friction handout
TRANSCRIPT
-
8/11/2019 Friction Handout
1/9
Friction
The force of friction is a common but complex force. The exact method
by which friction works is still a topic of great scientific interest but wecan make some general statements about it. We do know that it arises
from the electromagnetic forces between atoms and molecules at the
surfaces of objects.
We can build a simple model of the friction force that is useful in many
situations. The model friction force has the following properties:
There are two types of frictional force. The force of static friction and
the force of kinetic friction:
The direction of the static frictional force is along the contact surfaceand opposite in direction of any applied force.
The magnitude of the static friction force is given by
Fs sRn
Where sis the coefficient of static friction and Rnis the normal reaction.
The direction of the kinetic frictional force is opposite the directionof motion of the object it acts on.
The magnitude of the kinetic friction force is given by
FkkRn
Where kis the coefficient of static friction and Rnis the normal reaction.
The coefficients of friction depend on the nature of the surface.
The frictional force is nearly independent of the contact area betweenthe objects.
The kinetic friction force is usually less than the maximumstaticfriction force.
Mathematics made simple M1 Friction
- 1 -
-
8/11/2019 Friction Handout
2/9
Friction
The plot below of the frictional force vs. the applied force illustratessome of the features of the frictional force.
ote that the frictional force e!uals the applied force "in magnitude#
until it reaches the maximum possible value ms. Then the object begins
to move as the applied force exceeds the maximum frictional force. When
the object is moving the frictional force is kinetic and roughly constant
at the value mk which is below the maximum static friction force.
The table below summarizes the main characteristics of the frictional
force.
Static Friction Kinetic Friction
Symbol fs fk
Direction opposite direction ofapplied force
opposite direction ofobjects motion
!a"nitude #ms$ mk$
$ote% &n the !' !odule it will be assumed that s= k=
Mathematics made simple M1 Friction
- 2 -
-
8/11/2019 Friction Handout
3/9
Friction
(imitin" e)uilibrium
$f the object is at rest and the forces are in e!uilibrium with the limitingfriction% the object is said to be in limitin" e)uilibrium. &t this point% the
friction force is at its maximum value% called the limitin" friction.
*oefficient of friction +,The magnitude of the maximum frictional force is a fraction of the
normal reaction "Rn#. This fraction is called the coefficient of friction for the two surfaces in contact.
Fmax= RnFor a perfectly smooth surface% ='.$ote%
The maximum force will only act if:
there is a state of limiting e!uilibrium ormotion is taking place
$n other cases the frictional force is e!ual and opposite the applied force
-"i.e. F =- where- # Rn#
xample
& block of mass (' kg rests on a hori)ontal plane% the coefficient of the
friction between the block and the plane being '.*. +alculate the friction
force acting on the block when a hori)ontal force -is applied to the block
and the magnitude of -is:
"i# ,' "ii# -' "iii# /./
Solution
Mathematics made simple M1 Friction
- 3 -
10 kg
10g N
RnN
F P
Resolve upwards : Rn10g =0 Rn=9
Ma!imum "riction "orce Fma!=Rn Fma!=0#$ 9=%# N
&i' ( ) %# so F=20 N *here is no motion#
&ii' ( ) %# so F =+0 N *here is no motion#
&iii' ( =%# F =Fma!=%# *he ,lock is on
the point o" moving &euili,rium'#
-
8/11/2019 Friction Handout
4/9
Frictionxample
+alculate the maximum frictional force which can act when a block of
mass kg rests on a rough hori)ontal surface% the coefficient of friction
between the surfaces being
"i# '.* "ii# '.0
Solution
xample
& block of mass , kg rests on a rough hori)ontal surface. & hori)ontal
force of (- is applied to the block. $f the block is on the point of
moving% find the coefficient of friction between the block and the
surface.
Solution
xample
& block of mass kg rests on a rough hori)ontal surface. & hori)ontal
force of ,-. is applied to the block. $f the block is on the point of
moving% find the coefficient of friction between the block and the
surface.
Solution
Rn=g = 1./ =-'
Mathematics made simple M1 Friction
- + -
% kg
%g N
RnN
Fma!=Rn
2 kg
2g N
RnN
F 1+
&i' Resolve upwards : Rn%g =0 Rn=+9
Ma!imum "riction "orce Fma!=Rn Fma!=0#$ +9 =29#+ N
&ii' Resolve &as ,e"ore': Rn=+9Fma!=Rn =0#3 +9 =1+#. N
Resolve upwards : Rn2g =0 Rn=19#$
Ma!imum "riction "orce Fma!=Rn
1+0 .1+
19 $
F.
R .= = =
-
8/11/2019 Friction Handout
5/9
2+ %0 %
+9
F ..
R= = =
Friction
xample
Solution
+omponents diagram
xample
Mathematics made simple M1 Friction
- % -
$ kg
$g N
RnN
F
P
30
( cos 30
Rn ( sin30
$g N
F
& block of mass * kg rests on a rough
hori)ontal surface. The coefficient of friction
between block and the surface is '.. & force
- is applied at an angle of 0'to thehori)ontal. Find the value of -when the block
is about to move.
Resolving : ( cos 30 F =0F =( cos 30 /1
Resolving : R ( sin 30 $g=
0 R =$g ( sin30 /2F =R /3u, /1 and /2 into /3
( cos30 =0#%&%#% 0#%(' ( cos 30 =29#+ 0#2%( (&cos 30 0#2%' =29#+
( =2$#3 N
-
8/11/2019 Friction Handout
6/9
Solution
+omponents diagram
Frictionxample
Solution
*omponents dia"ram
xample
Mathematics made simple M1 Friction
- $ -
$ kg
$g N
RnN
F
P
30
( cos 30
Rn
$g ( sin30
F
& block of mass * kg rests on a rough
hori)ontal surface. The coefficient of
friction between block and the surface
is '.. & force - is applied at an angle
of 0'to the hori)ontal. Find the valueof -when the block is about to move.
Resolving : ( cos 30 F =0F =( cos 30 /1
Resolving : R ( sin 30 $g =0 R =$g ( sin30 /2F =R /3u, /1 and /2 into /3
( cos30 =0#%&%#% 0#%('
( cos 30=
29#+ 0#2%( (&cos 30 0#2%' =29#+ ( =+.#. N
& block of mass * kg rests on a rough
hori)ontal surface. The coefficient of
friction between block and the surface is
'.. The forces - and ,-are applied at an
angle of 0'and *'to the hori)ontal.Find the value of -when the block isabout to move.
2
* kg
R , 2
F$030
Resolve: 2( cos$0 ( cos30 F 0 F 2( cos$0 ( cos30
Resolve: R 2( sin$0 ( sin30 $g 0 R $g 2( sin$0 ( sin30
F =R /3u, /1 and /2 into /3
( 0#$$0( =0#%&%# (&1#.320#%''1#$$0( =29#+ 0#$1$(2#+2( =29#+
(=11# N
& block of mass * kg rests on a rough
hori)ontal surface. & force of , is applied
at an angle of 0'to the hori)ontal. $f theblock is about to move% find the coefficient of
friction between the block and the surface.
-
8/11/2019 Friction Handout
7/9
Solution
+omponents diagram
Frictionxample
& block of mass m kg rests on a rough plane inclined at 0'to thehori)ontal. The coefficient of friction between the block and the plane is
'.. & force -acts on the block up the plane along the line of greatest
slope. Find the possible values of -if the block remains stationary.
Solution*here are two possi,ilities: &a' *he ,lock is a,out to move down the slope4
&,' *he ,lock is a,out to move up the slope#
&a'
Resolving parallel to the plane : ( F mg sin 30 =0(=+#9m F /1
Resolving perpendicular to the plane: Rnmg cos 30 =0Rn=#+. m /2
F =R 3042 =-.1m '."/.-/5m#
2 ='.*5 m
Mathematics made simple M1 Friction
- . -
2(cos$0
( cos30F
( sin30 $g
$ kg
$g N
F
30
2% cos 30
Rn 2% sin30
$g N
F
m N
30
RnN
( N
30
Rn
mg cos30
mg sin30
(
Force diagram Parallel and Perpendicular components
F
Resolving : 2% cos 30 F =0F =2% cos 30 /1F =21#$%1
Resolving : R 2% sin 30 $g=
0 R =+$#3 /2
21 $%10 +$
+$ 3
F ..
R .= = =
-
8/11/2019 Friction Handout
8/9
"b#
So/ the ran"e of 0alues of - is% 1.234m - 5.'6 m.
Frictionxample
& block of mass m kg rests on a rough plane inclined at 0'to thehori)ontal. The coefficient of friction between the block and the plane is
'.. & hori)ontal force -acts on the block. Find the possible values of -if
the block remains stationary.
Solution
There are two possibilities: "a# The block is about to move down the slope%
"b# The block is about to move up the slope.
+a,
+b,
Mathematics made simple M1 Friction
- -
30
Rn
mg cos30
mg sin30
(
Parallel and Perpendicular components
F
30
( N
mg N
Rn
(cos30
mg cos30 (sin30
Rn F
mg sin30
2arallel: 2 F 6 mg sin0' 7 '
2 7 F 8 -.1m
2erpendicular: Rn9 mg cos0' ='
Rn=/.-/5 mF =R 2 ='. /.-/5 m 8 -.1 m 2 =1.(- m
Resolving parallel:
( cos30 F mg sin30 =0 /1Resolving perpendicular:
Rn- mg cos 30 (sin30 =0 /2
F =R( cos30 mg sin30 0#%&mg cos30 (sin30' =0
0#$$0( 0#%mg 0#+330mg 0#2%(=0
1#11$(=0#0$.mg (=0#0$ mg
-
8/11/2019 Friction Handout
9/9
So/ the ran"e of 0alues of - is% 1.12 m" - '.3' m".
Mathematics made simple M1 Friction
- 9 -
(cos30
mg cos30 (sin30
Rn
Fmg sin30
Resolving parallel:
( cos30 F mg sin30 =0 /1
Resolving perpendicular:
Rn-
mg cos 30 (sin30=
0 /2
F =R( cos30 mg sin30 0#%&mg cos30 (sin30' =0
0#$$0( 0#%mg 0#+330mg 0#2%( =0
0#$1$( =0#933mg (=1#%1 mg