# Friction Fulton County Schools Home by nikeborome

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```									           Physics 152
Fall 2004
Tuesday, Sep 14

Lecture #7 - Chapter 5.1: FRICTION
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Properties of Friction
1. fs between any two surfaces
in contact is opposite the
applied force that is parallel
to the surface and equal in
magnitude

fs< µsN
µs - coefficient of static friction
N - Normal force

2. When the object is on the
verge of slipping
fs,max = µsN
Properties of Friction
3. fk is opposite the applied
force that is parallel to the
surface and is

fk = µkN

µk - coefficient of kinetic friction
N - Normal force

For constant velocity:
F = fk
Properties of Friction
4. µs and µk depend on the
nature of the surface.
However:

µk < µs

µs - coefficient of static friction
µk - coefficient of kinetic friction

5. fs and fk do not depend on
area because weight and µ
do not depend on area.
(empirical finding)
(1) In three experiments, three different horizontal forces are applied to
the same block lying on the same countertop. The force magnitudes
are F1= 12 N, F2= 4 N and F3= 8 N. In each experiment, the block
remains stationary inspite of the applied force.
(a) Rank the forces according to the magnitude fs of the static frictional
force on the block from the countertop, greatest first.
(1) F2, F 3, F 1
(2) F1, F 3, F 2
(3) F3, F 2, F 1
(b) Rank the forces according to the maximum value fs,max of that force,
greatest first.
(1) F2, F 3, F 1
(2) F1, F 2, F 3
(3) F3, F 2, F 1
(4) all tie
(2) In the figure (a), a 'Batman' thermos is sent sliding
leftwards across a long plastic tray. What are the
directions of kinetic frictional forces on...
(a) the thermos
(a) downward
(b) upward
(c) leftward
(d) rightward
(2) In the figure (a), a 'Batman' thermos is sent sliding
leftwards across a long plastic tray. What are the
directions of kinetic frictional forces on...
b) the tray (from each other)
1. upward
2. leftward
3. rightward
4. downward
(2) In the figure (a), a 'Batman' thermos is sent sliding
leftwards across a long plastic tray.
(c) Does the kinetic frictional forces on the thermos increase
or decrease the speed of the thermos relative to the
floor?
1. decrease
2. increase
2. In figure (b), the tray is now sent sliding leftward beneath
the thermos. What now are the directions of the kinetic
frictional forces on...
(d) the thermos
1. upward
2. rightward
3. downward
4. leftward
2. In figure (b), the tray is now sent sliding leftward beneath
the thermos. What now are the directions of the kinetic
frictional forces on...
(e) the tray (from each other)
1. rightward
2. downward
3. leftward
4. upward
2 (f) Does the kinetic frictional forces on the thermos
increase or decrease the speed of the thermos relative to
the floor?
decrease
increase

(g) Do the kinetic frictional forces always slow the
objects?
yes
no
3) If you press an apple crate against a wall so hard that the
crate cannot slide down the wall, what is the direction of...
(a) the static frictional force fs on the crate from the wall?
1. upward
2. leftward
3. rightward
4. downward
(b) the normal force N on the crate from the wall?
1. horizontal, left of you
2. horizontal, towards you
3. horizontal, right of you
4. horizontal, towards wall
If you increase your push, what happens to...
(c) Fs
1. no change
2. decreases
3. increases

(d) N
1. decreases
2. no change
3. increases
(e) fs,max
1. no change
2. increases
3. decreases
Example: Static Friction
A box of mass m =10.21 kg is at rest on a floor. The
coefficient of static friction between the floor and the box is
µs = 0.40 . A rope is attached to the box and pulled at an
angle of q = 30o above horizontal with tension T = 40 N.

Does the box move?

T

q
static friction (= 0.40 )    m
*Select axes & draw FBD of box:
y

*Apply FNET = ma
x
y: N + T sin q - mg = maY = 0
N
N = mg - T sin q = 80 N
T
x: T cos q - fs = maX
fs    m            q

The box will move if:
T cos q - fs > 0                  mg
y: N = 80 N                                        y

x: T cos q - fs = maX
x
The box will move if: T cos q - fs > 0
N
T cos q = 34.6 N
T
fMAX = N = (0.40)(80N)
= 32 N                     fMAX = N
m        q

So T cos q > fMAX and the box does move            mg
Example: Static Friction
A detached block of rock with m =1.8 x 107 kg is at rest on a
“bedding plane fracture”. The coefficient of static friction
between the rock and the fracture µs = 0.63 . The bedding
plane fracture is inclined q = 24o above horizontal.
Does the block slide?

y
fs   x
q
q

W
y
N
fs      x
q
q

W

Fy = 0=N - Wcosq = N - mgcosq
N = mg cosq = 160 x 106 Newtons
y
N
fs      x
q
q

W

Fy = 0=N - Wcosq = N - mgcosq
N = mg cosq = 160 x 106
Newtons
fs = µs N = 102 x 106 Newtons
Wx= mg sinq = 72 x 106 Newtons

Block will not slide because Wx < fs !
y
N
fs   x
q
q

W

Alternative Method:
From example 5-2 in textbook:
µs = tan q max           Arctan(µs) = q max
q max= 32   o

Will not slide because
q = 24 o < q max
Example: Dynamic Friction
A box of mass m1 = 1.5 kg is being pulled by a
horizontal string having tension T = 90 N. It slides
with friction (µk = 0.51) on top of a second box
having mass m2 = 3.0 kg, which in turn slides on a
frictionless floor.

What is the acceleration of the second box ?

T                          m1    slides with friction (=0.51 )

a=?                   m2              slides without friction
*First draw FBD of the top box:
N1,2
N1,2

T             m1                  T          f k= N1 = m1g
fk

W1=m1g
W1

(Note: a1 ~ 55 m/s2)
*Newton’s 3rd law says the force box 2 exerts on box 1 is
equal and opposite to the force box 1 exerts on box 2.

Action-Reaction Pair from Frictional Forces:

f1,2
m1

f2,1               m2

Action-Reaction Pair from Contact Forces:
N1,2

m1

m2
N2,1
*Now consider the FBD of box 2:
N2,
g
N2,
g

f2,1
f2,1 = km1g
m2

N2,1
N2,1                           =m1g
W2=m2g
W2
*Finally, solve F = ma in the horizontal direction:

N2,
g                m1g = m2a

a = 2.5 m/s2
f2,1 = m1g

N2,1
=m1g
W2=m2g

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