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Application of Newton’s Laws Involving Friction*
Uniform circular motion – Kinematic*
Nonuniform circular motion*
Velocity – Dependent Forces :
Drag and Terminal Velocity*
Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces
Friction is always present when two solid surfaces slide along each other.The microscopic details are not yet fully understood.
The Friction are nearly independent of the area of contact between surfaces.
5-1 Application of Newton’s Laws Involving Friction*
Friction (AF_0516)
Force of static friction
Force of kinetic friction kfrF N : depending on the nature of the surface s k
As long as the box is not moving, Ffr = FA
N : Normal force sfrF N
Note that, in general, μs > μk.
P. 113
Conceptual Example : A box against a wall.You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. What is the free-body diagram of this box?
Conceptual Example : A box against a wall.You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. What is the free-body diagram of this box?
x
yx
y
mg
fFN
What is the minimum force you need to apply to keep the box from sliding down?(static friction and kinetic coefficients are 𝝁𝒔 and 𝝁𝒌 ,respectively.)
𝒔𝒔
𝒔
For an object in motion on a surface,
,s sf N
k kf N
,s sf N
,max !s sf N
For an object rest on a surface,Static Friction Force is usually unknown!
Only the maximum Static friction can be predicted
Important Properties of Friction force for Analysis
A block is pulled by a force Fp on a table top,if the angle the Force Fp is pulling the weight is variable, what would be the angle such that the block moves with maximum acceleration? Assume the kinetic friction coefficient is k.
pF
Kinetic Friction
Use Newton’s second law to solve for the acceleration of the block.
0sin mgFFF pNy
maFFF frpx cos
sin pN FmgFfr k NF F
cos sinp p k kF F mga
m
NF
mg
PF
frF
To find the maximum, we differentiate a with respect to
sin cosp p kF Fdad m
sin cos0p p kF Fda
d m
tan k
At maximum acceleration,
2
2
cos sinp p kF Fd ad m
Since 0 sin ,cos 0k
02
2
d
ad cos sinp p kF F mga
m
is a maximum.
To confirm maximum acceleration condition,
m
M
FT
CCU Physics 5 - 3
?a
Kinetic Friction
m
M
FT
?a
1: mx F T f ma 0: 1 mgNy
For m,
For M,1 2: Mx T f f Ma
2 1: 0y N N Mg
mgNf 11
gMmNf )(22
Physical relation: m Ma a a
mamgTF MagMmmgT )(
MmgMmFa
)3(
Fmg
1NT
1f
2N
1N
1fT
M g2f
x
y
A ramp, a pulley, and two boxes.Box A, of mass mA, rests on a surface inclined at to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. If the coefficient of static friction is s, determine what range of values for mass B will keep the system at rest.
Static Friction
A AF m a
: cos 0 Ay N m g A: sin 0 Sx T m g f
A AF m a
: cos 0 Ay N m g A: sin 0 Sx T m g f
B BF m a
: 0 By T m g BT m g
A
A
sinsin
S
B
f T m gm g m g
A
A
sinsin
S
B
f T m gm g m g
Sf N Sf NA Asin cosBm g m g m g A Asin cosBm g m g m g
A (sin cos )Bm m A (sin cos )Bm m
Box A almost slides downward Box A almost slides upward
T
Bm g
N
Am gSf
x
y
x
y
T
B BF m a
: 0 By T m g BT m g
A AF m a
: cos 0 Ay N m g A: sin 0 Sx T m g f
A
A
sinsin
S
B
f T m gm g m g
A cosN m g
Sf N
SN f N
Sf N
A Acos cosSm g f m g
A A Acos sin cosBm g m g m g m g
A A A Asin cos cos sinBm m m m m
A Asin cos sin cosBm m m
(A)(B)
(C)
(D)(E)
(F)
(G)
(H)
x
y
ABF
/4 x
y
If F = 3 mg, which of the following is the direction of the frictional force applied to block B?
s = 0.9 ; k = 0.8MA =MB = m
ABF
/4 x
y
x
yBF
mB gfS
N
x
yA
NmA g
N1fS
,4 ,
3
B A
B A
a a am m m
F mg
B BF m a
: cos sin 0 S By N f m g : sin cos S B Bx F N f m a (1)
(2)
A AF m a
: sin cos S A Ax N f m a
1: N cos sin 0 S By N f m g (3)
(4)
: 2 2 0 Sy N f mg : 3 2 2 Sx mg N f ma
: 2 2 Sx N f ma
1: N 2 2 0 Sy N f mg
(5)(6)
(7)
(8)(5)+(6)3 2 2 Smg f ma mg
(7) -(6) 2 2 Sf ma mg (9)
(10)
(10) -(9) 4 2 Sf mg
2 4 Sf mg
(A)(B)
(C)
(D)(E)
(F)
(G)
(H)
x
y
ABF
/4 x
y
If F = 3 mg, which of the following is the direction of the frictional force applied to block B?
s = 0.9 ; k = 0.8MA =MB = m
Ans (F)
F
Now the static friction coefficient between all surfaces is , what is the range of force F that can keep the box m from sliding up or down the inclined surface?
F
Now the static friction coefficient between all surfaces is , what is the range of force F that can keep the box m from sliding up or down the inclined surface?
N
Mg
x
y
mN
Sf F
mg
x
y
mN
Sf
Circular Motion (AF_0602)
Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that
5-2 Uniform Circular Motion
.
P. 120
v Sv r
v t S tv r
0 0( 0) ( 0)
lim limt t
v t S tv r
a vv r
2va
r
Uniform circular motion
v
a
2 ,rvT
2var
An object that executes uniform circular motion is under the influence of a net force that is always normal to its instantaneous velocity and always pointing to the center of the orbit.
2
cos , sinmvF mar
F v / /and F a
Uniform circular motion
sin,cos rrr
t
dttdr
dttdr
dtrdv sin,cos
trtr cos,sin ttr cos,sin
trtrdtd
dtrda cos,sin2
2
dttdr
dttdr cos,sin
trtr sin,cos 22
ttr sin,cos2
cos , sinr r t r t
2 2 fT
cos , sinr r t r t
ttrv cos,sin
ttra sin,cos2
Uniform circular motion
v
a
2 ,rv rT
Tf 1 2 f
Assume the period of motion is T
,v r
A motion with constant speed!
2T
Define,
22 va r
r
