1
Theoretical Mechanics
2
3
1,The basic concepts and the difference between them,
① The difference between a resultant R and the principal vector R',
Resultant force R is the sum of the translational effect of the
principal vector and the rotational effect of the principal
moment (principal vector and principal moment reduce to
a resultant force),(It depends on the point of application)
Principal vector R' only causes only a translational effect,
Additionally a principal moment may exist,(it does not
depend on the point of application)
② The difference between a force moment and a force couple moment
Force moment is the moment of a force about a point,it depends
on the center point,Thereby,it can not be transferred
arbitrarily in a plane,
Force couple moment can be transfered freely in the plane or in
another plane which is parallel to the plane,It does not
depend on the center point,
Review of Statics
4
一, 基本概念及概念的区分
① 合力 R 和主矢 R' 的区别
合力 R, 是主矢的移动效果和主矩转动效果之和 (主矢,主矩可
简化为合力 )。 (与作用点有关 )
主矢 R',只代表移动效果，且有主矩存在时 (与作用点无关 )
② 力矩和力偶矩的区别,
力矩,是力对那一点的转矩,与取矩点有关,且不能在平面内
任意移动。
力偶矩, 它是在平面内和平行它自身平面内可任意移动,与取矩
点无关。
静力学复习
5
③ The invariants of the reduction of a
general coplanar force system to
a given center,
The invariants of the reduction of a
general force system in space to
a given center are
in a plane,
in space,
'R
// ; ' MR ? ?RRM ????,
④ The way to judge the direction of the force of friction,
Force of Friction is a sort of constraint reaction force,
Its direction is always opposite to the relative motion of the bodies,
'
6
③ 平面任意力系向某点简化的不变量,
空间任意力系向某点简化的不变量
平面中,
空间中,
'R
// ; ' MR ? ?RRM ????,
④ 摩擦力的方向判定
摩擦力是一种约束反力,方向总是与物体相对运动方向 (趋势
方向 )相反
'
7
[Example]
The back wheel has driving moment,The back wheel has no driving moment,
If the back wheel leaves the ground If the back wheel leaves the ground
the direction of motion is backward,the direction of motion is forward,
A running car A braking car
8
[例 ]
后轮有主动力矩，后轮离地 后轮无主动力，后轮离地
运动方向是向后的 运动方向是向前的
9
⑤ The way to deal with sign of inequality in problems with friction,
∵ Nf≥F,But in general,we only consider the conditions when the
motion is impending (i.e.,Nf=F ),and then append a sign of
inequality on the result or judge the range of equilibrium,Therefore
we can avoid troubles and miscalculations which would occur in
solving inequations,
(1) Basic theories (in common use)
① Three coplanar equivalent forces must intersect at one point,
(to determine the direction of the unknown forces)
② The law of projection of the resultant force,RX=?X
③ The law of the moment of a resultant force,
projection equation,
?? )()( iOO FmRm
?? )()( izz FmRm
2,Basic equations and theories,
10
⑤ 摩擦问题中对不等号的处理
∵ Nf≥F，但一般的情况下是选临界状态代入 ( 即 Nf=F ) 计
算，得出结果后再加上不等号，或判断出平衡区间，以减少不
等式运算所带来的麻烦和由此出现的误算。
(一 ),基本定理 (常用的 )
① 三力平衡必汇交，必共面 (用于确定未知力的方向 )
② 合力投影定理,RX=?X
③ 合力矩定理,
投影式,
?? )()( iOO FmRm
?? )()( izz FmRm
二、基本方程和基本定理
11
(2) Basic equations
in a plane in space
? ? 0X
? ? 0Y
? ? 0Z
0?? Xm
0?? Ym
0?? Zm
? ? 0X
? ? 0Y
? ? 0Am
3,The steps,the skills and noticeable problems in solving questions
(1) The steps of solution,
① Select a body to be investigated;
② Draw the force diagram (draw all the forces),
③ Select a coordinate system,write down the equations;
④ Solve the equations,determine the unknown quantities,
12
(二 ) 基本方程
平面 空间
? ? 0X
? ? 0Y
? ? 0Z
0?? Xm
0?? Ym
0?? Zm
? ? 0X
? ? 0Y
? ? 0Am
三, 解题步骤，解题技巧，解题注意问题
(一 )解题步骤,①选研究对象
②作受力图（有用，没有用的力均画上 ）
③选坐标列方程
④解方程，求出未知数
13
(3) Noticeable problems,
① A force couple has no projection on a coordinate
axis,
② Generally,the direction of the force of friction can
not be supposed,
③ Notice ≥ or ≤ in the additional equation,
④ All forces in the diagram should be drawn,
(2) Skills of the solution,
① Firstly,find out the double-force equivalent rods,
② Select the coordinate ? unknown quantities,
③ Select the point of intersection of unknown forces
(the line parallel to the unknown forces) as the
center point (axis) of the force moment,
④ The method to deal with problems of systems
and bodies is to start with the known forces,from
the whole --> parts,
14
(三 ),注意问题,①力偶在坐标轴投影不出现
②摩擦力的方向一般不能假设
③附加方程中的 ≥或 ≤号别最后忘记
④受力图中力要画全
(二 ),解题技巧,①先找二力杆
②选坐标轴 ?未知力
③选取矩点（轴）与未知力相交或平行
④从已知力下手，物系问题由整体 -->局部
15
4,Example,
[Example 1]
Draw the force diagram,
16
四、例题,
[例 1] 画受力图
17
Solution,Investigate the whole system,
the force diagram is given below,
( N )14140
02145s i n,0 F r o m
??
????????
FG
FGC
S
QSm
( N )10000
045c o s0
( N )5000
0110 F r o m
??
?????
???
?????
?
?
C
FGC
C
CG
X
S XX
Y
Q Ym
[Example 2] Q=5000N,the weight of the rod can be neglected,
Determine SDE,SFG and the reaction forces at point C,
( N )14140
02145s i n,0 F r o m
??
???????
DE
DEA
S
QSm
18
解,研究整体，受力如图
( N )14140
02145s i n,0
??
????????
FG
FGC
S
QSm由
(N)1 0 0 0 0
045c o s0
(N)5 0 0 0
0110
??
?????
???
?????
?
?
C
FGC
C
CG
X
S XX
Y
Q Ym由
[例 2] 已知, Q=5000N，杆重不计。求 SDE,SFG和 C点的反力。
( N )1 4 1 4 0
02145s i n
0
??
?????
??
DE
DE
A
S
QS
m由
19
Secondly,investigate the whole system,
From
?
?
? c t g
2tg2
0tg2-,0 F r o m QQNaQaNm CCB ?????????
? ? 0X 0?? CA NX
? ?0Y 0)( ??? PQY A
? ? 0Am 0)(tg2 ?????? aQPaNM
CA ?
?c t g2QNX CA ?????
aPQM A )2( ???
[Example 3] AB=2a,the weight of AB is P,the weight of BC is Q,
∠ ABC=?,Determine the reaction forces at points A and C,
Solution,First,investigate the rod BC,its force diagram is below,
)( PQY A ??
20
再研究整体：由 ??? c t g2tg2 0tg2-,0
QQNaQaNm
CCB ?????????由
? ? 0X 0?? CA NX
? ?0Y 0)( ??? PQY A
? ? 0Am 0)(tg2 ?????? aQPaNM
CA ?
?c t g2QNX CA ?????
aPQM A )2( ???
[例 3] 已知,AB=2a,重为 P,BC重为 Q,∠ ABC=?
求,A,C两点的反力。
解,先研究 BC，受力如图
)( PQY A ??
21
According to the judgments of the
internal forces on special elements
we get
S1=P1 S2=0 S3=P2
[Example 4]In the fig below a truss is shown whose weight can be
neglected,Determine the internal forces on all elements,
Statically indeterminate Statically determinate Statically indeterminate
[Example 5] Are the problems below statically determinate or
statically indeterminate,
22
由零杆判断方法,
S1=P1
S2=0
S3=P2
[例 4] 已知桁架，不计各杆自重，求下列指定杆的内力。
静不定 静定 静不定
[例 5] 判断下列静定与静不定问题。
23
Solution,
zOz
yOy
z
y
x
FmFm
FmFm
aFFm
cFFm
Fm
)]([)(
)]([)( A g a i n
m)(N20)(
)mN(5.12)(
0)( F r o m
?
?
????
????
?
?
[Example 6]The bottom O of the right-angled bent lever OABC is a
fixed support,A force F acts on point C as shown in the fig,F=100N,
a=200mm,b=150mm and c=125mm,Determine the moment of the
force F about the fixed point O? (the line of action of force F parallel
to the axis x)
58 6.1
)(
)(
t g
)mN(6.23)]([)]([)( 22
?????
????
??
Fm
Fm
FmFmFm
y
z
zyO
24
解,
zOz
yOy
z
y
x
FmFm
FmFm
aFFm
cFFm
Fm
)]([)(
)]([)(
m)(N20)(
)mN(5.12)(
0)(,
?
?
????
????
?
?又
由
[例 6] 直角曲杆 OABC的 O端为固定端,C端受到力 F的作用,如图。
已知,F=100N,a=200mm,b=150mm,c=125mm
求：力 F对固定端 O点的矩？（力 F平行于 x轴）
58 6.1
)(
)(
t g
)mN(6.23)]([)]([)( 22
?????
????
??
Fm
Fm
FmFmFm
y
z
zyO
25
[Example 7] Two homogeneous rods AB and BC are
attached to each other at point B and to a wall at point A
by means of joins,But BC is attached to the wall at point
C by friction,The coefficient of friction between the
wall and the rod BC is f=0.5,The two rods are equal in
length and weight,Determine the maximum angle ? for
which the system is in equilibrium,
26
[例 7] 匀质杆 AB和 BC在 B端铰接,A端铰接在墙上，
C端则由墙阻挡，墙与 C端接触处的摩擦系数 f=0.5，
试求平衡时最大角度 ?，已知两杆长相等、重量相同。
27
)3( ????? CNfF
????? 1.28 42c t g f i n d w e( 3 ) i n t o t h e mngs u b s t i t u t i ??
( 2 )--- 02c o s22c o s2s i n ??????? ??? lPlFlN C
Solution,from the whole system analysis
( 1 )------- 02co s222s i n2 ?????? ?? lPlN C
? ? 0Am
? ? 0f r o m, a n a l y z eTh e n BmBC
,h a v e w e( 2 ) f r o m,2c t g2o b t a i n w e( 1 ) F r o m PFPN C ?? ?
28
)3( ????? CNfF
??????? 1.28 42ct g )3(,( 2 ),2ct g2,)1( ??? 得代入得由得由 P FPN C
( 2 )--- 02c o s22c o s2s i n ??????? ??? lPlFlN C
解：由整体
( 1 )------- 02co s222s i n2 ?????? ?? lPlN C
? ? 0Am
? ? 0,BmBC 由再分析
29
[Example 8] The weight of the board ABCD is P=50N,
the length of every side is a=30cm,A is a ball pivot,B is
a hinge,AB is horizontal,the board locates on a sharp
point E,the force F=100N acts on point H and CE=ED,
BH=10cm,? =30o,
Determine the reaction forces at point A,B and E,
30
[例 8]已知板 ABCD重 P=50N，每边长 a=30cm,A为球
铰,B为蝶铰,AB边是水平的，板的 E点搁在一尖端
E点上,H点作用 F=100N，且,CE=ED,BH=10cm,?
=30o
求,A,B,E的
反力？
31
Solution,investigate
the whole system
From
)N(3.32,02c o s2s i n,0 ?????????????? BEBy ZaNaPBHFa - Zm ??
,N
a
Pa,Nm
E
Ex
( N )721
0c o s
2
0
??
????? ?
( N )2340c o s2s i n0,Y,BHFaNa,Ym BEBz ?????????????? ??
32
解,研究整体
)N(3.32,02c o s2s i n,0 ?????????????? BEBy ZaNaPBHFa - Zm ??
,N
a
Pa,Nm
E
Ex
( N )721
0c o s
2
0
??
????? ?
由
( N )2340c o s2s i n0,Y,BHFaNa,Ym BEBz ?????????????? ??
33
( N )110c o s0
( N )4230 s i n0
)N(100 100 0
.Z PNZ Z Z
.Y N Y YY
FXFXX
AEBA
AEBA
AA
???????
???????
??????
?
?
?
?
?
34
( N )110c o s0
( N )4230 s i n0
)N(100 100 0
.Z PNZ Z Z
.Y N Y YY
FXFXX
AEBA
AEBA
AA
???????
???????
??????
?
?
?
?
?
35
36
Theoretical Mechanics
2
3
1,The basic concepts and the difference between them,
① The difference between a resultant R and the principal vector R',
Resultant force R is the sum of the translational effect of the
principal vector and the rotational effect of the principal
moment (principal vector and principal moment reduce to
a resultant force),(It depends on the point of application)
Principal vector R' only causes only a translational effect,
Additionally a principal moment may exist,(it does not
depend on the point of application)
② The difference between a force moment and a force couple moment
Force moment is the moment of a force about a point,it depends
on the center point,Thereby,it can not be transferred
arbitrarily in a plane,
Force couple moment can be transfered freely in the plane or in
another plane which is parallel to the plane,It does not
depend on the center point,
Review of Statics
4
一, 基本概念及概念的区分
① 合力 R 和主矢 R' 的区别
合力 R, 是主矢的移动效果和主矩转动效果之和 (主矢,主矩可
简化为合力 )。 (与作用点有关 )
主矢 R',只代表移动效果，且有主矩存在时 (与作用点无关 )
② 力矩和力偶矩的区别,
力矩,是力对那一点的转矩,与取矩点有关,且不能在平面内
任意移动。
力偶矩, 它是在平面内和平行它自身平面内可任意移动,与取矩
点无关。
静力学复习
5
③ The invariants of the reduction of a
general coplanar force system to
a given center,
The invariants of the reduction of a
general force system in space to
a given center are
in a plane,
in space,
'R
// ; ' MR ? ?RRM ????,
④ The way to judge the direction of the force of friction,
Force of Friction is a sort of constraint reaction force,
Its direction is always opposite to the relative motion of the bodies,
'
6
③ 平面任意力系向某点简化的不变量,
空间任意力系向某点简化的不变量
平面中,
空间中,
'R
// ; ' MR ? ?RRM ????,
④ 摩擦力的方向判定
摩擦力是一种约束反力,方向总是与物体相对运动方向 (趋势
方向 )相反
'
7
[Example]
The back wheel has driving moment,The back wheel has no driving moment,
If the back wheel leaves the ground If the back wheel leaves the ground
the direction of motion is backward,the direction of motion is forward,
A running car A braking car
8
[例 ]
后轮有主动力矩，后轮离地 后轮无主动力，后轮离地
运动方向是向后的 运动方向是向前的
9
⑤ The way to deal with sign of inequality in problems with friction,
∵ Nf≥F,But in general,we only consider the conditions when the
motion is impending (i.e.,Nf=F ),and then append a sign of
inequality on the result or judge the range of equilibrium,Therefore
we can avoid troubles and miscalculations which would occur in
solving inequations,
(1) Basic theories (in common use)
① Three coplanar equivalent forces must intersect at one point,
(to determine the direction of the unknown forces)
② The law of projection of the resultant force,RX=?X
③ The law of the moment of a resultant force,
projection equation,
?? )()( iOO FmRm
?? )()( izz FmRm
2,Basic equations and theories,
10
⑤ 摩擦问题中对不等号的处理
∵ Nf≥F，但一般的情况下是选临界状态代入 ( 即 Nf=F ) 计
算，得出结果后再加上不等号，或判断出平衡区间，以减少不
等式运算所带来的麻烦和由此出现的误算。
(一 ),基本定理 (常用的 )
① 三力平衡必汇交，必共面 (用于确定未知力的方向 )
② 合力投影定理,RX=?X
③ 合力矩定理,
投影式,
?? )()( iOO FmRm
?? )()( izz FmRm
二、基本方程和基本定理
11
(2) Basic equations
in a plane in space
? ? 0X
? ? 0Y
? ? 0Z
0?? Xm
0?? Ym
0?? Zm
? ? 0X
? ? 0Y
? ? 0Am
3,The steps,the skills and noticeable problems in solving questions
(1) The steps of solution,
① Select a body to be investigated;
② Draw the force diagram (draw all the forces),
③ Select a coordinate system,write down the equations;
④ Solve the equations,determine the unknown quantities,
12
(二 ) 基本方程
平面 空间
? ? 0X
? ? 0Y
? ? 0Z
0?? Xm
0?? Ym
0?? Zm
? ? 0X
? ? 0Y
? ? 0Am
三, 解题步骤，解题技巧，解题注意问题
(一 )解题步骤,①选研究对象
②作受力图（有用，没有用的力均画上 ）
③选坐标列方程
④解方程，求出未知数
13
(3) Noticeable problems,
① A force couple has no projection on a coordinate
axis,
② Generally,the direction of the force of friction can
not be supposed,
③ Notice ≥ or ≤ in the additional equation,
④ All forces in the diagram should be drawn,
(2) Skills of the solution,
① Firstly,find out the double-force equivalent rods,
② Select the coordinate ? unknown quantities,
③ Select the point of intersection of unknown forces
(the line parallel to the unknown forces) as the
center point (axis) of the force moment,
④ The method to deal with problems of systems
and bodies is to start with the known forces,from
the whole --> parts,
14
(三 ),注意问题,①力偶在坐标轴投影不出现
②摩擦力的方向一般不能假设
③附加方程中的 ≥或 ≤号别最后忘记
④受力图中力要画全
(二 ),解题技巧,①先找二力杆
②选坐标轴 ?未知力
③选取矩点（轴）与未知力相交或平行
④从已知力下手，物系问题由整体 -->局部
15
4,Example,
[Example 1]
Draw the force diagram,
16
四、例题,
[例 1] 画受力图
17
Solution,Investigate the whole system,
the force diagram is given below,
( N )14140
02145s i n,0 F r o m
??
????????
FG
FGC
S
QSm
( N )10000
045c o s0
( N )5000
0110 F r o m
??
?????
???
?????
?
?
C
FGC
C
CG
X
S XX
Y
Q Ym
[Example 2] Q=5000N,the weight of the rod can be neglected,
Determine SDE,SFG and the reaction forces at point C,
( N )14140
02145s i n,0 F r o m
??
???????
DE
DEA
S
QSm
18
解,研究整体，受力如图
( N )14140
02145s i n,0
??
????????
FG
FGC
S
QSm由
(N)1 0 0 0 0
045c o s0
(N)5 0 0 0
0110
??
?????
???
?????
?
?
C
FGC
C
CG
X
S XX
Y
Q Ym由
[例 2] 已知, Q=5000N，杆重不计。求 SDE,SFG和 C点的反力。
( N )1 4 1 4 0
02145s i n
0
??
?????
??
DE
DE
A
S
QS
m由
19
Secondly,investigate the whole system,
From
?
?
? c t g
2tg2
0tg2-,0 F r o m QQNaQaNm CCB ?????????
? ? 0X 0?? CA NX
? ?0Y 0)( ??? PQY A
? ? 0Am 0)(tg2 ?????? aQPaNM
CA ?
?c t g2QNX CA ?????
aPQM A )2( ???
[Example 3] AB=2a,the weight of AB is P,the weight of BC is Q,
∠ ABC=?,Determine the reaction forces at points A and C,
Solution,First,investigate the rod BC,its force diagram is below,
)( PQY A ??
20
再研究整体：由 ??? c t g2tg2 0tg2-,0
QQNaQaNm
CCB ?????????由
? ? 0X 0?? CA NX
? ?0Y 0)( ??? PQY A
? ? 0Am 0)(tg2 ?????? aQPaNM
CA ?
?c t g2QNX CA ?????
aPQM A )2( ???
[例 3] 已知,AB=2a,重为 P,BC重为 Q,∠ ABC=?
求,A,C两点的反力。
解,先研究 BC，受力如图
)( PQY A ??
21
According to the judgments of the
internal forces on special elements
we get
S1=P1 S2=0 S3=P2
[Example 4]In the fig below a truss is shown whose weight can be
neglected,Determine the internal forces on all elements,
Statically indeterminate Statically determinate Statically indeterminate
[Example 5] Are the problems below statically determinate or
statically indeterminate,
22
由零杆判断方法,
S1=P1
S2=0
S3=P2
[例 4] 已知桁架，不计各杆自重，求下列指定杆的内力。
静不定 静定 静不定
[例 5] 判断下列静定与静不定问题。
23
Solution,
zOz
yOy
z
y
x
FmFm
FmFm
aFFm
cFFm
Fm
)]([)(
)]([)( A g a i n
m)(N20)(
)mN(5.12)(
0)( F r o m
?
?
????
????
?
?
[Example 6]The bottom O of the right-angled bent lever OABC is a
fixed support,A force F acts on point C as shown in the fig,F=100N,
a=200mm,b=150mm and c=125mm,Determine the moment of the
force F about the fixed point O? (the line of action of force F parallel
to the axis x)
58 6.1
)(
)(
t g
)mN(6.23)]([)]([)( 22
?????
????
??
Fm
Fm
FmFmFm
y
z
zyO
24
解,
zOz
yOy
z
y
x
FmFm
FmFm
aFFm
cFFm
Fm
)]([)(
)]([)(
m)(N20)(
)mN(5.12)(
0)(,
?
?
????
????
?
?又
由
[例 6] 直角曲杆 OABC的 O端为固定端,C端受到力 F的作用,如图。
已知,F=100N,a=200mm,b=150mm,c=125mm
求：力 F对固定端 O点的矩？（力 F平行于 x轴）
58 6.1
)(
)(
t g
)mN(6.23)]([)]([)( 22
?????
????
??
Fm
Fm
FmFmFm
y
z
zyO
25
[Example 7] Two homogeneous rods AB and BC are
attached to each other at point B and to a wall at point A
by means of joins,But BC is attached to the wall at point
C by friction,The coefficient of friction between the
wall and the rod BC is f=0.5,The two rods are equal in
length and weight,Determine the maximum angle ? for
which the system is in equilibrium,
26
[例 7] 匀质杆 AB和 BC在 B端铰接,A端铰接在墙上，
C端则由墙阻挡，墙与 C端接触处的摩擦系数 f=0.5，
试求平衡时最大角度 ?，已知两杆长相等、重量相同。
27
)3( ????? CNfF
????? 1.28 42c t g f i n d w e( 3 ) i n t o t h e mngs u b s t i t u t i ??
( 2 )--- 02c o s22c o s2s i n ??????? ??? lPlFlN C
Solution,from the whole system analysis
( 1 )------- 02co s222s i n2 ?????? ?? lPlN C
? ? 0Am
? ? 0f r o m, a n a l y z eTh e n BmBC
,h a v e w e( 2 ) f r o m,2c t g2o b t a i n w e( 1 ) F r o m PFPN C ?? ?
28
)3( ????? CNfF
??????? 1.28 42ct g )3(,( 2 ),2ct g2,)1( ??? 得代入得由得由 P FPN C
( 2 )--- 02c o s22c o s2s i n ??????? ??? lPlFlN C
解：由整体
( 1 )------- 02co s222s i n2 ?????? ?? lPlN C
? ? 0Am
? ? 0,BmBC 由再分析
29
[Example 8] The weight of the board ABCD is P=50N,
the length of every side is a=30cm,A is a ball pivot,B is
a hinge,AB is horizontal,the board locates on a sharp
point E,the force F=100N acts on point H and CE=ED,
BH=10cm,? =30o,
Determine the reaction forces at point A,B and E,
30
[例 8]已知板 ABCD重 P=50N，每边长 a=30cm,A为球
铰,B为蝶铰,AB边是水平的，板的 E点搁在一尖端
E点上,H点作用 F=100N，且,CE=ED,BH=10cm,?
=30o
求,A,B,E的
反力？
31
Solution,investigate
the whole system
From
)N(3.32,02c o s2s i n,0 ?????????????? BEBy ZaNaPBHFa - Zm ??
,N
a
Pa,Nm
E
Ex
( N )721
0c o s
2
0
??
????? ?
( N )2340c o s2s i n0,Y,BHFaNa,Ym BEBz ?????????????? ??
32
解,研究整体
)N(3.32,02c o s2s i n,0 ?????????????? BEBy ZaNaPBHFa - Zm ??
,N
a
Pa,Nm
E
Ex
( N )721
0c o s
2
0
??
????? ?
由
( N )2340c o s2s i n0,Y,BHFaNa,Ym BEBz ?????????????? ??
33
( N )110c o s0
( N )4230 s i n0
)N(100 100 0
.Z PNZ Z Z
.Y N Y YY
FXFXX
AEBA
AEBA
AA
???????
???????
??????
?
?
?
?
?
34
( N )110c o s0
( N )4230 s i n0
)N(100 100 0
.Z PNZ Z Z
.Y N Y YY
FXFXX
AEBA
AEBA
AA
???????
???????
??????
?
?
?
?
?
35
36
