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

(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。

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

§ 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

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

1.力在空间的表示,

?
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）

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）

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,

zyx FFF,,
zyx FFFF ???
222 ZYXF ????
F
Z
F
Y
F
X ??? g?? c o s,c o s,c o s
kZFjYFiXF zyx ???,,

kZjYiXF ???

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,
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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

in FFFFFR ??????? ?321
2、解析法,

kZjYiXF iiii ???
kZjYiXR iii ??? ???
? iX
?? ix XR
?? iy YR
?? iz ZR

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
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3、合力投影定理,

??? ?????? 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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? ?0X
? ?0Y
? ?0Z

∴ 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

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,
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§ 5-2 空间力偶系

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,
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[证 ] ① 作 II//Ⅰ, cd // ab
② 作一对平衡力 R,R' (在 E点,且

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

④ 在 I内的力偶（ F1,F1'）等效变成 II内的（ F2,F2' ）

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,
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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

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 ?

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

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

[例 ] 汽车反镜的球铰链
§ 5-3 力对点的矩与力对轴的矩

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

dFFrFrFmFrFm OO ??????? ),s i n ()(,)(
21 ???
kZjYiXF ???由于 kzjyixr ???
ZYX
zyx
kji
FrFm O ???? )(
kFmjFmiFm
kyXxYjxZzXizYyZ
zOyOxO )]([)]([)]([
)()()(
???
??????

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

)()()()( 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

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

)(c o s)( FmFm zO ??? g )()]([ FmFm zzO ?

[证 ]
''2)()( BOAFmFm xyzz ???

''co s BOAO A B ?g? ??

''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

)(
)(
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 )()()( ???

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

§ 5-4 空间一般力系向一点简化
nFFFF ?321,,

（ 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
[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

[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

'',',' 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

??? ?????? 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

§ 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

0,0' ?? OMR
3,若 则力系可合成为一个 合力,主矢 等

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

OMR ?'
OMR //'
0,0' ?? OMR

??????? iOOO FRR
M
R
M
ddRM 合力,',
① 若 时
OMR ?'
)( dRM O ??

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 ?

<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
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 ??? ?

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 ]

?
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

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

0)( ?? ? iOO FmM00' ??? ? FR
??? ??? 222 )()()(' ZYXR?又
222 ))(())(())(( ??? ??? FmFmFmM zyxO

??
??
??
??
??
??
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

0
0
0
?
?
?
?
?
?
Z
Y
X

?
?
?
?
?
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

65
Spherical gemel
66

67
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

§ 5-7 平行力系的中心 物体的重心

1、平行力系的中心

?? )()( 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

iiC xPxP ??? ?
P
Pz
z
P
Py
y
P
Px
x
i
C
i
C
i
C
?
?
?
?
?
?
?
?
?
,
,

(n? ),常用积分法求物体的重心位置。 ?

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

P
dVz
zP
dVy
yP
dVx
x VCVCVC ??? ???
ggg
,,
iii VP ?g? ?
?? V dVP g

V
dVz
zV
dVy
yV
dVx
x VCVCVC ???
?
?
?
?
?
?,,

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
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

P
zPzzPPz ii
CiiC
?? ??? ??,

P
zPz
P
yPy
P
xPx ii
C
ii
C
ii
C
??? ??????,,

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

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

?dRdL ??
? c o s ?? Rx
R
dR
L
dLx
x
L
C
?
??
?
?
2
c o s
2
?
?
?
???
?
?
??
?
?s inRx
C ?

[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

cm4.6
21
2211
?
?
?
?
?
?
?
SS
ySyS
A
yA
y
ii
C

cm248 cm4 21,8 0 cm 212 221 )πR(y,y,π RSS ?????

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

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

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

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

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 ?

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
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

1,空间力系的平衡方程

?
?
?
?
?
?
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

x

? ?
? ?
? ?
0
0
0
z
y
m
m
X
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
z
y
x
m
m
m
Y
X

xoy

? ?
? ?
? ?
? ?
? ?
? ?
0
0
0
0
0
0
'x
z
y
x
m
m
m
m
Y
X

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

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

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）

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。

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

?????
?????
???
???
60c o s45c o s
60s i n45c o s
45c o s
45s i n
PP
PP
PP
PP
y
x
xy
z

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;

)kN(5 4 6
,045s i n15s i n',0
1
1
??
??????
T
QTY

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
??
??
??
??

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

① 力偶不出现在投影式中
② 力偶在力矩方程中出现

③ 力争一个方程求一个支

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
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

0,0
,0,0
,0,0
,0,0
,0,0
0,0
1
2
3
3
3
2
2
1
?????
???????
???????
??????
???????
??
?
?
?
?
?
?
DDx
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,
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

TB及支座 A,B的反力。

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