静力学摩擦

motion.
Method of Analysis
1. Assume equilibrium
2. Solve the equilibrium equations ⇒ friction force.
3. Check the assumption. If F ≤Fmax=µSN , Answer. Otherwise, F=µkN
∑ Fy=0 FN -P·cosα-F·sinα =0
∑ Fx=0 F·cosα-P·sinα -FS=0
补充方程:
FS= µS ·FN
∴
Fmax
=
sinα + µ S cosα cosα − µ S sinα
P
α
(FS)
(2) F=Fmin
∑Fy=0 FN -P·cosα-F·sinα =0
∑Fx=0 F·cosα-P·sinα +FS=0
a. Static case (no P) No friction force, no relative motion, no tendency of relative motion.
b. Static case (P <Fmax) No relative motion, but having the tendency of relative motion and a friction force existing. 0≤F =P<Fmax=µSN µS: coefficient of static friction
of sliding up. Fs acts toward down .
(FS)
α
(2) F=Fmin, Block is on the verge of sliding down. FS acts toward up.
2. Assuming block is in equilibrium
(1) F=Fmax
to the general conclusion that the
block can be in equilibrium only if the line of action of
FR stays within the sector (bounded by ±ϕ). For more general loadings, the line of action of FR must lie within the cone, called cone of static friction, that is formed by
2. Solve for the unknowns using equilibrium equations
Type III The problem statement implies impending sliding, but the surfaces at which sliding impends are not known. Method of Analysis 1
µs , µk 的大小可由实验测定，它们与接触物体的材料和表 面状态有关。常用材料的µs , µk见表5－1。
5.3 Angle of Friction
a.Angle of friction
As shown in Fig. a, the reaction of rough surface is
FR .FR=FN＋FS . We let α be the angle between the contact force FR and the normal n to the contact surface.The upper limit of α, denoted by ϕ , is reached at
∑ Fy=0
F=W=100N
N
x
∑ Fx=0
N=P=500N
3. Check
Fmax=N·µS=0.3×50=150N F=100N<Fmax 4. Comment
F Answer
What is the result if P is still 500N, but the weight of the body changes to 200N?
tg-1µS<α . Determine the value of the push force F acting in horizon that makes block in equilibrium.
Solution: (Impending sliding)
1. The FBD. (1) F=Fmax, Block is on the verge
3. Choose the correct answer by inspection of the solution.
Method of Analysis 2
1.Determine all possible ways in which sliding can impending.
2. For one of the cases, set F=Fmax at the surface where sliding impends and solve the equilibrium equations.
Dry friction refers to the friction force that exists between two unlubricated solid surface. 5.2 Coulomb’s Theory of Dry Friction
Let us observe an experiment:
补充方程:
FS= µS ·FN
∴
Fmin
=
sinα − µ S cosα cosα + µ S sinα
P
sinα − µ S cosα P < F < sinα + µ S cosα P
cosα + µ S sinα
cosα − µ S sinα
Sample problem 5-3 The spool in Fig. (a) weights 25N, and its center of gravity is located at the geometric center. The weight of block C is 50N. The coefficients of static friction at the two points of contact are as shown. Determine the largest horizontal force P that can be applied without disturbing the equilibrium of the system.
c. Case of impending sliding (P=Fmax ) The surfaces on the verge of sliding. F =Fmax=µSN
d. Dynamic case (P>Fmax ) Surfaces sliding relative to each other. F=Fk=µkN µk: coefficient of kinetic friction
Sample problem 5-1 The weight of body is W, and pushing
force is P. W=100N, P=500N. µS=0.3. Determine the friction.
Байду номын сангаас
Solution: 1. The FBD
y
2. Assuming the block in equilibrium. So we have
1. Determine all possible ways in which sliding can impend.
2. For one of the cases, set F=Fmax at the surfaces where sliding impends and solving the equilibrium equations. In general, a different solution is obtained for each mode of impending sliding.
3. Check the solution by comparing the friction force at each of the other surfaces with its limiting value. If all these forces are less than or equal to their maximum permissible value, then the solution is correct. If a friction force exceeds it limiting value, the solution is invalid and another mode of impending sliding must be analyzed. This must be continued until the correct solution is found.
Chapter 5 Dry Friction
5.1 Introduction In most of equilibrium problems that we have
analyzed up to this point, the surface of contact have been smooth (frictionless). The reactive forces were, therefore, normal to the contact surface. The concept of a smooth surface is, of course, an idealization. All real surface also provide a force component that is tangent to the surface, called the friction force, that resists sliding.
In Fig.(a) , the friction force F
opposes the tendency of P to slide F
the block to the left. If the direction
P
of P is reversed, the direction of F
would also be reversed. This leads
impending sliding when FS=Fmax=µSFN (Fig.b).
Therefore, we have
tgϕ = Fmax
FN
= µs FN
FN
= µs
The angle ϕ is called the angle of static friction.
b. Self-locking
rotating the sector about the normal n. Observe that the
vertex angle of the cone of static friction is 2ϕ.
If the block is in equilibrium, then
α≤ϕ
5.4 Problem Classification and Analysis Type I The problem statement dose not specify impending
∵ W=200>Fmax=150N ∴ Block is not in equilibrium.
In this case, the friction force F= N·µk. The body will move down at an accelerate.
Sample problem5-2
The weight of the block is P.
Type II The problem statement implies impending sliding, and the surfaces where sliding impends are known.
Method of Analysis
1. Set F=Fmax= µSN at surfaces where sliding impends