Direction of Frictional Force
Friction
Push an object across a table at
a constant velocity.
Can we always determine the direction
of the kinetic frictional force?
Magnitude of Kinetic Friction
Forces on the block:
+y
+x
v
P
+x
Try a simpler situation.
P
v
Block sliding on table with no one pushing.
T
Fk
Motion in the x direction: ax = 0,
v
The object always slows down
There is a force P in the +x direction
The kinetic frictional force can never
make an object go faster.
There must be an additional force
in the -x direction.
The only contact of our object with another
object is the contact with the table.
The kinetic frictional force is always
in the direction opposite to the
velocity of the object.
The additional force must be caused by the
table and be parallel to its surface.
What determines the magnitude of this force?
T
The simplest dependence is linear.
+x
N
P is in the y direction and cannot affect
motion in the x direction
Increasing P must increase a
force in the x direction
The kinetic frictional force increases.
How?
Static Friction
Kinetic Friction
+y
Fk = k N
W
W
If you push down on an object, it takes a
larger force T to move the object at a
constant velocity.
This is an experimental question which you
will work on in the lab.
We name this force "kinetic friction", Fk.
Fk
N
a
Σ Fx = m ax = 0
P
+y
Pull a block across the table
An object is at rest relative to the
surface supporting it.
You push on it and it still doesn’t move
Fk is in the direction opposite to v.
v=0
P
Fk is not in the direction of N
Motion in the y direction: ay = 0 , Σ Fy = m ay
Σ Fy =N + (-W ) + (-P ) = 0
N=W+P
Increasing P, increases N
N is an interaction acting at the between the
table surface and the block.
Same place that Fk is acting.
It is reasonable that the kinetic frictional force
increases as the normal force increases.
+y
How something slides depends on the
material on the two surfaces that slide.
k = µ k the coefficient of kinetic friction
P
N
W
P
+x
N
W
The theory of the behavior of kinetic friction:
Fk = µ k FN in the direction opposite to v.
If the object doesn't accelerate,
there must be another force in the - x direction.
A force between the surface on the object
and the surface of the table.
Static frictional force.
Direction and Magnitude
Direction
Opposite to P in direction parallel
to surface
Example Problem
Static Friction
Simple theory of static friction:
Fs (maximum) = µs N
While cross country skiing, you find you are
on the top of a small hill. The hill side is a
Direction is opposite to the acceleration that
would occur without Fs
Magnitude
gentle slope at an angle of 20 degrees to the
horizontal. You are new at this and are a bit
Varies
You push harder,
the block still doesn't move
afraid of going too fast since you are not
This theory is a simplified
generalization of experience.
good at stopping. You stop at the top of the
hill and estimate that the hill side goes 100
If you push hard enough,
the block does accelerate.
Check its applicability in your lab
feet before it gets to the bottom. How fast
will you be going at the bottom of the hill?
The static frictional force has a maximum.
Maximum depends on:
Materials on the contact surface.
Coefficient of static friction, µs
The coefficient of kinetic friction between
your skis and the dry snow is 0.04.
How hard the surface pushes
on the object
Normal force, N
N
v − vo v f
a=a= f
=
t f − to t f
N
a
W
θ
f
f
20o
+x
N
Question:
What is the speed at the bottom of the hill?
Approach:
Get final speed from the acceleration.
Get acceleration from the forces.
Perpendicular components of
motion are independent.
Interested in motion along
the hill's slope.
Choose one axis of coordinate system
along the slope of the hill. Call it x.
for constant acceleration
+y
µk = 0.04
100 ft
Relevant equations
Force diagram
Free-body diagram
v
W
f
W
Since the forces on
the skier do not change
with time, ao = af= a
v0 = 0
x0
t0
Target : vf
Σ Fx = m ax
Wx − f = ma
N − Wy = 0
a
f=µN
W=mg
a
xf
tf
θ = 20°
µk = 0.04
1 2
at
2 f
Σ Fy = m ay
Motion diagram
xo = 0
to = 0
v0 = 0
a=?
xf =
+y
N
vf
xf = 100 ft
tf = ?
vf = ?
θ
components
f
+x
θ
W
W
sin θ = x
W
cos θ =
Wy
W
tan θ =
Wx
Wy
Plan:
unknown
Find vf
vf
a=
vf
tf
1
a , tf
vf
2xf
(g sin θ − µg cos θ )
g sin θ − µg cos θ =
Do the algebra and see if it does
N − mg cos θ = 0
Find tf
xf =
Perhaps m cancels out.
N = mg cos θ
1 2
at
2 f
(g sin θ − µg cos θ )
2
f = µ mg cosθ
θ
Find a
mg sin θ − f = ma
3
f, m
f=µN
4
N
Find N
N − mg cos θ = 0
5
6 unknowns but only 5 equations
g sin θ − µg cos θ = a
2xf g(sin θ − µ cos θ ) = v f
Check units
1
x f = (g sin θ − µg cos θ )t f2
2

s
2xf
=t
(g sin θ − µg cos θ ) f
 m
  = vf
 s
• Picture with quantities and coordinate system
Situation, geometry, motion, forces.
Coordinate system x axis along acceleration
vf = 44 ft/s
Evaluate:
• Free-body and force diagrams
• Final velocity from constant acceleration
kinematics
A 20-kg crate sitting on a horizontal floor is
attached to a rope that pulls 37 degrees
above the horizontal. The coefficient of static
friction between the crate and floor is 0.50.
Determine the least rope tension that will
cause the crate to start sliding.
20 kg
v= 0
T
Units of velocity are correct, ft/s
A fast runner goes 100 m in 10 sec.
which is 10 m/s
The skier goes about 14 m/s
It's fast but possible
The is the speed of the skier at the bottom
of the hill. The question is answered
ok
EXAMPLE
Review
ft
(sin20o −0. 04cos20o ) = vf
s2

[m] m2  = vf
Is there another equation for m??
2(100ft)(32
= vf
mg sin θ − µmg cos θ = ma
Yes! m cancels out
Find f
2xf
(g sin θ − µg cos θ )
37o
f
• Acceleration from forces in x direction
Σ Fx = m ax
• Add forces in x direction
W = mg
Fk = µk FN
• Normal force from motion in y direction
ay = 0
Σ Fy = m ay
W
N
µ = 0.50
Question:
What is force of rope on crate
necessary to start the crate sliding?
Approach:
Least rope tension to start sliding when its
horizontal component equals
the maximum static frictional force
Components of forces
horizontal
vertical to get normal force
Consider crate just before it starts moving
Free-body diagram
N
f
T
f
θ
+x
N
W
T
1
T sinθ
θ = Ty
Tx
Tx - f = 0
Σ Fx = m ax
Tx - f = 0 (since ax = 0)
f max = Tx
Σ FY = m aY
2
f max
Find f max
N - W -Ty = 0 (since ay =0)
f max = µs N
W = mg
3
N - mg -Ty = 0
Ty
T
cos θ = x
T
T
f max = µs (- T sinθ
θ + mg)
3
µs (- T sinθ
θ + mg) = Tx
2
cosθ
θ = µs (- T sinθ
θ + mg)
T
1
T cosθ
θ = µs ( - T sinθ
θ + m g)
T cosθ
θ + µs T sinθ
θ = µs m g
Forces in the y direction:
Components
4
T (cosθ
θ + µs sinθ
θ ) = µs m g
Ty
Find Ty
Ty
sinθ
θ= T
Target : T
T = µs mg
(cosθ
θ + µs sinθ
θ)
4
N
Find N
f ≤ µs N
5
- T sinθ
θ + mg = N
Find Tx
forces in the x direction
Relevant equations
sinθ
θ=
unknown
Find T
cosθ
θ = Tx
T
+y
T
W
The plan:
Force diagram
T=
5
5 unknowns, 5 equations
Can be solved.
Example - Chap 6 -55
10°
µs m g
cosθ
θ + µs sinθ
θ
What is the speed
of the object?
0.5 m
0.2 kg
r
2
T = (.50) (20 kg) (9.8 m/s )
o
[ (cos 37 ) + (.50) (sin 37 o )]
2
T = 98 (kg m/s ) = 89 kg m/s
(1.1)
2
T = 89 N
Evaluate:
Correct units: force is in Newtons
Force required to lift box is
(20 kg)(10 m/s2 ) = 200 N
89 N is less than that which
is reasonable
The least rope tension for the box to
slide is when the static frictional force
is at its maximum. The question is
answered.
A conical pendulum is an object which is
free to move at the end of a very light string
with the other end of the string fixed.
Given the proper push, the object can
swing in a horizontal circle such that the
string is always at the same angle to the
vertical and the object maintains the same
distance from the fixed point. If the mass
of the object is 0.20 kg, the length of the
string is 0.50 m, and the angle from the
vertical is 10°, what is the speed of the
object?
v
Kinematics for circular motion relates the
speed of an object to its acceleration.
Can get acceleration of the object from
the forces on the object.
Choose one coordinate along the radius of
the circle.
Ignore air resistance and any friction of
the string rubbing against the support.
Free body diagram
of object
Force diagram
of object
+y
T
T
unknowns
θ
+x
a
W
Geometry
a, r
1
Tx
sin θ =
T
m = 0.20 kg
L
cos θ =
θ = 10°
r
sin θ = L
r
tan θ =
Ty
T
Tx
Ty
Forces in x direction: Tx = ma
Forces in y direction:
g tan θ = a
Tx = ma
Tx
3
2
g tan θ = v
r
Find Tx
Tx
Ty
4
Find Ty
Ty - W = may = 0
v2
a=
r
W = mg
mg tan θ = ma
2
Find a
tan θ =
Ty - mg = 0
Check Units:
sin θ
Lg
=v
cos θ
[m]
m m
=
 s2  s
ok
(0.50m ) 9.8 m2 

s  = v = 0.39 m
s
cos 10o
m/s are correct units for a speed.
The question was answered by finding the
speed of the object.
Circle has a radius of L sin θ = r = 0.09 m
Circumference = 0.55m
Time to go around circle =Period = 1.4 second
A reasonable time
0.4 m/s is not an unreasonable speed for
an object swinging at the end of a string.
Ty
L sin θ = r
g tan θ =
5
5 unknowns, 5 equations ok
Target: v
Tx
= mg
tan θ
Tx = mg tan θ
r
sin θ = L
L = 0.50 m
θ
sin 10
v
Find r
W
o
Find v
v2
a=
r
Ty = mg
v2
L sin θ
L g sin θ tan θ = v2
sin θ
Lg
=v
cos θ