
)(
????
???
???
P
P
PmPmPm
Pm
z
zyyyxy
y
114
)mN(2.3860c o s45c o s560s i n45c o s6
0)5(6)()()()(
????????
?????????
PP
PPPmPmPmPm yxzzyzxzz
)mN(8.8445s i n6
600)()()()(
????
??????
P
PPmPmPmPm zzxyxxxx
)mN(7.7045s i n5
500
)()()(
)(
????
???
???
P
P
PmPmPm
Pm
z
zyyyxy
y
115
[Example 2]Let AB=3m,AE=AF=4m and Q=20kN,Determine the
tensile forces of rope BE,BF and the internal force of rod AB,
)kN(5 4 6
,045s i n15s i n',0
1
1
??
??????
T
QTY
At point C,
Solution,Study point C and point B
respectively,draw the force diagram,
116
[例 2] 已知,AB=3m,AE=AF=4m,Q=20kN;
求,绳 BE,BF的拉力和杆 AB的内力
)kN(5 4 6
,045s i n15s i n',0
1
1
??
??????
T
QTY
由 C点,
解：分别研究 C点和 B点作受力图
117
)kN( 230
,)kN( 419
5
3
s i n,
5
4
43
4
c o s
0 s i n s i n60c o s,0
045c o s c o s45c o s c o s60s i n,0
045c o s c o s45c o s c o s,0
2
32
22
3212
321
32
?
???
??
?
?
??????
???????
??????
?
?
?
N
TT
TTTNZ
TTTY
TTX
??
??
??
??
At point B,
118
)kN( 230
,)kN( 419
5
3
s i n,
5
4
43
4
c o s
0 s i n s i n60c o s,0
045c o s c o s45c o s c o s60s i n,0
045c o s c o s45c o s c o s,0
2
32
22
3212
321
32
?
???
??
?
?
??????
???????
??????
?
?
?
N
TT
TTTNZ
TTTY
TTX
??
??
??
??
由 B点,
119
This problem tells us,
① The force couple does not
appear in the projection equations,
② If the force couple appear in the
moment equations we treat it as a
vector,just like the projection of a
force in the projection equations,
③ Try to make every equation
dependent only on one unknown
quantity,
④ Be aware of the reaction force
of the constraints in space,
[Example 3] Curved lever ABCD,∠ ABC=∠ BCD=900,AB=a,
BC=b,CD=c,m2,m3,Determine the reaction force of the
constraints and m1?
120
此题训练,
① 力偶不出现在投影式中
② 力偶在力矩方程中出现
是把力偶当成矢量后，类
似力在投影式中投影
③ 力争一个方程求一个支
反力
④ 了解空间支座反力
[例 3] 曲杆 ABCD,∠ ABC=∠ BCD=900,AB=a,BC=b,
CD=c,m2,m3 求：支座反力及 m1=?
121
Solution,
0,0
,0,0
,0,0
,0,0
,0,0
0,0
1
2
3
3
3
2
2
1
?????
???????
???????
??????
???????
??
?
?
?
?
?
?
DDx
ADDA
ADDA
AADz
AADy
D
YcbZmm
a
m
ZZZZZ
a
m
YYYYY
a
m
YaYmm
a
m
ZaZmm
XX
32
32
1 )()( ma
cm
a
b
a
mc
a
mbcYbZm
DD ????
?
??
? ????????
122
解,
0,0
,0,0
,0,0
,0,0
,0,0
0,0
1
2
3
3
3
2
2
1
?????
???????
???????
??????
???????
??
?
?
?
?
?
?
DDx
ADDA
ADDA
AADz
AADy
D
YcbZmm
a
m
ZZZZZ
a
m
YYYYY
a
m
YaYmm
a
m
ZaZmm
XX
32
32
1 )()( ma
cm
a
b
a
mc
a
mbcYbZm
DD ????
?
??
? ????????
123
[Example 4] We know the
weight of the rod AB is 80N,
AD and CB are strings and
there are two point A and C on
the same vertical line,The
bottom of A and B are smooth,
∠ ABC=∠ BCE=600,
Moreover AD is horizontal,
AC is a plumb line,Determine
TA,TB and the reaction forces
at point A and B,
Solution,Select the axis of
projection and the axis of
moment skillfully,make every
equation dependent on one
unknown quantity,
124
[例 4] 已知,AB杆,AD,CB为
绳,A,C在同一垂线上,AB
重 80N,A,B光滑接触，
∠ ABC=∠ BCE=600,且 AD水
平,AC铅直。求平衡时,TA，
TB及支座 A,B的反力。
解,思路：要巧选投影轴和取
矩轴，使一个方程解出一个未
知数。
125
0N8,0 F r o m ????? PNZ B
0
2
160c o s
,0'
????????
??
CEPACT
m
B
DD
N)( 1.2380
6
3
3
3
2
60c t g
2
60c o s60c t g
2
1
60c o s
????????
??????????
PP
T
ACPACT
B
B
CEAC ????? 60co s60ct gA g ai n ?
126
0N8,0 ????? PNZ B由
0
2
160c o s
,0'
????????
??
CEPACT
m
B
DD
N)( 1.2380
6
3
3
3
2
60c t g
2
60c o s60c t g
2
1
60c o s
????????
??????????
PP
T
ACPACT
B
B
CEAC ????? 60c o s60c t g?又
127
)N( 5.11
2
1
80
6
3
60c o s
060c o s,0
???????
???????
BA
BA
TT
TTX
)N( 20
2
3
80
6
3
060s i n,0
?????
??????
A
BA
N
TNY
128
)N( 5.11
2
1
80
6
3
60c o s
060c o s,0
???????
???????
BA
BA
TT
TTX
)N( 20
2
3
80
6
3
060s i n,0
?????
??????
A
BA
N
TNY
129
130
Theoretical Mechanics
2
3
Practically,there exists many force systems in which the action
lines of the forces are not coplanar,these are force systems in space,
They are the most general force systems,
Diagram (a) is a concurrent forces system in space,
Diagram (b) is an arbitrary force system in space,if neglecting the
forces of the wind from the side it is a parallel force system in space,
The force of
wind from
the front
The force of
wind from
the side b
4
工程中常常存在着很多各力的作用线不在同一平面内的力
系，即空间力系，空间力系是最一般的力系。
(a)图为空间汇交力系； (b)图为空间任意力系；
(b)图中去了风力为空间平行力系。
迎 面
风 力
侧 面
风 力
b
5
Chapter 5,Force system in space
§ 5–1 Concurrent force system in space
§ 5–2 System of force couples in space
§ 5–3 The moment of a force about a point or an axis
§ 5–4 Reduction of a force system in space to a given
point
§ 5–5 Discussion of the result of the reduction of a
force system in space
§ 5–6 Equilibrium equations of a force system in
space and their applications
§ 5–7 The center of a system of parallel forces and the
center of gravity of an object
Class of exercises
6
第五章 空间力系
§ 5–1 空间汇交力系
§ 5–2 空间力偶系
§ 5–3 力对点的矩与力对轴的矩
§ 5–4 空间一般力系向一点的简化
§ 5–5 空间一般力系简化结果的讨论
§ 5–6 空间一般力系的平衡方程及应用
§ 5–7 平行力系的中心与物体的重心
习题课
7
§ 5-1 Concurrent force system in space
1,The projection and decomposition of a force on the axes in space,
(1)The representation of force in space,
a force is given by its
magnitude,its direction and its
point (line) of application
magnitude,
point of application,the point at
which the force acts onto the
object,
direction,it can be determined by
the three angles ?,?,g or by the angle of inclination ? and the angle
of depression ?,
?
g
?
Fxy
O
FF ?
8
一、力在空间轴上的投影与分解,
1.力在空间的表示,
力的三要素,
大小、方向、作用点 (线 )
大小,
作用点,在物体的哪点就是哪点
方向,
由 ?,?,g三个方向角确定
由仰角 ? 与俯角 ? 来确定。
?
g
?
Fxy
O
FF ?
§ 5-1 空间汇交力系
9
(2) The method of direct projection
From the fig we see
g
?
?
c o s
,c o s
,c o s
??
??
??
FZ
FY
FX
????g c o sc o sc o sc o ss i n ???????? FFFX xy
????g s i nc o ss i ns i ns i n ???????? FFFY xy
?g s i nco s ???? FFZ
(3) The method of indirect projection
When the angles ?,? and g are not
easy to determine we can project F
onto the plane xy firstly and then
project the result Fxy onto the axes x
and y,
10
2、一次投影法（直接投影法）
由图可知,
g
?
?
c o s
,c o s
,c o s
??
??
??
FZ
FY
FX
????g c o sc o sc o sc o ss i n ???????? FFFX xy
????g s i nc o ss i ns i ns i n ???????? FFFY xy
?g s i nco s ???? FFZ
3、二次投影法（间接投影法）
当力与各轴正向夹角不易
确定时，可先将 F 投影到 xy
面上，然后再投影到 x,y轴上,
即
11
(4) Decomposition of a force along the coordinate axes,
If the three perpendicular components
of a force along the coordinate axes
are
then,zyx
FFF a n d,
zyx FFFF ???
222 ZYXF ????
F
Z
F
Y
F
X ??? g?? c o s,c o s,c o s
kZFjYFiXF zyx ???,,
due to
kZjYiXF ???
We get
Fx
Fy
Fz
12
4、力沿坐标轴分解,
若以 表示力沿直角
坐标轴的正交分量，则,
zyx FFF,,
zyx FFFF ???
222 ZYXF ????
F
Z
F
Y
F
X ??? g?? c o s,c o s,c o s
kZFjYFiXF zyx ???,,
而,
kZjYiXF ???
所以,
Fx
Fy
Fz
13
(1) The graphical method,
It is same as in the case of the composition of a coplanar system of
concurrent forces,It can be solved by the force polygon,
The resultant force is equal to the geometrical sum of the
components,
in FFFFFR ??????? ?321
(2)The analytical method,
Because,substitute it into the equation above
We get
is the projection of the resultant force onto the
axis x,hence
kZjYiXF iiii ???
kZjYiXR iii ??? ???
? iX
?? ix XR
?? iy YR
?? iz ZR
2,The composition of a concurrent force system in space,
14
1、几何法,与平面汇交力系的合成方法相同，也可用力多
边形方法求合力。
即：合力等于各分力的矢量和
in FFFFFR ??????? ?321
2、解析法,
由于 代入上式
合力
由 为合力在 x轴的投影,∴
kZjYiXF iiii ???
kZjYiXR iii ??? ???
? iX
?? ix XR
?? iy YR
?? iz ZR
二、空间汇交力系的合成,
15
(3) The law of projection of the resultant force,
The projection of the resultant force of a force system in space onto
an axis is equal to the algebraic sum of the projections of the
component forces onto the same axis,
The resultant force is given by
??? ?????? 222222 )()()( ZYXRRRR zyx
R
R
R
R
R
R zyx ???? g?? c o s,c o s,c o s
16
3、合力投影定理,
空间力系的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。
??? ?????? 222222 )()()(,ZYXRRRR zyx合力
R
R
R
R
R
R zyx ???? g?? c o s,c o s,c o s
17
3,The equilibrium of a concurrent force system in space:
? ?0X
? ?0Y
? ?0Z
They are called equilibrium equations
of a concurrent force system in space,
Hence the necessary and sufficient condition of equilibrium in
the analytical method is given by the equations
Hence the necessary and sufficient condition of equilibrium in
the graphical method is that the forces form a closed polygon,
0?? ? iFR
The necessary and sufficient condition of equilibrium is that
the resultant force is zero,
18
三、空间汇交力系的平衡,
? ?0X
? ?0Y
? ?0Z
称为平衡方程
空间汇交力系的平衡方程
∴ 解析法 平衡充要条件为,
∴ 几何法 平衡充要条件为该力系的 力多边形封闭 。
0?? ? iFR
空间汇交力系平衡的充要条件是,力系的合力为零,即,
19
§ 5–2 System of force couples in space
Beside the magnitude and the turning direction,the action plane of a
force couple in space has to be determined,therefore a force couple
is represented by a vector,
1,A force couple can be represented by a vector
The turning direction of the force
couple is determined by the right-
hand rule,Looking from the bottom
of the vector,the clockwise rotation is
positive,
Force couple in space is a free vector,
20
§ 5-2 空间力偶系
由于空间力偶除大小、转向外，还必须确定力偶的作用面，
所以空间力偶矩必须用矢量表示。
一、力偶矩用矢量表示,
力偶的转向为右手螺旋定则。
从力偶矢末端看去，顺时针转动
为正。
空间力偶是一个自由矢量。
21
Proof,① Draw II//Ⅰ,cd // ab
② Draw a couple of equilibrium
forces R and R' at point E,
(-R=R' should be satisfied)
③ Composing F1 and R we get
F2 at the point d composing
F1'and R' we get F2' at point c,R-F1=F2,R'- F1'= F2 ',
④ The force couple (F1,F1') in plane I is replaced by (F2,F2') in
plane II,
2,The law of equivalent law of a force couple in space
A force couple acting on a rigid body can be replaced by any other
force couple of the same moment and turning direction lying in a
parallel plane without altering the external effect on that body,
22
[证 ] ① 作 II//Ⅰ, cd // ab
② 作一对平衡力 R,R' (在 E点,且
使 -R=R')
③ 由反向平行力合成得,
F1与 R合成得 F2，作用在 d点
F1'与 R'合成得 F2'，作用在 c点
且 R-F1=F2, R'- F1'= F2'
④ 在 I内的力偶（ F1,F1'）等效变成 II内的（ F2,F2' ）
二、空间力偶的等效定理
作用在同一刚体的两平行平面的两个力偶，若它们的转向相
同，力偶矩的大小相等，则两个力偶等效。
23
Because the force couple in space is a free vector it has three elements
① the magnitude of the moment of the force couple
② the direction of the the moment of the force couple is the
normal of the action plane of the force couple,
③ the turning direction,given by the right-hand rule,
m
3,Composition and equilibrium of a system of force couple in space
Because the moment of a force couple in space is a free vector,it
can be translated to any point without changing its direction,
Especially,it can be translated to the concurrent point,Because the
moment are vectors they can be composed according to usual
vector rules,The resultant moment of force couple is the sum of the
moments of the component force couples,
24
由此可得出,空间力偶矩是自由矢量,它有三个要素,
①力偶矩的大小 =
② 力偶矩的方向 ——与力偶作用面法线方向相同
③转向 ——遵循右手螺旋规则。
m
三、空间力偶系的合成与平衡
由于空间力偶系是自由矢量，只要方向不变，可移至任意
一点，故可使其滑至汇交于某点，由于是矢量，它的合成符合
矢量运算法则。 合力偶矩 = 分力偶矩的矢量和
25
?
?
??????
n
i
in mmmmmm
1
321 ?
The projections are
0?? xm
0?? ym
0?? zm
m
m
m
m
m
mmmmm zyx
zyx ?????? g?? c o s,c o s,c o s;
222
0?? ? imm
Obviously the equilibrium condition of the force couple system in
space is
26
?
?
??????
n
i
in mmmmmm
1
321 ?
投影式 为,
0?? xm
0?? ym
0?? zm
m
m
m
m
m
mmmmm zyx
zyx ?????? g?? c o s,c o s,c o s;
222
0?? ? imm
显然空间力偶系的平衡条件是,
27
In the plane it is an algebraic quantity,
In space,it is a vector,
[Example] the spherical hinges on the
reflective mirror of a car
§ 5–3 The moment of a force about a point or an axis
1,The moment of a force about a point
)a r e a (2)( A O BdFFm O ?????
If r is the radius vector at point A then
28
在平面中：力对点的矩是代数量。
在空间中：力对点的矩是矢量。
[例 ] 汽车反镜的球铰链
§ 5-3 力对点的矩与力对轴的矩
一、力对点的矩的矢量表示
面积A O BdFFm O ???? 2)(
如果 r 表示 A点的矢径，则,
29
The moment of the force about the
point o is equal to the vector
product of the radius vector of the
force at the point and the force,
dFFrFrFmFrFm OO ??????? ),s i n ()(,)(
21 ???
a n d B e c a u s e kZjYiXF ??? kzjyixr ???
ZYX
zyx
kji
FrFm O ???? )(
kFmjFmiFm
kyXxYjxZzXizYyZ
zOyOxO )]([)]([)]([
)()()(
???
??????
The angle between
the two vectors is
O
30
即,力对点的矩等于矩心到该力
作用点的矢径与该力的矢量积。
dFFrFrFmFrFm OO ??????? ),s i n ()(,)(
21 ???
kZjYiXF ???由于 kzjyixr ???
ZYX
zyx
kji
FrFm O ???? )(
kFmjFmiFm
kyXxYjxZzXizYyZ
zOyOxO )]([)]([)]([
)()()(
???
??????
两矢量夹角为
O
31
Definition,
It is an algebraic quantity,Direction is + –
)''a r e a (2)()( BOAdFFmFm xyxyOz ??????
2,The moment of a force about an axis
Conclusion,the moment of a
force about an axis which is
parallel to the force is zero,
In other words,if the force F
and the axis are coplanar the
moment is zero,
)()()()( xyOxyzzzz FmFmFmFm ???
[Proof]
32
定义,
它是代数量，方向规定 + –
的面积''2)()( BOAdFFmFm xyxyOz ??????
二、力对轴的矩
结论, 力对 //它的轴的
矩为零。即力 F与轴共
面时，力对轴之矩为零。
)()()()( xyOxyzzzz FmFmFmFm ???
[证 ]
33
The moment of a force about an axis which is parallel to the force is
zero,In other words,if the force F and the axis are coplanar the
moment is zero,
34
力对 //它的轴的矩为零。即力 F与轴共面时，力对轴之矩为零。
35
)(c o s)(:H e n c e FmFm zO ?? g )()]([ FmFm zzO ?
3,The relationship between the moment of the force about a point
and the moment of the force about an axis through the point
)a r e a (2)( A O BFm O ??
Proof,
''2)()( BOAFmFm xyzz ???
Draw an arbitrary axis Z through the
point O then
''co s BOAO A B ?g? ??
From the geometrical relationship,
''2c o s2 BOAO A B ?g? ??Hence,
36
即,
)(c o s)( FmFm zO ??? g )()]([ FmFm zzO ?
三、力对点的矩与力对通过该点的轴之矩的关系
面积由于 A O BFm O ?2)( ?
[证 ]
''2)()( BOAFmFm xyzz ???
通过 O点作任一轴 Z，则,
''co s BOAO A B ?g? ??
由几何关系,
''2c o s2 BOAO A B ?g? ??所以,
37
Theorem,the projection of the vector of the moment of a force about a
point on an axis through this point is equal to the moment of the force
about the axis,This is the required relationship between the moment of
a force about a point and the moment of the force about an axis
through this point,
)(
)(
c o s,
)(
)(
c o s,
)(
)(
c o s
Fm
Fm
Fm
Fm
Fm
Fm
O
z
O
y
O
x ??? g??
222 ))(())(())(()( FmFmFmFm zyxO ???
kFmjFmiFm zyx )()()( ???
kFmjFmiFmFrFm zOyOxOO )]([)]([)]([)( ?????
Due to
The moment of the force about point O is given by
38
定理, 力对点的矩矢在通过该点的任意轴上的投影等于这力
对于该轴的矩。这就是力对点之矩与对通过该点轴之矩的关系。
)(
)(
c o s,
)(
)(
c o s,
)(
)(
c o s
Fm
Fm
Fm
Fm
Fm
Fm
O
z
O
y
O
x ??? g??
222 ))(())(())(()( FmFmFmFm zyxO ???
kFmjFmiFmFrFm zOyOxOO )]([)]([)]([)( ?????
kFmjFmiFm zyx )()()( ???
又由于
所以力对点 O的矩为,
39
Te method of reduction of a force system in space is the same as
in the planar case,only the planar coordinates have to be extended
to space coordinates,
§ 5–4 Reduction of a force system in space to a given point
nFFFF ?321,,
Suppose there is a force
system in space acting on a
rigid body
Reduce the force
system to an arbitrary
point O,
40
把研究平面一般力系的简化方法拿来研究空间一般力系的
简化问题，但须把平面坐标系扩充为空间坐标系。
§ 5-4 空间一般力系向一点简化
nFFFF ?321,,
设作用在刚体上有
空间一般力系
向 O点简化
（ O点任选）
41
① According to the Theorem of Translation of a Force we
translate every force to the point O,We get a
concurrent force system in space and
an additional force couple system,
[notice] are the moments of the forces about point O,
② Because the force couple in space is a free vector it can be
always translated to the point O,
'',',' 321 nFFFF ?
nmmm ?,,21
nmmm ?,,21
42
① 根据力线平移定理，将各力平行搬到 O点得到一空
间汇交力系,和附加力偶系
[注意 ] 分别是各力对 O点的矩。
② 由于空间力偶是自由矢量，总可汇交于 O点。
'',',' 321 nFFFF ? nmmm ?,,21
nmmm ?,,21
43
③ Composing,we get the principle vector,
In other words,(the principle vector passes
the point O,and it does not depend on the position of the
point O)
Composing,we get the principle moment,
(the principle moment depends on
the position of point O)
'',',' 321 nFFFF ? 'R
?? ?? ii FFR '' 'R
nmmm ?,,21 OM
?? ?? )( iOiO Fmmm OM
44
③ 合成 得主矢
即 （主矢 过简化中心 O,
且与 O点的选择无关）
合成 得主矩
即,（主矩 与简化中心 O有关）
'',',' 321 nFFFF ? 'R
?? ?? ii FFR '' 'R
nmmm ?,,21 OM
?? ?? )( iOiO Fmmm OM
45
If the given center O is the origin of the coordinate system,then
the magnitude of the principle vector is
The direction is
According to the relationship between the moment of the force about
a point and the moment of the force about an axis through the point
The magnitude of the principle moment is
and the direction is,
??? ?????? 222222 )()()('''' ZYXRRRR zyx
'c o s,'c o s,'c o s R
Z
R
Y
R
X ??? ??? g??
????
??
????
??
)(])([ ;)(])([;)(])([
FmFmmFmFmm
mFmFm
zzOOzyyOOy
OxixxiO
222
OzOyOxO MMMM ???
O
Oz
O
Oy
O
Ox
M
M
M
M
M
M ??? 'c o s,'c o s,'c o s g??
46
若取 简化中心 O点为坐标原点，则,
主矢大小
主矢方向
根据力对点之矩与力对轴之矩的关系,
则 主矩大小 为,
主矩方向,
??? ?????? 222222 )()()('''' ZYXRRRR zyx
'c o s,'c o s,'c o s R
Z
R
Y
R
X ??? ??? g??
????
??
????
??
)(])([ ;)(])([;)(])([
FmFmmFmFmm
mFmFm
zzOOzyyOOy
OxixxiO
222
OzOyOxO MMMM ???
O
Oz
O
Oy
O
Ox
M
M
M
M
M
M ??? 'c o s,'c o s,'c o s g??
47
If a force system in space is reduced to a given point,we get a
principle vector and a principle moment,It follows,
§ 5–5 Discussion of the result of
the reduction of a force system in space
1,If the force system is in equilibrium,
0,0' ?? OMR
2,If the force system is equivalent to a force couple,
the moment is equal to the sum of the moments MO of the forces
about the given point,The principle moment does not depend on
the position of the given center,
0,0' ?? OMR
3,If the force system is equivalent to a resultant
force,The principle vector which is equal to the resultant
force vector of the force system,the principle vector passes
the given center O (it depends on the position of the given
center,If the point O is changed,the principle moment is not
zero),
0,0' ?? OMR
'R
R R
48
空间一般力系向一点简化得一主矢和主矩，下面针对主
矢、主矩的不同情况分别加以讨论。
§ 5-5 空间一般力系简化结果的讨论
1,若,则该力系 平衡 （下节专门讨论）。
0,0' ?? OMR
2、若 则力系可合成一个 合力偶,其矩等于原力
系对于简化中心的主矩 MO。 此时主矩与简化中心的位置无关。
0,0' ?? OMR
3,若 则力系可合成为一个 合力,主矢 等
于原力系合力矢,合力 通过简化中心 O点。 （此时
与简化中心有关，换个简化中心，主矩不为零）
0,0' ?? OMR 'R
R R
49
Because
?????? iOOO FRR
M
R
M
ddRM is f o r c er e s u l t a n t t h ea n d
'
g e t w e
4.If,we can discuss from ①
two kinds of situations separately,②
OMR ?'
OMR //'
0,0' ?? OMR
① if
OMR ?'
)( dRM O ??
Turn MO into(R'',R),where R' counteracts R'',so we get R,
50
4,若 此时分两种情况讨论。即,①
②
OMR ?'
OMR //'
0,0' ?? OMR
由于做
??????? iOOO FRR
M
R
M
ddRM 合力,',
① 若 时
OMR ?'
)( dRM O ??
可进一步简化，将 MO变成 ( R'',R)使 R'与 R''抵消只剩下 R。
51
② If,——it is the condition of force screw,
[Example]① Twist the screw ② Shell helix
OMR //'
③ R' is neither parallel with nor perpendicular to M0,there is
an arbitrary angle ? between them,
In this case <1> decompose MO into M// and M?,
<2> deal with M// and M? by ① and ②,respectively,
' '
52
② 若 时,——为力螺旋的情形 （新概念，又移动又转动）
[例 ] ①拧螺丝 ②炮弹出膛时炮弹螺线
OMR //'
③ R′ 不平行也不垂直 M0,最一般的成任意角 ?
在此种情况下,<1>首先把 MO 分解为 M//和 M?
<2>将 M//和 M? 分别按①、②处理。
' '
53
M? make the principle vector
R' translate,the distance is,'
s in
'' R
M
R
MOO O ??? ?
Therefore,there is a force screw
about point O',
Because M// is a free vector it
can be translated to the point O',
M//a is invariant,?
54
M? 使主矢 R'搬家，搬家的矩离,
'
s in
'' R
M
R
MOO O ??? ?
所以在 O'点处形成一个力螺旋 。
因为 M// 是自由矢量,
可将 M//搬到 O'处
M//不变,
?
55
[notice]
The invariant quantities after reduction are R'and M// (they do not
depend on the given center),
If the given center is the point O we get M?,But if the given center
is the point O'it is changed to M?',M// is an invariant quantity (it
does not depend on the given center and the force couple of the
force system)
?
56
[注意 ]
力系简化中的不变量（不随简化中心改变）有,R′, M//
简化中心为 O时：为 M? 当简化中心为 O′ 时，为 M?′
但 M//总是不变的（ 它是
原力系中的力偶与简化
中心无关 ）
?
57
)('' RmROOMM OO ???? ?
?? )(
ism o m e n t p r i n c i p l e t h eB e c a u s e
iOO FmM
??? )()( iOO FmRM
f o r m, in t h e u s e dn o r m a lly is
)()( ?? izz FmRM
After the reduction of the force system in space to the point O,we
get a principle vector R‘ and a principle moment MO, If MO?R' we
can get a new resultant R which acts on the new given center O',
The law of the resultant moment of the force system in space,
58
)('' RmROOMM OO ???? ?
?? )( iOO FmM主矩又 ?
??? )()( iOO FmRM
?? )()( izz FmRM常用投影式
空间力系向 O点简化后得主矢 R'和主矩 MO,若 MO?R'，可
进一步合成为一个 作用在新简化中心 O'点的 合力 R 。
空间力系的合力矩定理,
59
1,The necessary and sufficient conditions for equilibrium of a
force system in space are
0)( ?? ? iOO FmM00' ??? ? FR
??? ??? 222 )()()(':b e c a u s ea g a i n ZYXR
222 ))(())(())(( ??? ??? FmFmFmM zyxO
So the equilibrium equations of a force system in space are
They can also be expressed by 4
moment equations,5 moment
equations or 6 moment equations,
but each case has some limiting conditions,
??
??
??
??
??
??
0)(,0
0)(,0
0)(,0
FmZ
FmY
FmX
z
y
x
§ 5–6 Equilibrium equations of
a force system in space and their applications
60
一、空间任意力系的平衡充要条件是,
0)( ?? ? iOO FmM00' ??? ? FR
??? ??? 222 )()()(' ZYXR?又
222 ))(())(())(( ??? ??? FmFmFmM zyxO
所以 空间任意力系的平衡方程 为,
还有四矩式，五矩式和六矩式,
同时各有一定限制条件。
??
??
??
??
??
??
0)(,0
0)(,0
0)(,0
FmZ
FmY
FmX
z
y
x
§ 5-6 空间一般力系的平衡方程及应用
61
The equilibrium equations of a concurrent force system in space are
Because all action lines of the forces are
concurrent at a point and all axes are
through this point the moment equations are
identities,0
0
0
?
?
?
?
?
?
Z
Y
X
The equilibrium equations of a parallel force system in space,
supposing all the action lines are parallel to the axis z are
they are all identities,?
?
?
?
?
?
0)(
0)(
0
Fm
Fm
Z
y
x ?
?
?
?
?
?
0
0
0)(
Y
X
Fm z
62
空间汇交力系的平衡方程为,
因为各力线都汇交于一点，各轴都通过
该点，故各力矩方程都成为了恒等式。
0
0
0
?
?
?
?
?
?
Z
Y
X
空间平行力系的平衡方程，设各力线都 // z 轴。
因为
均成为了恒等式。 ?
?
?
?
?
?
0)(
0)(
0
Fm
Fm
Z
y
x ?
?
?
?
?
?
0
0
0)(
Y
X
Fm z
63
1,Spherical gemel
2,Constraints in space
There are,in general,six possible motions of a body in space
(translations along three axes and rotations about three axes),
How many kinds of motion are hindered by the constraints in the
following cases,If there are constraints there are reaction forces of
the constraints,too,The impediment of a translation is a reaction
force,the impediment of a rotation is a reaction force couple,
[Examples]
64
1、球形铰链
二、空间约束
观察物体在空间的六种（沿三轴移动和绕三轴转动）可能
的运动中，有哪几种运动被约束所阻碍，有阻碍就有约束反力。
阻碍移动为反力，阻碍转动为反力偶。 [例 ]
65
Spherical gemel
66
球形铰链
67
2,Radial bearing,hinges,ball bearing
68
2、向心轴承，蝶铰链，滚珠 (柱 )轴承
69
3,Sliding bearing
70
3、滑动轴承
71
4,Thrust bearing
72
4、止推轴承
73
5,Splint with a pin
74
5、带有销子的夹板
75
6,Fixed support in space
76
6、空间固定端
77
If a system of parallel forces in
space has a resultant force and the
point of its application is the point
C,point C is called the center of
the system of parallel forces,The
center of gravity of an object is a
special example of the center of a
system of parallel forces,
§ 5–7 The center of a system of parallel
forces and the center of gravity of an object
1,The center of a system of parallel forces and the center of gravity
of object
(1) The center of a system of parallel forces,
From the law of the resultant moment we get,
?? )()( iOO FmRm nnC FrFrFrRr ???????? ?2211
78
空间平行力系，当它有合
力时，合力的作用点 C 就是此
空间平行力系的中心 。而物体
重心问题可以看成是空间平行
力系中心的一个特例。
§ 5-7 平行力系的中心 物体的重心
一、空间平行力系的中心、物体的重心
1、平行力系的中心
由合力矩定理,
?? )()( iOO FmRm nnC FrFrFrRr ???????? ?2211
79
0110,L e t PFFPRR ??
nnC rFrFrFrR ????? ?2211
?
???????
i
iinn
C F
rF
R
rFrFrFr ?2211
R
zF
z
R
yF
y
R
xF
x iiCiiCiiC ??? ???,,a r e e q u a t i o n s p r o j e c t i o n T h e
80
0110,PFFPRR ??令
nnC rFrFrFrR ????? ?2211
?
???????
i
iinn
C F
rF
R
rFrFrFr ?2211
R
zFz
R
yFy
R
xFx ii
C
ii
C
ii
C
??? ???,,:投影式
81
If we treat the gravity on a body
as a system of parallel forces the
problem of determination of the
center of gravity is just the
problem of the determination of
the center of a system of parallel
forces,From the law of the
resultant moment we have
iiC xPxP ??? ?
P
Pz
z
P
Py
y
P
Px
x
i
C
i
C
i
C
?
?
?
?
?
?
?
?
?
,
,
If the body is divided into smaller and smaller
parts the determination of the center of gravity
becomes more accurate,In the limiting case,
(n- ),we used to use an integration to
determine the position of the center of gravity,
?
2,The formula for the center of gravity,
82
如果把物体的重力都看成为
平行力系，则求重心问题就
是求平行力系的中心问题。
由合力矩定理,
iiC xPxP ??? ?
P
Pz
z
P
Py
y
P
Px
x
i
C
i
C
i
C
?
?
?
?
?
?
?
?
?
,
,
物体分割的越多，每一小部分体积越小，
求得的重心位置就越准确。在极限情况下，
(n? ),常用积分法求物体的重心位置。 ?
二、重心坐标公式,
83
?? V dVP g
If gI denote the weight of part No,I and ⊿ Vi is the volume of it,
then, Substitution into the equations above gives in the
limiting case,
Where,this is the accurate formula to determine the center of
gravity,
P
dVz
zP
dVy
yP
dVx
x VCVCVC ??? ???
ggg
,,
iii VP ?g? ?
For a homogeneous body,g =constant,the formula is simplified to,
V
dVz
zV
dVy
yV
dVx
x VCVCVC ???
?
?
?
?
?
?,,
Using the same arguments,we can fine out the formulae for thin plane
and a thin-long rod,
84
设 gi表示第 i个小部分每单位体积的重量, ⊿ Vi第 i个小体积，则
代入上式并取极限，可得,
式中,上式为 重心 C 坐标的精确公式 。
P
dVz
zP
dVy
yP
dVx
x VCVCVC ??? ???
ggg
,,
iii VP ?g? ?
?? V dVP g
对于均质物体,g =恒量，上式成为,
V
dVz
zV
dVy
yV
dVx
x VCVCVC ???
?
?
?
?
?
?,,
同理对于薄平面和细长杆均可写出相应的公式。
85
According to the fact that the center
of the system of parallel forces does
not depend on the direction of the
forces,the action line can be chosen
to be parallel to the axis y,Using the
law of the resultant moment and
determining the moment about the
axis x we find
P
zPzzPPz ii
CiiC
?? ??? ??,
In sum,the equations of the center of gravity are,
P
zP
z
P
yP
y
P
xP
x iiCiiCiiC ???
?
?
?
?
?
?,,
Substituting △ Pi= △ mig and P=Mg into the equations above we
get the equations of center of mass,
M
zmz
M
ymy
M
xmx ii
C
ii
C
ii
C
??? ??????,,
86
根据平行力系中心位置与各平
行力系的方向无关的性质，将力
线转成与 y轴平行，再应用合力矩
定理对 x 轴取矩得,
P
zPzzPPz ii
CiiC
?? ??? ??,
综合上述得 重心坐标公式 为,
P
zPz
P
yPy
P
xPx ii
C
ii
C
ii
C
??? ??????,,
若以△ Pi= △ mig,P=Mg 代入上式可得质心公式
M
zmz
M
ymy
M
xmx ii
C
ii
C
ii
C
??? ??????,,
87
By the same arguments,we can find the equations for a
homogeneous body,a homogeneous board and a homogeneous rod,
They are
V
zV
z
V
yV
y
V
xV
x iiCiiCiiC ???
?
?
?
?
?
?,,:b o d y
A
zA
z
A
yA
y
A
xA
x iiCiiCiiC ???
?
?
?
?
?
?,,:p l a t e
l
zl
z
l
yl
y
l
xl
x iiCiiCiiC ???
?
?
?
?
?
?,,:r o d
88
同理：可写出均质体，均质板，均质杆的形心（几何中心）
坐标分别为,
V
zVz
V
yVy
V
xVx ii
C
ii
C
ii
C
??? ??? ???,,:立体
A
zAz
A
yAy
A
xAx ii
C
ii
C
ii
C
??? ??? ???,,:平板
l
zlz
l
yly
l
xlx ii
C
ii
C
ii
C
??? ??? ???,,:细杆
89
Solution,Due to symmetry the center of this arc must be on the
axis Ox,hence yC=0,Select a small part,
?dRdL ??
? c o s ?? Rx
R
dR
L
dLx
x
L
C
?
??
?
?
2
c o s
2
?
?
?
???
?
?
??
?
?s inRx
C ?
The following is an example of the determination of the center of
gravity of a body by integration,
[Example] Determine the center of gravity of the homogeneous arc
with the radius R and the top angle 2?,
O
90
解,由于对称关系，该圆弧重心必在 Ox轴，即 yC=0。取微段
?dRdL ??
? c o s ?? Rx
R
dR
L
dLx
x
L
C
?
??
?
?
2
c o s
2
?
?
?
???
?
?
??
?
?s inRx
C ?
下面用积分法 求物体的重心实例,
[例 ] 求半径为 R，顶角为 2? 的均质圆弧的重心。
O
91
cm4.6
21
2211
?
?
?
?
?
?
?
SS
ySyS
A
yA
y
ii
C
3,The ways to determine the center of gravity of an object,
① Method of combination,
Solution,
cm248 cm4 21,8 0 cm 212 221 )πR(y,y,π RSS ?????
Determine the center of gravity
of this composed unit,
We know
92
三、重心的求法, ① 组合法
cm4.6
21
2211
?
?
?
?
?
?
?
SS
ySyS
A
yA
y
ii
C
由
解,
cm248 cm4 21,8 0 cm 212 221 )πR(y,y,π RSS ?????
求：该组合体的重心？
已知,
93
0)( F r o m ?? Fm B 01 ????? CW xPlP P
lPx W
C
1???
The area and formulae of the center of gravity of some simple
graphs can be found in the table,
② Experimental method,
<1>hanging method <2> weighting method
Horizontal
line
94
0)( ?? Fm B由 01 ????? CxPlP 称 P
lPx
C
1??? 称
简单图形的面积及重心坐标公式可由表中查出。
② 实验法,
<1>悬挂法 <2>称重法
95
[Example] We know RC=100mm,RD=50mm,Px=466N,Py=352N
and Pz=1400N,When the system is in equilibrium (rotation with
uniform speed) determine the force Q (acting at the nadir point of
roller C) and the reaction forces of bearing A and B,
Solution,① select the object under study ② draw the force diagram
③ select a coordinate system and write out the equations,
If there is only one unknown quantity in every
equation,it is easy to get the solution,
96
[例 ] 已知, RC=100mm,RD=50mm,Px=466N,Py=352N,Pz=1400N
求：平衡时 (匀速转动 )力 Q=？ (Q力作用在 C轮的最低点）
和轴承 A,B的约束反力？
解,①选研究对象 ②作受力图 ③选坐标列方程
最好使每一个方程有一个未知数，方便求解。
97
)N(746,010050;0
)N(352,0;0 F r o m
????????
?????
?
?
QQPm
PYPYY
xzy
yAyA ?
98
)N(746,010050;0
)N(352,0;0
????????
?????
?
?
QQPm
PYPYY
xzy
yAyA ?由
99
)N(385,020s i n ;0
)N(2040,020s i n 50300200 ;0
)N(729,020c o s ;0
)N(437,020c o s 5020050300 ;0
0
0
0
0
???????
???????
???????
???????
?
?
?
?
AzBA
BzBx
AxBA
BByxz
ZQPZZZ
ZQPZm
XQPXXX
XQXPPm
A
A
100
)N(385,020s i n ;0
)N(2040,020s i n 50300200 ;0
)N(729,020c o s ;0
)N(437,020c o s 5020050300 ;0
0
0
0
0
???????
???????
???????
???????
?
?
?
?
AzBA
BzBx
AxBA
BByxz
ZQPZZZ
ZQPZm
XQPXXX
XQXPPm
A
A
101
Method 2,Project the force system in space on the three axes,
change the problem into that of a force system in a plane,and
solve it,Please,practice it by yourselves after the class,
102
方法 (二 ),将空间力系投影到三个坐标平面内，转化为平面力
系平衡问题来求解，请同学们课后自己练习求解。
103
1,Concepts and content,
1)The force couple or the moment of a force about a point is a
vector,
2)The moment of force about an axis,a force couple in plane and
the moment of a force in plane about a point are algebraic quantities,
3)The law of projection of the resultant force of a force system in
space is
4)The law of the resultant moment of the force system in space is
5)The relationship between the moment of the force about a point
and the moment of a force about an axis which passes the point is
)(,FmM o
?????? iziyix ZRYRXR,,
)()( ioo FmRm ??
)()(:e q u a t i o n p r o j e c t i o n izz FmRm ??
? ? )()( FmFm zZo ?
Exercises to chapter 5 the force system in space
104
一、概念及内容,
1、空间力偶及空间力对点之矩是矢量,
2、空间力对轴之矩和平面力偶、平面力对点之矩是代数量。
3、空间力系 合力投影定理,
4、空间力系的 合力矩定理,
5,空间力对点之矩与对轴之矩的关系,
)(,FmM o
?????? iziyix ZRYRXR,,
)()( ioo FmRm ??
)()(,izz FmRm ??投影式
? ? )()( FmFm zZo ?
第五章, 空间力系, 习题课
105
2,Basic Equations
1) The equilibrium equations of a force system in space are
?
?
?
?
?
?
0
0
0
Z
Y
X
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
z
y
x
m
m
m
Z
Y
X
? ?
? ?
? ?
0
0
0
z
y
x
m
m
m
? ?
? ?
? ?
0
0
0
z
y
m
m
X
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
z
y
x
m
m
m
Y
X
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
'x
z
y
x
m
m
m
m
Y
X
The additional conditions of the
equations of four moments,the
equations of five moments and
the equations of six moments
make the equations independent,
A general
force
system in
space
Concur-
-rent
force
system
in space
System
of force
couples
in space
Force
system
in space
on x axis
only
Force
system
in space
on xoy
plane
only
106
二、基本方程
1,空间力系的平衡方程
空
间
一
般
力
系
空
间
汇
交
力
系
?
?
?
?
?
?
0
0
0
Z
Y
X
空
间
力
偶
系
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
z
y
x
m
m
m
Z
Y
X
? ?
? ?
? ?
0
0
0
z
y
x
m
m
m
空
间∥
x
轴
力
系
? ?
? ?
? ?
0
0
0
z
y
m
m
X
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
z
y
x
m
m
m
Y
X
空
间∥
xoy
面
的
力
系
? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
'x
z
y
x
m
m
m
m
Y
X
四矩式, 五矩式和六矩式的附加条件均
为使方程式独立。
107
3,Steps,skills and remarkable problems in solution,
1) Steps,① Select the object to study (as in the case of a plane),
② Draw the force diagram,
③ Select a coordinate system,write down the equations,
④ Solve the equations,determine the unknown quantities,
2) Some problems about the force systems in space,
① x,y,z may not be the three axes of the moment,We can select
six axes arbitrarily,
② The equations for the moments can be less than three (because
MO is a vector),
③ There are six independent equations for the force system in space
(because the body in space has six degree of freedom),there are
only three degree of freedom in a plane,
④ The problem of friction is involved,in general,
108
三、解题步骤、技巧与注意问题,
1,解题步骤, ①选研究对象
(与平面的相同 ) ② 画受力图
③选坐标、列方程
④解方程、求出未知数
2、空间力系的几个问题,
① x,y,z (三个取矩轴和三个投影轴可以不重合 )可以任选的
六个轴。
② 取矩方程不能少于三个（ ∵ MO是矢量）
③ 空间力系独立方程六个（ ∵ 空间物体六个自由度）
平面三个自由度
④ 空间力系中也包括摩擦问题。
109
2) Skills,
① It is always convenient to replace the projection axis by the axis
of the moment,
② It’s better,if the coordinate axis ? unknown quantities or if
the axis of the moment is parallel to the unknown quantities
or intersects with the unknown quantities,
③ It is normal to study a system from the whole to the parts,
④ The friction force is F = N f,its direction is opposite to the
tendency of motion,
3) Remarkable problems,
① A force couple does not appear in the projection axes equations,
② A force couple in space is a vector,the force couple in a plane is
an algebraic quantity,
③ The usual way to determine the center of gravity of a body is the
method of combination,The center of gravity,the center of mass
and the centroid of area are the same in the case of a
homogeneous body,
110
2,解题技巧,
①用取矩轴代替投影轴，解题常常方便
②投影轴尽量选在与未知力 ?，力矩轴选在与未知力平行或相交
③一般从整体 —>局部的研究方法。
④摩擦力 F = N f,方向与运动趋势方向相反。
3,注意问题,
① 力偶在投影轴中不出现（即在投影方程中不出现）
② 空间力偶是矢量，平面力偶是代数量。
③ 求物体重心问题常用组合法。
对于均质物体，重心、中心、形心为同一点。
111
[Example 1] Let P=2000N,point C is in the plane Oxy,
Determine the moment of force P about the three axes,
?????
?????
???
???
60c o s45c o s
60s i n45c o s
45c o s
45s i n
PP
PP
PP
PP
y
x
xy
z
Solution,① Select a body to investigate; ② Draw the force diagram,
③ Select a coordinate system and write down the equations,
112
[例 1] 已知,P=2000N,C点在 Oxy平面内
求：力 P对三个坐标轴的矩
?????
?????
???
???
60c o s45c o s
60s i n45c o s
45c o s
45s i n
PP
PP
PP
PP
y
x
xy
z
解,①选研究对象；②画受力图；③选坐标列方程。
113
)mN(2.3860c o s45c o s560s i n45c o s6
0)5(6)()()()(
????????
?????????
PP
PPPmPmPmPm yxzzyzxzz
)mN(8.8445s i n6
600)()()()(
????
??????
P
PPmPmPmPm zzxyxxxx
)mN(7.7045s i n5
500
)()()(
)(
????
???
???
P
P
PmPmPm
Pm
z
zyyyxy
y
114
)mN(2.3860c o s45c o s560s i n45c o s6
0)5(6)()()()(
????????
?????????
PP
PPPmPmPmPm yxzzyzxzz
)mN(8.8445s i n6
600)()()()(
????
??????
P
PPmPmPmPm zzxyxxxx
)mN(7.7045s i n5
500
)()()(
)(
????
???
???
P
P
PmPmPm
Pm
z
zyyyxy
y
115
[Example 2]Let AB=3m,AE=AF=4m and Q=20kN,Determine the
tensile forces of rope BE,BF and the internal force of rod AB,
)kN(5 4 6
,045s i n15s i n',0
1
1
??
??????
T
QTY
At point C,
Solution,Study point C and point B
respectively,draw the force diagram,
116
[例 2] 已知,AB=3m,AE=AF=4m,Q=20kN;
求,绳 BE,BF的拉力和杆 AB的内力
)kN(5 4 6
,045s i n15s i n',0
1
1
??
??????
T
QTY
由 C点,
解：分别研究 C点和 B点作受力图
117
)kN( 230
,)kN( 419
5
3
s i n,
5
4
43
4
c o s
0 s i n s i n60c o s,0
045c o s c o s45c o s c o s60s i n,0
045c o s c o s45c o s c o s,0
2
32
22
3212
321
32
?
???
??
?
?
??????
???????
??????
?
?
?
N
TT
TTTNZ
TTTY
TTX
??
??
??
??
At point B,
118
)kN( 230
,)kN( 419
5
3
s i n,
5
4
43
4
c o s
0 s i n s i n60c o s,0
045c o s c o s45c o s c o s60s i n,0
045c o s c o s45c o s c o s,0
2
32
22
3212
321
32
?
???
??
?
?
??????
???????
??????
?
?
?
N
TT
TTTNZ
TTTY
TTX
??
??
??
??
由 B点,
119
This problem tells us,
① The force couple does not
appear in the projection equations,
② If the force couple appear in the
moment equations we treat it as a
vector,just like the projection of a
force in the projection equations,
③ Try to make every equation
dependent only on one unknown
quantity,
④ Be aware of the reaction force
of the constraints in space,
[Example 3] Curved lever ABCD,∠ ABC=∠ BCD=900,AB=a,
BC=b,CD=c,m2,m3,Determine the reaction force of the
constraints and m1?
120
此题训练,
① 力偶不出现在投影式中
② 力偶在力矩方程中出现
是把力偶当成矢量后，类
似力在投影式中投影
③ 力争一个方程求一个支
反力
④ 了解空间支座反力
[例 3] 曲杆 ABCD,∠ ABC=∠ BCD=900,AB=a,BC=b,
CD=c,m2,m3 求：支座反力及 m1=?
121
Solution,
0,0
,0,0
,0,0
,0,0
,0,0
0,0
1
2
3
3
3
2
2
1
?????
???????
???????
??????
???????
??
?
?
?
?
?
?
DDx
ADDA
ADDA
AADz
AADy
D
YcbZmm
a
m
ZZZZZ
a
m
YYYYY
a
m
YaYmm
a
m
ZaZmm
XX
32
32
1 )()( ma
cm
a
b
a
mc
a
mbcYbZm
DD ????
?
??
? ????????
122
解,
0,0
,0,0
,0,0
,0,0
,0,0
0,0
1
2
3
3
3
2
2
1
?????
???????
???????
??????
???????
??
?
?
?
?
?
?
DDx
ADDA
ADDA
AADz
AADy
D
YcbZmm
a
m
ZZZZZ
a
m
YYYYY
a
m
YaYmm
a
m
ZaZmm
XX
32
32
1 )()( ma
cm
a
b
a
mc
a
mbcYbZm
DD ????
?
??
? ????????
123
[Example 4] We know the
weight of the rod AB is 80N,
AD and CB are strings and
there are two point A and C on
the same vertical line,The
bottom of A and B are smooth,
∠ ABC=∠ BCE=600,
Moreover AD is horizontal,
AC is a plumb line,Determine
TA,TB and the reaction forces
at point A and B,
Solution,Select the axis of
projection and the axis of
moment skillfully,make every
equation dependent on one
unknown quantity,
124
[例 4] 已知,AB杆,AD,CB为
绳,A,C在同一垂线上,AB
重 80N,A,B光滑接触，
∠ ABC=∠ BCE=600,且 AD水
平,AC铅直。求平衡时,TA，
TB及支座 A,B的反力。
解,思路：要巧选投影轴和取
矩轴，使一个方程解出一个未
知数。
125
0N8,0 F r o m ????? PNZ B
0
2
160c o s
,0'
????????
??
CEPACT
m
B
DD
N)( 1.2380
6
3
3
3
2
60c t g
2
60c o s60c t g
2
1
60c o s
????????
??????????
PP
T
ACPACT
B
B
CEAC ????? 60co s60ct gA g ai n ?
126
0N8,0 ????? PNZ B由
0
2
160c o s
,0'
????????
??
CEPACT
m
B
DD
N)( 1.2380
6
3
3
3
2
60c t g
2
60c o s60c t g
2
1
60c o s
????????
??????????
PP
T
ACPACT
B
B
CEAC ????? 60c o s60c t g?又
127
)N( 5.11
2
1
80
6
3
60c o s
060c o s,0
???????
???????
BA
BA
TT
TTX
)N( 20
2
3
80
6
3
060s i n,0
?????
??????
A
BA
N
TNY
128
)N( 5.11
2
1
80
6
3
60c o s
060c o s,0
???????
???????
BA
BA
TT
TTX
)N( 20
2
3
80
6
3
060s i n,0
?????
??????
A
BA
N
TNY
129
130
