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
第四章 摩擦
§ 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
习题课
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
前几章我们把接触表面都看成是绝对光滑的，忽略了物体
之间的摩擦，事实上完全光滑的表面是不存在的，一般情况下
都存在有摩擦。
[例 ]
第四章 摩 擦
§ 4-1 引言
平衡必计摩擦 ?
7
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
一、为什么研究摩擦？
二、怎样研究摩擦，掌握规律
利用其利，克服其害。
三、按接触面的运动情况看,
摩擦分为 滑动摩擦
滚动摩擦 ?
9
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
产生阻止物体运动的力叫滑动摩擦力。
（ 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
一、静滑动摩擦力
（翻页请看动画）
所以增大摩擦力的途径为：①加大正压力 N,
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
13
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 ?? ''
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二、动滑动摩擦力,（与静滑动摩擦力不同的是产生了滑动）
大小,（无平衡范围）
动摩擦力特征,方向：与物体运动方向相反
定律,（ f '只与材料和表面情况有
关，与接触面积大小无关。）
m a x0 FF ?? 0?? X
NfF ??m a x
NfF ?? ''
NfF ?? ''
3,特征,大小,（平衡范围）满足
静摩擦力特征,方向：与物体相对滑动趋势方向相反
定律,（ f 只与材料和表面情况有
关，与接触面积大小无关。）
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)
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maxF
m?
三、摩擦角,
① 定义：当摩擦力达到最大值 时其全反力
与法线的夹角 叫做 摩擦角 。
翻
页
请
看
动
画
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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四、自锁
① 定义：当物体依靠接触面间的相互作用的摩擦 力 与正
压力（即全反力），自己把自己卡 紧，不会松开
（无论外力多大），这种现象称为自锁。
当 时，永远平衡（即自锁）
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
摩擦系数的测定, OA绕 O 轴转动使物块刚开始下滑时测出
?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
考虑摩擦时的平衡问题，一般是对临界状态求解，这时可
列出 的补充方程。其它解法与平面任意力系相同。
只是平衡常是一个范围
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,
水平力 Q的平衡范围。②当水平力 Q = 60N时，物体能否平衡？
（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
解, ① 先求使物体不致于上滑的 图 (1)
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
补充方程
由
??
??
?
?
tg1
tg,
m a x f
fGQ
?
??解得
??
??
tgtg1
tgtg
m
m
?
?
? G
)(tg m?? ??? G
??
??
??
t gtg1
tgtg
)(tg
:
m
m
m ?
?
??
应用三角公式
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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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同理,
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
解得,
平衡范围应是
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
擦系数 f =0.5,求 ? 多大时，梯子能处于平衡？
解, 考虑到梯子在临界平衡状
态有下滑趋势，做 受力图。
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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
注意,由于 ?不可能大于,
所以梯子平衡倾角 ? 应满足
?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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由实践可知，使滚子滚动比使它滑动省力，下图的受力分析
看出一个问题，即此物体平衡，但没有完全满足平衡方程。
)(0,0
0,0
0,0
不成立???
???
???
?
?
?
rQM
NPY
FQX
A
Q与 F形成主动力偶使前滚
§ 4-4 滚动摩擦
出现这种现象的原因是，
实际接触面并不是刚体，它们
在力的作用下都会发生一些变
形，如图,
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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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此力系向
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 ;
滚动 摩擦 ③ 与滚子半径无关 ;
④ 滚动摩擦定律：, d 为滚动 摩 擦系数。
m a x0 MM ??
maxM
NM ?? dm a x
滚阻力偶与主动力偶（ Q,F）相平衡 （翻页请看动画）
' 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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滚动摩擦系数 d 的说明,
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。
根据力线平移定理，将 N和 M合成一个力 N',
N'=N
'N
Md ? NdNdM ????? '
d?? d
从图中看出，滚阻力偶 M的力偶臂正是 d（滚阻系数），
所以,d 具有长度量纲 。
由于滚阻系数很小，所以在工程中大多数情况下滚阻力
偶不计，即滚动摩擦忽略不计。
' 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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第四章, 摩擦, 习题课
本章小结
一、概念,
1、摩擦力 ----是一种切向约束反力，方向总是
与物体运动趋势方向相反。
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,
物体不动（平衡）。
3,自锁
当 时自锁。 m???
m???
51
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 ??
52
二、内容,
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）
三、解题中注意的问题,
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
力、尺寸或角度的一个平衡范围。（原因是
和 ）
NfF ??m a x
m???
NfF ??
53
4,Example
[Example 1]
Draw the force diagrams
of the bodies in the figure,
54
四、例题
[例 1] 作出下列各物体
的受力图
55
[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
56
[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
状态受力图； 状态受力图
57
[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,
58
[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,
求能自锁的倾斜角 ? 。
解：研究楔块，受力如图
0c o s)c o s (,0 1 ?? ???? ??? RRX由
1,RR ?由二力平衡条件
时能自锁即当
极限状态
又
'26112
)( '26112
'4351.0tg,1.0tg
2,
0
0
01
??
???
?????
?????
?
??
??
??
?????
f?
59
[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
??
??
??
??
??
??
??
60
[例 4] 已知,B块重 Q=2000N，与斜面的摩擦角 ? =15°,A块与
水 平面的摩擦系数 f=0.4，不计杆自
重。 求：使 B块不下滑，物块 A最小
重量。
解,① 研究 B块，若使 B块不下滑
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)
(the body already moves)
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
64
[练习 1] 已知,Q=10N,f '动 =0.1 f 静 =0.2
求,P=1 N; 2N,3N 时摩擦力 F？
解,
N2,0,N 2 ???? ? PFXP 由时
所以物体运动：此时 N11.010'' ????? fNF
动
（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
66
[练习 2] 已知 A块重 500N，轮 B重 1000N,D轮无摩擦,E
点的摩擦系数 fE=0.2,A点的摩擦系数 fA=0.5。
求：使物体平衡时块 C的重量 Q=？
解,① A不动（即 i点不产
生 平移）求 Q
N2505005.0
' 11
???
??? NfFT A
由于
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
分析轮有
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
即
化简
???
????
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 ?
显然,如果 i,E两点均不产生运动,Q必须小于 208N,即
)N(2 0 8m a x ?Q
73
Additional equation ③
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
①
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 ????
②
74
补充方程 ③ fNF ??
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。
求：在大球滚过小球时,f=?
解, ① 研究整体 FPX ???,0由 ①
fQQP ??? )( 21将①、②代入③得,
要保证大球滚过小球，必须使大球与小球之间不打滑
21,0 QQNY ????
②
75
② Determine f between the balls,investigate the bigger ball,
0c o s)90c o s (,0 10 ??????? ?? NFPX
PFDPDFm O ????????,0
22
,0
①
0c o ss i n 1 ????? ?? NPP
②
Additional equation,③
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
①
② 求大球与小球之间的 f,研究大球
0c o ss i n 1 ????? ?? NPP
②
补充方程 ③
f
FNfNF ????
11,
将③代入②得,④
0c o ss i n ????? ?? fPPP
又
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 ?
结论,当 和 时能保证大球能滚过小
球的条件。
D
df ?
21 QQ
Pf
??
D
df ?
?? ?
?
s in1
c o s
解④得,
[注 ]大球与小球间的 f
又一种求法,
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
解,① 作法线 AH和 BH
② 作 A,B点的 摩 擦角 ?
交 E,G两点
③ E,G两点间的水平距
离 l为人的 活 动范围
[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
面间的摩擦系数（摩擦角均为 ?），如人在梯子上走动，试分析
不使梯子滑动，人的活动应限制在什么范围内？
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
所以人在 AC和 BD段活动
都不能满足三力平衡必汇
交的原理,只有在 CD段
活动时,才能满足三力
平衡必汇交，能交上
（有交点）
证明,由几何关系
83
84
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
第四章 摩擦
§ 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
习题课
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
前几章我们把接触表面都看成是绝对光滑的，忽略了物体
之间的摩擦，事实上完全光滑的表面是不存在的，一般情况下
都存在有摩擦。
[例 ]
第四章 摩 擦
§ 4-1 引言
平衡必计摩擦 ?
7
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
一、为什么研究摩擦？
二、怎样研究摩擦，掌握规律
利用其利，克服其害。
三、按接触面的运动情况看,
摩擦分为 滑动摩擦
滚动摩擦 ?
9
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)
10
1,定义,相接触物体，产生相对滑动（趋势）时，其接触面
产生阻止物体运动的力叫滑动摩擦力。
（ 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
一、静滑动摩擦力
（翻页请看动画）
所以增大摩擦力的途径为：①加大正压力 N,
② 加大摩擦系数 f
（ f — 静滑动摩擦系数）
（ f '— 动摩擦系数）
11
12
13
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
二、动滑动摩擦力,（与静滑动摩擦力不同的是产生了滑动）
大小,（无平衡范围）
动摩擦力特征,方向：与物体运动方向相反
定律,（ f '只与材料和表面情况有
关，与接触面积大小无关。）
m a x0 FF ?? 0?? X
NfF ??m a x
NfF ?? ''
NfF ?? ''
3,特征,大小,（平衡范围）满足
静摩擦力特征,方向：与物体相对滑动趋势方向相反
定律,（ f 只与材料和表面情况有
关，与接触面积大小无关。）
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?
三、摩擦角,
① 定义：当摩擦力达到最大值 时其全反力
与法线的夹角 叫做 摩擦角 。
翻
页
请
看
动
画
fN NfNFm ???? m a xtg ?
② 计算,
17
18
19
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,
20
四、自锁
① 定义：当物体依靠接触面间的相互作用的摩擦 力 与正
压力（即全反力），自己把自己卡 紧，不会松开
（无论外力多大），这种现象称为自锁。
当 时，永远平衡（即自锁）
m?? ?
m?? ?
② 自锁条件,
21
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
摩擦系数的测定, OA绕 O 轴转动使物块刚开始下滑时测出
?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
23
24
25
26
27
§ 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)
28
§ 4-3 考虑滑动摩擦时的平衡问题
考虑摩擦时的平衡问题，一般是对临界状态求解，这时可
列出 的补充方程。其它解法与平面任意力系相同。
只是平衡常是一个范围
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,
水平力 Q的平衡范围。②当水平力 Q = 60N时，物体能否平衡？
（翻页请看动画）
29
30
31
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
解, ① 先求使物体不致于上滑的 图 (1)
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
补充方程
由
??
??
?
?
tg1
tg,
m a x f
fGQ
?
??解得
??
??
tgtg1
tgtg
m
m
?
?
? G
)(tg m?? ??? G
??
??
??
t gtg1
tgtg
)(tg
:
m
m
m ?
?
??
应用三角公式
33
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 ??
34
同理,
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
解得,
平衡范围应是
m a xm i n QQQ ??
35
[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,
36
[例 2] 梯子长 AB=l，重为 P，若梯子与墙和地面的静摩
擦系数 f =0.5,求 ? 多大时，梯子能处于平衡？
解, 考虑到梯子在临界平衡状
态有下滑趋势，做 受力图。
37
)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,
38
)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
注意,由于 ?不可能大于,
所以梯子平衡倾角 ? 应满足
?90
0'0 908736 ?? ?
39
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.,
40
由实践可知，使滚子滚动比使它滑动省力，下图的受力分析
看出一个问题，即此物体平衡，但没有完全满足平衡方程。
)(0,0
0,0
0,0
不成立???
???
???
?
?
?
rQM
NPY
FQX
A
Q与 F形成主动力偶使前滚
§ 4-4 滚动摩擦
出现这种现象的原因是，
实际接触面并不是刚体，它们
在力的作用下都会发生一些变
形，如图,
41
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)
42
此力系向
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 ;
滚动 摩擦 ③ 与滚子半径无关 ;
④ 滚动摩擦定律：, d 为滚动 摩 擦系数。
m a x0 MM ??
maxM
NM ?? dm a x
滚阻力偶与主动力偶（ Q,F）相平衡 （翻页请看动画）
' d
43
44
45
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,
46
滚动摩擦系数 d 的说明,
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。
根据力线平移定理，将 N和 M合成一个力 N',
N'=N
'N
Md ? NdNdM ????? '
d?? d
从图中看出，滚阻力偶 M的力偶臂正是 d（滚阻系数），
所以,d 具有长度量纲 。
由于滚阻系数很小，所以在工程中大多数情况下滚阻力
偶不计，即滚动摩擦忽略不计。
' d
47
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 )
48
第四章, 摩擦, 习题课
本章小结
一、概念,
1、摩擦力 ----是一种切向约束反力，方向总是
与物体运动趋势方向相反。
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 )
49
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?? ?
50
2,全反力与摩擦角
a.全反力 R（即 F 与 N 的合力）
b,当 时,
物体不动（平衡）。
3,自锁
当 时自锁。 m???
m???
51
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 ??
52
二、内容,
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）
三、解题中注意的问题,
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
力、尺寸或角度的一个平衡范围。（原因是
和 ）
NfF ??m a x
m???
NfF ??
53
4,Example
[Example 1]
Draw the force diagrams
of the bodies in the figure,
54
四、例题
[例 1] 作出下列各物体
的受力图
55
[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
56
[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
状态受力图； 状态受力图
57
[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,
58
[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,
求能自锁的倾斜角 ? 。
解：研究楔块，受力如图
0c o s)c o s (,0 1 ?? ???? ??? RRX由
1,RR ?由二力平衡条件
时能自锁即当
极限状态
又
'26112
)( '26112
'4351.0tg,1.0tg
2,
0
0
01
??
???
?????
?????
?
??
??
??
?????
f?
59
[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
??
??
??
??
??
??
??
60
[例 4] 已知,B块重 Q=2000N，与斜面的摩擦角 ? =15°,A块与
水 平面的摩擦系数 f=0.4，不计杆自
重。 求：使 B块不下滑，物块 A最小
重量。
解,① 研究 B块，若使 B块不下滑
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)
(the body already moves)
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
64
[练习 1] 已知,Q=10N,f '动 =0.1 f 静 =0.2
求,P=1 N; 2N,3N 时摩擦力 F？
解,
N2,0,N 2 ???? ? PFXP 由时
所以物体运动：此时 N11.010'' ????? fNF
动
（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
65
[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
66
[练习 2] 已知 A块重 500N，轮 B重 1000N,D轮无摩擦,E
点的摩擦系数 fE=0.2,A点的摩擦系数 fA=0.5。
求：使物体平衡时块 C的重量 Q=？
解,① A不动（即 i点不产
生 平移）求 Q
N2505005.0
' 11
???
??? NfFT A
由于
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
分析轮有
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
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
QQF
即
化简
???
????
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 ?
显然,如果 i,E两点均不产生运动,Q必须小于 208N,即
)N(2 0 8m a x ?Q
73
Additional equation ③
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
①
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 ????
②
74
补充方程 ③ fNF ??
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。
求：在大球滚过小球时,f=?
解, ① 研究整体 FPX ???,0由 ①
fQQP ??? )( 21将①、②代入③得,
要保证大球滚过小球，必须使大球与小球之间不打滑
21,0 QQNY ????
②
75
② Determine f between the balls,investigate the bigger ball,
0c o s)90c o s (,0 10 ??????? ?? NFPX
PFDPDFm O ????????,0
22
,0
①
0c o ss i n 1 ????? ?? NPP
②
Additional equation,③
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
①
② 求大球与小球之间的 f,研究大球
0c o ss i n 1 ????? ?? NPP
②
补充方程 ③
f
FNfNF ????
11,
将③代入②得,④
0c o ss i n ????? ?? fPPP
又
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 ?
结论,当 和 时能保证大球能滚过小
球的条件。
D
df ?
21 QQ
Pf
??
D
df ?
?? ?
?
s in1
c o s
解④得,
[注 ]大球与小球间的 f
又一种求法,
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
解,① 作法线 AH和 BH
② 作 A,B点的 摩 擦角 ?
交 E,G两点
③ E,G两点间的水平距
离 l为人的 活 动范围
[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
面间的摩擦系数（摩擦角均为 ?），如人在梯子上走动，试分析
不使梯子滑动，人的活动应限制在什么范围内？
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
所以人在 AC和 BD段活动
都不能满足三力平衡必汇
交的原理,只有在 CD段
活动时,才能满足三力
平衡必汇交，能交上
（有交点）
证明,由几何关系
83
84
