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Chapter 6
Force and Motion II
Forces of Friction

When an object is in motion on a
surface or through a viscous medium,
there will be a resistance to the motion


This is due to the interactions between the
object and its environment
This is resistance is called the force of
friction
More About Friction





Friction is proportional to the normal force
The force of static friction is generally greater
than the force of kinetic friction
The coefficient of friction (µ) depends on the
surfaces in contact
The direction of the frictional force is opposite
the direction of motion
The coefficients of friction are nearly
independent of the area of contact
Static Friction, ƒs
Static friction acts to
keep the object from
moving
 If F increases, so does ƒs
 If F decreases, so does ƒs
 ƒs  µ n

Kinetic Friction
The force of
kinetic friction
acts when the
object is in
motion
 ƒk = µ n

Example:

A Block of mass m1 on a
rough, horizontal surface
is connected to a ball of
mass m2 by a lightweight
cord over a lightweight
frictionless pulley. A force
of magnitude F at an angle q with the horizontal is
applied to the block as shown. The coefficient of
Kinetic friction between block and surface is mk.
Determine the acceleration of the objects.
Solution:
Free Body diagrams:
Newton’s 2nd Law:
X-axis: F cosq – T – fk = m1a
(1)
Y-axis: F sinq + n – m1g = 0
(2)
(2)  n = m1g – F sinq  fk = mkn = mk(m1g – F sinq)
(1)  T = Fcosq - mk(m1g – F sinq) - m1a
But Also,
T = m2a – m2g
Solution Cont’d:

T=T
Fcosq - mk(m1g – F sinq) - m1a = m2a – m2g
F(cosq  μk sinq ) - g(m2  μk m1 )
a
m1  m2
Knowing the value of a one can easily deduce
the value of T = m2a – m2g.
Connected Objects




Apply Newton’s Laws
separately to each
object
The acceleration of both
objects will be the same
The tension is the same
in each diagram
Solve the simultaneous
equations
More About Connected
Objects

Treating the system as one object
allows an alternative method or a check

Use only external forces



Not the tension – it’s internal
The mass is the mass of the system
Doesn’t tell you anything about any
internal forces
Newton’s Law applied to uniform
circular motion:

A particle moving with uniform speed v in a circular path
of radius r, experiences a centripetal acceleration ac
that has a magnitude:
v2
| ac |
r
ac is directed toward the
center of the circle and is
always  to v.
Newton’s Law applied to uniform
circular motion Cont’d:

If we apply Newton’s law along the radial
direction,

2
v
F  mac  m
r
F acts toward the center
of the circle.
Newton’s Law applied to uniform
circular motion Cont’d Ex:

The ball is put in a
circular motion on a
frictionless table.
If the string breaks at
some point how the
ball moves afterward?
Tangential (at) and Radial (aN)
Acceleration

In the case when the speed changes
along a circular path the total Acceleration
vector is given by:
a = at + aN
Where,
|aN| = v2/r (directed toward the center of the circle)
d|v|
|at| =
(directed tangent to the circle)
dt
Tangential (at) and Radial (aN)
Acceleration Cont’d:
aN
at
a
a = at + aN
|aN| = v2/r
|at| =
d|v|
dt
Example
The above figure shows the path of a park ride that travels at
constant speed through five circular arcs of radii R0, 2R0, and
3R0. Rank the arcs according to the magnitude of the
centripetal force on a rider traveling in the arcs, greatest first.
423.15
Application:

A ball of mass m= 0.500 Kg is attached to the
end of a cord 1.50 m long. The ball is whirled
in a horizontal circle.
If the ball can withstand a maximum tension of
50.0 N, what is the maximum speed the ball
can attain before the cord breaks.
x
Solution:
Applying Newton’s Law:

v2
F  mac  m
r
v2
T m
r

Maximum tension correspond to max v:
VMax 
r  TMax
1.5  50

 150  12.2m/s
m
0.5
N
x
T
mg
Example:

A small object of mass
m is suspended from a
string of length L. The
object revolves with
constant speed v in a
horizontal circle of
radius r, Find an
expression for v as
function of T, r and m?
Solution:

ApplyingNewton’s
Law:


mg  T  ma
Projection along the
horizontal axis:
2
v
(1)
Tsin(q )  m
Ac on x-axis
r
Projection along the vertical
axis:
(2)
Tcos(q )  mg
v2
Dividing (1) by (2)  tanq 
 v  r.g.tanθ
rg
Example:

A 1500-kg car moving on a
flat, horizontal road negotiates
a curve, (See Figure). If the
radius of the curve is 35.0 m
and the coefficient of static
friction between the tires and
dry pavement is 0.500, find
the maximum speed the car
can have and still make the
turn successfully.
Solution:

Force of friction enables the
car to remains in its circular
path.
  

mg  fs  n  ma
Projection along the radius:
2
v
But, fs, max  μsn  μsmg
fs  m
r
2
vmax
 vmax  μs .g.r
 μsmg  m
r
 0.500 9.80 35.0  13.1m/s
More Applications:

A small sphere of mass m
is attached to the end of
a cord of length R and set
into motion in a vertical
circle about a fixed point
O. Determine the tension
in the cord at any instant
when the speed of the
sphere is v and the cord
makes an angle q with
the vertical.
Solution:

Note that along the path
the speed is not constant.
 

mg  T  ma
Projection along the radius:
v2
- mgcos q  T  m
R
 v2

 T  m
 gcosq 
 R



Application Cont’d
What speed would the ball have
as it passes over the top of the
circle if the tension in the cord
goes to zero instantaneously?
 v2


T m
 gcosq 
 R



At the top q = 180°
 v2

 T  m
 g
 R



T = 0, (i.e. cord breaks)  v top  gR
Motion in the Presence of
Resistive Forces:
The medium through which the object moves
exerts a resistive force R on it.
 The medium can be either a liquid or a gas.
 Some examples are the air resistance
associated with moving vehicles (called also
air Drag) and the viscous forces that act on a
object moving through a liquid.

Resistive Forces:
The direction of the resistive force R is always
opposite to the direction of motion.
 The Magnitude of R is proportional to the speed
of the object:
R = -b v
v is the velocity of the object and b is a constant
whose value depends on the properties of the
medium and on the shape and dimensions of the
object.

Terminal Speed
When the Drag force ( upward air
resistance for example) equals the
driving force (downward force of
gravity), the net force on the object is
zero.
 The constant speed of the object is the

terminal speed
Application:

Consider a small
sphere of mass m
released from rest in
a liquid as shown in
the adjacent figure,
assuming that the
buoyant force is
negligible, describe
the motion of the
sphere.
Application Cont’d:
The forces acting
on the sphere are:
1) Resistive Force:
R =-b v
2) The gravitational
Force:
Fg = mg

Application Cont’d:


Applying Newton’s law:
R
dv
dv
b
mg - b v  m

 g v
dt
dt
m
mg
We can obtain the terminal
speed by putting dv/dt = 0.
mg – b vT = 0
 vT 
mg
b
(+) y
Application Cont’d:

The Equation
dv
b
 g v
dt
m
is called a differential equation.