1
Theoretical Mechanics
2
3
Chapter 4,Friction
§ 4–1 Introduction
§ 4–2 Sliding friction
§ 4–3 Equilibrium problems with friction
§ 4–4 Rolling friction
Class of exercises
4

§ 4–1 引言
§ 4–2 滑动摩擦
§ 4–3 考虑摩擦时的平衡问题
§ 4–4 滚动摩擦

5
Until now,in solving problems of statics,we regarded the
surfaces of contact as absolutely smooth neglecting friction
between bodies,In fact surfaces without friction do not exist,
friction exists,in general,
[Example]
Chapter 4 Friction
§ 4-1 Introduction
The system is in equilibrium only
if friction is taken into account,
?
6

[例 ]

§ 4-1 引言

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1,why shall we study friction?
2,The way to study friction,to master its laws,
to use the advantages of friction and to
overcome its shortages,
3,According to the motion of the surfaces of contact
friction can be classified into Sliding Friction
Rolling Friction ?
8

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1) Concept,When two bodies can slide on each other a force of
friction is developed at the surface of contact which prevents the
motion of the bodies,It is called static force of friction,(it is equal
to the tangential constraint reaction force with which the surface of
contact acts on the bodies)
§ 4-2 Sliding Friction
1,Static force of friction
(Look for the animation on the next page)
The ways to increase the force of friction,
① increase of the force N,
② increase of the coefficient of friction f,
2) Status,① static,
② critical,(It is going to start to slide)
③ sliding,
PF ? ) v al u eu n f i x edan is w i t h g r o w i n g ( PF
NfF ??m a x
NfF ?? ''
(f — the coefficient of static friction)
(f '— the coefficient of kinetic friction)
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1,定义,相接触物体，产生相对滑动（趋势）时，其接触面

（ 就是接触面对物体作用的切向约束反力）
2,状态, ①静止,
②临界：（将滑未滑）
③滑动,
PF ? )( 不固定值??? FP
NfF ??m a x
NfF ?? ''
§ 4-2 滑动摩擦

（翻页请看动画）

② 加大摩擦系数 f
（ f — 静滑动摩擦系数）
（ f '— 动摩擦系数）
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12
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m a x0 FF ??
0?? X
NfF ??m a x
3)Characteristics,Magnitude,(range of equilibrium)
and should be satisfied,
Characteristics Direction,oppose to the relative motion of the
of the static bodies,
force of friction,Law,(f depends only on the materials
and the surface conditions,it does not depend on
the area of the surface of contact),
2,Slipping force of friction,(the difference with static one is that
motion has occurred)
Magnitude,(without equilibrium range)
Characteristics
of the slipping Direction,opposite to the motion,
force of friction,
Law,(f‘ only depends on the materials
and the surface conditions,it does not depend on
the area of the surface of contact),
NfF ?? ''
NfF ?? ''
14

m a x0 FF ?? 0?? X
NfF ??m a x
NfF ?? ''
NfF ?? ''
3,特征,大小,（平衡范围）满足

15
fN NfNFm ???? m a xtg ?
② Calculation,
m?
3,Angle of friction,
① Concept,If the magnitude of the static force of friction increases
from zero to the maximum value,the maximum angle which
the total reaction of a rough support makes with the normal to the
surface is called the angle of friction,
maxF
(Look for the animation on the next page)
16
maxF
m?

① 定义：当摩擦力达到最大值 时其全反力

fN NfNFm ???? m a xtg ?
② 计算,
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4,Self-locking
① Concept,A body can be at rest by the mutual action of the
force of friction and the normal pressure(I.e.,Total
Reaction Force),it will not slide (no matter how
great an applied force is),this phenomenon is
called self-locking,
When the system is always in
equilibrium (I.e.,it is self-locking),
m?? ?
m?? ?
② Conditions of self-locking,
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① 定义：当物体依靠接触面间的相互作用的摩擦 力 与正

（无论外力多大），这种现象称为自锁。

m?? ?
m?? ?
② 自锁条件,
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Measurement of the coefficient of static friction,Rotate OA
about the axis O, When the slipping of the block is impending,
measure the angle ?,f=tg?,(This is the coefficient of static
friction between these two kind of materials)
fN NfNFm ???? m a xtg ?
③ Example of application of self-locking
(Look for the animation
on the next page)
22

?角,f=tg?,(该两种材料间静 摩 擦系数 )
fN NfNFm ???? m a xtg ?
（翻页请看动画）
③ 自锁应用举例
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§ 4-3 Equilibrium problems with friction
The examination of the equilibrium conditions of a body taking
friction into account is usually limited to the consideration of the
conditions under which motion is impending and we can use the
additional equation, The others steps are same as in
the case of range the general coplanar force system,just consider an
equilibrium as a range,(it will be introduced in the examples.)
NfF ??m a x
[Example 1] Let ? =30o,G =100N,f =0.2,① Determine the
range of the horizontal force Q for the body to be at rest,② Is the
body in equilibrium if Q = 60N,
(Look for the animation
on the next page)
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§ 4-3 考虑滑动摩擦时的平衡问题

NfF ??m a x
（从例子说明）。
[例 1] 已知,? =30o,G =100N,f =0.2 求：①物体静止时,

（翻页请看动画）
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Solution,① Firstly,we shall determine the maximum force Qmax
under the action of which the body can not slip up to the inclined
plane,see fig.(1),
NfF
GQNY
FGQX
??
????
????
?
?
m a x
m a x
m a xm a x
,e q u a t i o n A d d i t i o n a l
0c o ss i n,0
0s i nc o s,0 F r o m
??
??
?
?
tg1
tg,f i n d We
m a x f
fGQ
?
??
??
??
tgtg1
tgtg
m
m
?
?
? G
)(tg m?? ??? G
??
????
t gtg1
tgtg)(tg
m
m
m ?
???
By using the trigonometric
function theorem
32

maxQ
NfF
GQNY
FGQX
??
????
????
?
?
m a x
m a x
m a xm a x
,
0c o ss i n,0
0s i nc o s,0

??
??
?
?
tg1
tg,
m a x f
fGQ
?
??解得
??
??
tgtg1
tgtg
m
m
?
?
? G
)(tg m?? ??? G
??
??
??
t gtg1
tgtg
)(tg
:
m
m
m ?
?
??

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In the same way,
② Secondly,we shall determine the minimum force Qmin under
the action of which the body can not slip down,see fig.(2),
) ( t g t g1 tgs i n c o s c o ss i n mm i n ?????? ?? ??? ?????? Gf fGGffQ
We find,
The range of equilibrium is
m a xm i n QQQ ??
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② 再求使物体不致下滑的 图 (2) minQ
) ( t g t g1 tgs i n c o s c o ss i n mm i n ?????? ?? ??? ?????? Gf fGGffQ

m a xm i n QQQ ??
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[Example 2] The length of the ladder is AB=l,its weight
is P,the coefficient of the static friction between ladder
and wall or ground is f =0.5,Determine the values of the
angle ?,when the ladder is in equilibrium?
Solution,Let us examine the position
of impending slip of the ladder,at
which it tends to fall down,Draw the
force diagram,
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[例 2] 梯子长 AB=l，重为 P，若梯子与墙和地面的静摩

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)2(0,0
)1(0,0 F r o m
????????
?????????
?
?
PFNY
FNX
BA
AB
)5(
)4(
??????????
??????????
BB
AA
NfF
NfF
)3( 0s i nco sco s2,0 m i nm i nm i n ????????????? ??? lNlFlPm BBA
222 1,1,1g e t we f
PPF
f
fPN
f
PN
BBA ???????
'0
22
m i n 87365.02
5.01a r c t g
2
1a r c t g ?
?
????
f
f?
Notice,As ? is smaller than the
angle of inclination of the ladder ?
should satisfy
?90
0'0 908736 ?? ?
Substituting the computed values into equ.(3),
we get,
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)2(0,0
)1(0,0
????????
?????????
?
?
PFNY
FNX
BA
AB由
)5(
)4(
??????????
??????????
BB
AA
NfF
NfF
)3( 0s i nco sco s2,0 m i nm i nm i n ????????????? ??? lNlFlPm BBA
)3(
1
,
1
,
1
,222 代入解得
f
PPF
f
fPN
f
PN
BBA ???????
'0
22
m i n 87365.02
5.01a r c t g
2
1a r c t g,?
?
????
f
f?得

?90
0'0 908736 ?? ?
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It is well known that it is easier to roll a roller than slip it,
Looking at the diagram of force analysis below,we may bring up a
question,although the body is in equilibrium it does not satisfy the
the equilibrium equations completely,
)s a t i s f i e dn o t is(0,0
0,0
0,0
???
???
???
?
?
?
rQM
NPY
FQX
A
The driving force couple
composed of Q and F tends to
drive the roller roll forward,
§ 4-4 Rolling Friction
The reason of this phenomenon
is that the actual surface of
contact is not rigid,under the
action of the force some
deformation will happen,as
shown in the following fig.,
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)(0,0
0,0
0,0

???
???
?
?
?
rQM
NPY
FQX
A
Q与 F形成主动力偶使前滚
§ 4-4 滚动摩擦

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① As the driving force couple (Q,F) increases,the force
couple of resistance M increases too;
Rolling ② there is a range of equilibrium;
friction ③ does not depend on the radius of the roller;
④ The law of rolling friction is,d is called
the coefficient of rolling friction,
m a x0 MM ??
maxM
NM ?? dm a x
The force couple of resistance balances the drive force couple (Q,F),
' d
Reduce this
force system
to point A,
(Look for the animation on the next page)
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A点简化
① 滚阻力偶 M随主动力偶（ Q,F）的增大而增大 ;
② 有个平衡范围 ;

④ 滚动摩擦定律：, d 为滚动 摩 擦系数。
m a x0 MM ??
maxM
NM ?? dm a x

' d
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According to theorem of translation of a force
the force N’ is the resultant of N and M,
N'=N
'N
Md ? NdNdM ????? '
d?? d
From the fig we can notice that the arm of the force couple of
resistance M is just d (the coefficient of rolling friction),
Consequently,d has the dimension of a length,
As d usually is very small,the force couple of resistance is
neglected in most engineering problems,hence the rolling
friction is neglected,
d
Explanation about the coefficient of rolling friction d,
① Its dimension is that of a length,Generally,the unit is mm or cm;
② It depends on the rigidity and temperature of the materials of the
roller and of the supporting surface;
③ The physical meaning of d is shown in the fig,
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①有长度量纲，单位一般用 mm,cm；
②与滚子和支承面的材料的硬度和温度有关。
③ d 的物理意义见图示。

N'=N
'N
Md ? NdNdM ????? '
d?? d

' d
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Chapter 4 Class of exercises
SUMMARY
1,Concepts
1) Friction Tangential constraint reaction force whose direction
is always opposite to the tendency of motion of the body,
a,When sliding does not occur F<f N
(F=P external force)
b,When sliding is impending Fmax=f ?N
c,When motion occurs F' =f '?N (Generally,f 'motion << f static )
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1、摩擦力 ----是一种切向约束反力，方向总是

a,当滑动没发生时 F<f N (F=P 外力 )
b,当滑动即将发生时 Fmax=f ?N
c,当滑动已经发生时 F' =f '?N (一般 f '动 << f 静 )
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3) Self-locking
When,the body is self-locking,m???
2) Total reaction force and angle of friction
a,The total reaction force R (the resultant of F and N)
b,When,the body does not move (equilibrium)
m?? ?
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2,全反力与摩擦角
a.全反力 R（即 F 与 N 的合力）
b,当 时,

3,自锁

m???
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2,Content,
1) Write down the equilibrium equations which take friction
into account
2) The methods to solve problems,① Analytical method
② Graphical method,
3) Beside the equilibrium equations there is the additional
equation (generally,it can be obtained when the
motion is impending),
4) The steps to solve problems are same as before,
NfF ??m a x
3,Problems in solution,
1) The direction of the force of friction can not be supposed,it
should be determined by the tendency of the motion of the body
(we can suppose the direction of the friction if it is an unknown
quantity which has to be determine),
2) Under the action of friction there is usually a range of
equilibrium,Therefore,the result is always a force,a dimension
or an angle in a certain range,(the reason is and ),m??? NfF ??
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1、列平衡方程时要将摩擦力考虑在内；
2、解题方法：①解析法 ② 几何法
3、除平衡方程外，增加补充方程 (一般在临界平衡
4、解题步骤同前。 状态计算）

1、摩擦力的方向不能假设，要根据物体运动趋势来判断。
（只有在摩擦力是待求未知数时，可以假设其方向）
2、由于摩擦情况下，常常有一个平衡范围，所以解也常常是

NfF ??m a x
m???
NfF ??
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4,Example
[Example 1]
Draw the force diagrams
of the bodies in the figure,
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[例 1] 作出下列各物体

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[Example 2]Draw the force diagrams of the bodies given in the figure,
① If P is minimum
and the system balances
② If P is maximum
and the system balances
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[例 2] 作出下列各物体的受力图
① P 最小维持平衡 ② P 最大维持平衡

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[Example 3] The component 1 is connected with the component 2 by
the wedge 3,the coefficient of friction between the components and
the wedge is f=0.1,Determine the angle of self-locking ?,
l o c k i n g,-s e l f isi t,'26112 w h e ni, e,,
c o n d i t i o n ) l i m i t i n g( '26112
'4351.0tg,1.0tga g a i n
2,
0
0
01
??
???
?????
?????
?
??
??
??
?????
f?
Solution,Investigate the
wedge,draw the force diagram,
1 RR ?
0co s)co s (,0 1 ?? ???? ??? RRX
From the conditions of double-
force equilibrium,
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[例 3] 构件 1及 2用楔块 3联结，已知楔块与构件间的摩擦系数 f=0.1,

0c o s)c o s (,0 1 ?? ???? ??? RRX由
1,RR ?由二力平衡条件

'26112
)( '26112
'4351.0tg,1.0tg
2,
0
0
01
??
???
?????
?????
?
??
??
??
?????
f?
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[Example 4] The weight of block B is Q=2000N,the angle of
friction is ? =15,the coefficient of
friction between block A and the
ground is f=0.4,neglect the self-
weight of the rod,
Determine the minimum weight of
the block A so that the block B does
not slip down,
Solution,① Investigate block B if it
can not slip down,
QQRS
RSX
Q
R
QRY
???
?
?
????
?????
?
?
?????
?
?
)(c t g
)s i n (
)c o s (
)c o s (
0)c o s (,0 ;
)s i n (
0)s i n (,0 F r o m
??
??
??
??
??
??
??
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[例 4] 已知,B块重 Q=2000N，与斜面的摩擦角 ? =15°,A块与

Q
QRS
RSX
Q
R
QRY
???
?
?
????
?????
?
?
?????
?
?
)(c t g
)s i n (
)c o s (
)c o s (
0)c o s (,0
)s i n (
0)s i n (,0
??
??
??
??
??
??
??由
61
)N(50002000
4.0
)1530(c t g)(c t g'
,0',0
??
?
?
?
???
????????
??
?
Q
ff
S
P
PfNfSFSX
??
② Then investigate block A
62
)N(50002000
4.0
)1530(c t g)(c t g'
,0',0
??
?
?
?
???
????????
??
?
Q
ff
S
P
PfNfSFSX
??
② 再研究 A块
63
[Exercise 1] Let Q=10N,f 'motion =0.1,f static =0.2,
Determine the force of friction F if P=1 N,2N and 3N,
Solution,
N2 g e t w e0f r o m N2 If ???? ? PFXP
Hence,when the body begin to move N11.010'' ????? fNF
m o t i o n
(no motion,F is equal to
the external force)
(critical equilibrium)
N2102.0 m a x ????? NfF st a t i c?
N1g e t w e0f r o m N1 If ????? ? PFXP
N,2N3 f r o m N3 If m a x ???? FPP
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[练习 1] 已知,Q=10N,f '动 =0.1 f 静 =0.2

N2,0,N 2 ???? ? PFXP 由时

（没动,F 等于外力）
（临界平衡）
（物体已运动）
N2102.0 m a x ????? NfF 静?
N1,0,N 1 ????? ? PFXP 由时
N2N3,N 3 m a x ???? FPP ?时
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[Exercise 2] Let the weight of block A be 500N,the weight of the
roller B is 1000N,the roller D is smooth,the coefficient of friction at
points E and A are fE=0.2,fA=0.5,
Determine the weight of the block C,Q=?,when the system is in
equilibrium,Solution,① A does not
move(i.e.,point i has no
translation),determine Q,
1
N2 5 05 0 05.0
' 11
???
??? NfFT A
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[练习 2] 已知 A块重 500N，轮 B重 1000N,D轮无摩擦,E

N2505005.0
' 11
???
??? NfFT A

1
67
N2 5 05 0 05.0 ???T
0)co s1010(co ss i n10s i n15 ?????? ???? QQT
Investigate
the roller,
0]c o sc o s[ s i n1015 22 ???? ???QT
)N(208
)
5
41(10
25015
]c o s1[10
15 ?
??
??
?
??
?
TQ
0?? EmFrom
68
N2 5 05 0 05.0 ???T
0)co s1010(co ss i n10s i n15 ?????? ???? QQT

0]c o sc o s[ s i n1015 22 ???? ???QT
)N(208
)
5
41(10
25015
]c o s1[10
15 ?
??
??
?
??
?
TQ
0?? Em由
69
② There is no horizontal displacement at
point E,hence
)
5
31000(2.02.0 ??????? QNfNF
E
? ? Qm i f i n d w e0 F r o m
)N(384
8.7
3000
a n d
068.13000,o b t a i n We
0)c o s5.0c o s( s i n10)6.01000(2.015
0)5c o s10(c o ss i n10s i n15
22
??
???
???????
?????
Q
QQ
QQ
QQF
???
????
70
② E 点不产生水平位移
)531 0 0 0(2.02.0,??????? QNfNF E即
? ? Qm i 可得由 0
)N(384
8.7
3 0 0 0
,
068.13 0 0 0,
0)c o s5.0c o s( s i n10)6.01 0 0 0(2.015
0)5c o s10(c o ss i n10s i n15
22
??
???
???????
?????
Q
QQ
QQ
QQF

???
????
71
③ The roller B does not move upwards,hence N≥0,;0s i n,0 F r o m ?????? ?QGNY B
)N(16706.01000,0531000s i n ????????? QQQGN B ?
Obviously,if there is no motion of both points i and E,Q
must be less than 208N,hence
)N(2 0 8m a x ?Q
72
③ B轮不向上运动，即 N≥0;0s i n,0 ?????? ?QGNY B由
)N(16706.01000,0531000s i n ????????? QQQGN B ?

)N(2 0 8m a x ?Q
73
fNF ??
21 QQ
Pf
???
∴ If,the bigger ball can roll across the smaller one,
21 QQ
Pf
??
[Exercise 3] P,D,d,Q1 and Q2 are known,P is horizontal,
Determine f when the bigger ball rolls across the smaller one,
Solution,① Investigate the whole system,
FPX ???,0f r o m

fQQP ??? )( 21
Substituting ① and ② into ③
we get,
The bigger ball can not roll across the smaller ball unless we
can ensure that the bigger ball can not slip on the smaller ball,
21,0 QQNY ????

74

21 QQ
Pf
???
∴ 当 时，能滚过去（这是小球与地面的 f 条件）
21 QQ
Pf
??
[练习 3] 已知,P,D,d,Q1,Q2,P为水平。

fQQP ??? )( 21将①、②代入③得,

21,0 QQNY ????

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② Determine f between the balls,investigate the bigger ball,
0c o s)90c o s (,0 10 ??????? ?? NFPX
PFDPDFm O ????????,0
22
,0

0c o ss i n 1 ????? ?? NPP

f
FNfNF ????
11,
Substituting ③ into ② we have,④
0c o ss i n ????? ?? fPPP
Again
dD
Dd
dD
dD
dD
dD
?
???
?
?
?
?
?
?
2
s i n1c o s,
22
22s i n 2 ????
76
0c o s)90c o s (,0 10 ??????? ?? NFPX
PFDPDFm O ????????,022,0由

② 求大球与小球之间的 f,研究大球
0c o ss i n 1 ????? ?? NPP

f
FNfNF ????
11,

0c o ss i n ????? ?? fPPP

dD
Dd
dD
dD
dD
dD
?
???
?
?
?
?
?
?
2
s i n1c o s,
22
22s i n 2 ????
77
Conclusion:When both and are
satisfied the bigger ball can roll across the smaller one,
∴ If,the bigger ball can roll across the small one,
D
df ?
D
df ?
21 QQ
Pf
??
D
df ?
?? ?
?
s in1
c o s
Solving ④ we find,
[note] the coefficient of
friction between the balls
can be obtained by
another way,
1
tg QPf ??? ?
78
∴ 当 时能滚过小球
D
df ?

D
df ?
21 QQ
Pf
??
D
df ?
?? ?
?
s in1
c o s

[注 ]大球与小球间的 f

1
tg QPf ??? ?
79
Solution,
① Draw the normal
AH and BH,
② Draw the angle of
friction ? at point A
and B,They intersect
at points E and G,
③ The horizontal distance
on which the man can
walk,between points E
and G is l,
[Exercise 4] A horizontal ladder lies on a V shape trough,the weight
of the ladder can be neglected,the coefficient of friction between the
ladder and the two inclined planes are the same (the angle of friction
is ? ),A man is walking on the ladder,Determine the range where the
man can walk without slipping of the ladder,
l
80

② 作 A,B点的 摩 擦角 ?

③ E,G两点间的水平距

[练习 4] 水平梯子放在直角 V形槽内，略去梯重，梯子与两个斜

l
81
Therefore,if the man walks
on AC or BD,the laws
cannot be satisfied (if a free
rigid body remains in
equilibrium under the
action of three nonparallel
coplanar forces the lines
of action of those forces
intersect at one point.),
but if he walks on CD
the forces can intersect at one point,
090???? A G BA E B
)60c o s ()30s i n ()60c o s (
)30c o s ()60s i n ()30c o s (
000
000
???
???
?????
?????
ABBGBD
ABAEAC
Proof,From the geometrical relationship we get
82
090???? A G BA E B
)60c o s ()30s i n ()60c o s (
)30c o s ()60s i n ()30c o s (
000
000
???
???
?????
?????
ABBGBD
ABAEAC

（有交点）

83
84