Applications of
Newton’s Laws
Mark Lesmeister
Dawson High School Physics
Acknowledgements
© 2013 Mark Lesmeister/Pearland ISD
 This work is licensed under the Creative Commons Attribution-
ShareAlike 3.0 Unported License. To view a copy of this license, visit
http://creativecommons.org/licenses/by-sa/3.0/ or send a letter to
Creative Commons, 444 Castro Street, Suite 900, Mountain View,
California, 94041, USA.
 Selected graphics and problems from OpenStax College. (2012, June
12). College Physics. Retrieved from the Connexions Web site:
http://cnx.org/content/col11406/1.7/
Inclined Planes
Part 1
Inclined Planes
 A box slides down two smooth rails with no friction. Find
the acceleration of the box.
q
FN
θ
Fg = mg
Example: Forces on an Incline
• A 40 kg wagon is
towed up a hill at an
18.5o incline. The tow
rope exerts a force of
140 N. The wagon
starts from rest.
• How fast is the wagon
going after 30 m?
Fapp=140N
q=18.5o
Givens, Unknowns and Models
•
•
•
•
•
m = 40.0 kg
Fapp = 140 N
q = 18.5o
Dx = 30 m
v0 = 0
Fn
Fapp
• v=?
• This is constant force in
the x direction (and
constant acceleration)
and equilibrium in y.
q mg cos(q)
mg
mg sin(q)
Method
v 2 = v02  2a x Dx
F
x
= max
Fn
Fapp
Fapp  mg sin( q ) = max
Fapp  mg sin( q )
m
q mg cos(q)
= ax
mg
mg sin(q)
Implement
Fapp  mg sin( q )
= ax
m
140 N - (40 kg)(9.81 m/s 2 )(sin(18.5 o ) )
= ax
40 kg
0.3872 m/s 2 = a x
v 2 = v02  2a x Dx
v 2 = 0  2(0.3872 m/s 2 )(30 m) = 23.232
v = 4.82 m/s
Evaluate the Solution
Fapp=140N
v = 4.82 m/s
q=18.5o
This is about 11 mi/hr, a
reasonable speed. It is positive,
so the wagon moves uphill, as
expected.
Friction
Part 2
Discuss this question and be prepared
to answer.
 What force pushes a car
forward?
 Gravity pulls down.
 The normal force
pushes upward.
 The car’s engine
pushes on the wheels,
but that is not an
outside force.
Resistive Forces
 Resistive forces, such as friction and
drag forces, are forces which oppose
the relative motion of two surfaces,
or a surface and a fluid.
 Friction occurs when two solid
surfaces interact.
 Drag forces occur in fluids, i.e.
liquids and gases.
Friction
 Friction is really just one
component of the force
between two surfaces.
 Friction is the force of
interaction parallel to the
surfaces.
 The perpendicular
component is the normal
force.
 We generally treat friction
and the normal force as
separate forces.
FN
FF
Static and Kinetic Friction
 The amount of friction between an object and a surface
depends on whether the object is moving relative to the
surface.
 Static friction applies when the object is at rest relative to the
surface.
 Kinetic friction applies when the object is moving relative to
the surface.
Static Friction
 The force is equal to the
applied force, until it exceeds
a certain maximum.

Fs  Fs,max
 Once the maximum force of
static friction is exceeded,
the object breaks free and
begins to slide.
Kinetic Friction
 Kinetic friction applies
when the object is moving
across the surface.
 The force of kinetic
friction is typically less
than the corresponding
maximum force of static
friction.
FN
 In other words, once
things are moving they are
easier to keep moving.
Fk
Fg
Accelerating a car.
 The force the road applies
to accelerate a car (speed
up, slow down or change
direction) is a frictional
force.
 If the wheels do not slip,
the friction can be treated
as static friction.
 If the wheels slip, kinetic
friction is more
appropriate.
FF
 This tire will accelerate to
the right because of the
force of friction applied by
the road.
Rolling Friction
 If the car moves along at
constant velocity, ideally no
forces are needed to
accelerate it.
 However, real tires deform,
and the road and tire must be
continually peeled apart as
the tire rolls.
 A small force is needed to
overcome the rolling friction
to keep the car moving with
constant velocity.
Frolling Friction
 This tire will slow down if
no other force is applied
because rolling friction
opposes the motion.
Mini-lab- Friction
 What factors affect the amount of the frictional force
between two surfaces?
 Using a force sensor and blocks of wood, investigate your
answers to the above question.
 Every group should investigate the relationship between
normal force and friction, and at least one other factor.
Lab Results
 Friction generally doesn’t depend on the surface area
between two objects.
 Objects with less surface area have more normal force / unit
area.
 Friction depends directly on the normal force between two
objects.
 The types of surfaces in contact make a difference in friction
observed.
Kinetic Friction and the Normal Force
 The force of kinetic
friction is given by the
formula
Fk =  k FN
 FN is the normal force
between the surfaces.
 μK is the coefficient of
kinetic friction, which
depends on the nature of
the surfaces in contact.
FN
Fk
Fg
Maximum Force of Static Friction
 The maximum force of
static friction is given by
FApp
Fs ,max =  s FN
 μs depends on the surfaces
in contact. It is usually
larger than the
corresponding μk.
FN
Fs
Fg
Kinetic Friction Example: Eins, Zwei,
Soffa
 A bartender slides a 0.45 kg beer stein horizontally along a
bar with an initial speed of 3.5 m/s. (The stein contains root
beer of course!) The stein comes to rest near the customer
after sliding 2.8 meters. Find the coefficient of kinetic
friction.
Static Friction Example: Limiting Angle
of Repose
FS
FN
Fg = mg
Given : q MAX = Maximum angle before sliding starts
Unknown : S = coefficient of static friction
Model : Equilibrium in both directions. Static friction.
 Fx = 0
 Fy = 0
Fg  x  Fs = 0
FN  Fg  y = 0
Fg  x = Fs
FN = Fg  y
mg sin q = Fs
FN = mg cos q
mg sin q max = Fs max
s =
Fs max
FN
mg sin q max
s =
mg cos q max
 s = tan q max
Practice Problems: Friction
 OpenStax College Physics textbook, “Friction”, Exercises 5
and 6.
Practice Problems: Friction
 P. 153, #37,44
Drag Forces
Part 3
Drag Forces
Source: Wikipedia commons, “B-52 landing with drogue chute”
What factors influence drag?
 Shape of the object.
 Cars and planes are given aerodynamic shape to reduce drag.
 Properties of the fluid.
 Denser fluids increase drag.
 Speed of the object relative to the fluid.
 The simplest case is when Fd is proportional to v.
Fd  bv
This gives the magnitude of Fd .
The direction is opposite the flow.
Terminal Speed
 Because the drag force
increases with increasing
speed, a falling object will
reach a speed where the
drag force balances the
force of gravity.
 This speed is called
terminal speed.
Terminal Speed
Fd  mg = ma y

bv  mg = ma y
 At terminal speed
bvT  mg = 0
mg
vT =
b
Inclined Planes with Friction
Practice 4D, #3
 A 75 kg box slides down a
25.0o ramp with an
acceleration of 3.60 m/s2.
 Find k between the box
and the ramp.
q=25o
Givens, Unknowns and Models
 m = 75 kg
 q = 25o
 ax= -3.60 m/s2
q=25o
 k = ?
 Constant force in x
FN
direction, equilibrium in y.
Fk
q mg cos(q)
mg
mg sin(q)
Method
F
F
x
= max
y
=0
Fn
Fk
q mg cos(q)
Use x equation to find Fk.
Use y equation to find FN.
Use Fk and FN to find 
Fk  mg sin( q ) = max
mg
mg sin(q)
FN  mg cos(q ) = 0
Implement
F
F
x
= max
y
=0
Fk  mg sin( q ) = max
FN
Fk
q mg cos(q)
mg
mg sin(q)
FN  mg cos(q ) = 0
FN
Fk
q mg cos(q)
mg
mg sin(q)
Fk  mg sin( q ) = max
Fk = max  mg sin( q )
FN  mg cos(q ) = 0
FN = mg cos(q )
Fk ma x  mg sin(q )
k =
=
FN
mg cos(q )
Fk m(a x  g sin(q )) (a x  g sin(q ))
k =
=
=
FN
mg cos(q )
g cos(q )
Fn
Fk
q
mg
k =
ax  g sin( q )
g cos(q )
 3.6 m/s 2  (9.81 m/s 2 sin( 25o ))
k =
9.8 m/s 2 cos( 25o )
k = .061
Homework 4c: 11-14
 A 3.00 kg block starts from rest at the top of a 30.0o incline
and accelerates uniformly down the incline, moving 2.00 m
in 1.50 s.
 What is the magnitude of the acceleration of the block?
 What is the coefficient of kinetic friction?
 What is the frictional force acting on the block?
 What is the speed of the block at the end?
Givens, Unknown and Model
 m = 3.00 kg
 q = 30.0o
 Dx = -2.00 m
 vi = 0
q
 Dt = 1.50 s
Fn
 ax = ?
Fk = ?
 k = ?
vf= ?
 Equilibrium in y, acceleration
in x.
FK
q mg cos(q)
mg
mg sin(q)
Finding a: Method, Implement and
Evaluate Solution
Dx = v0t  12 a x t 2
Dx = a x t
1
2
Fn
2
2 Dx = a x t 2
2 Dx
= ax
2
t
2(2 m)
= ax
2
(1.5 s)
 1.78 m/s 2 = a x
FK
q mg cos(q)
mg
mg sin(q)
Finding FK: Method, Implement and
Evaluate Solution
F
x
= max
Fn
Fk  mg sin( q ) = max
Fk = max  mg sin( q )
Fk = m(a x  g sin( q ))
q mg cos(q)
mg
Fk = (3.00 kg) (1.78 m/s 2  (9.81 m/s 2 ) sin( 30o ))
Fk = 9.38 N
FK
mg sin(q)
Finding μK: Method, Implement and
Evaluate Solution
Fk
k =
FN
 Fy = 0
Fn
FN  mg cos(q ) = 0
q mg cos(q)
FN = mg cos(q )
mg
FN = (3.00 kg)(9.81 m/s ) cos(30 )
2
FN = 25.5 N
k =
9.38 N
= 0.370
25.5 N
FK
o
mg sin(q)
Forces Applied At an Angle
Part 4
Practice 4D, #1
 A student moves a box of books down the hall by pulling
on a rope attached to the box. The student pulls with a
force of 185 N at and angle of 25.0o above the horizontal.
The box has a mass of 35.0 kg, and μK between the box
and the floor is 0.27. Find the acceleration of the box.
Identify G.U.M.
FN
Fapp
 m= 35 kg
 q = 25o
 Fapp = 185 N
FK
 K= 0.27
Fg
Y- direction- Equilibrium
X-direction- Constant Force
 ax = ?
Fapp, x = Fapp cos q
Fapp, y = Fapp sin q
Method
F
x
= max
Fapp, x  FK = max
Fapp cos q  FK = max
We need to find FK .
FK =  K FN
We need to find FN
FN
FK
Fg
Fapp
Method
F
y
=0
FN  Fapp, y  mg = 0
FN = mg  Fapp, y
FN
FK
FN = mg  Fapp sin q
Fg
Fapp
Implement
Using the y - direction equation above :
FN = (35 kg)(9.81 m/s 2 )  (185 N)sin(25 o )
FN = 265.166 N
Using the coefficien t of friction equation :
FK = (.27)(265.1 66 N) = 71.594 N
Using the x - direction equation above :
185 N(cos 25o )  71.594 = max
96.07 = max
96.07 N
= ax
35 kg
a x = 2.7 m/s 2
Evaluate the solution.
 Our answer is about 1/3 of g. This is a reasonable
acceleration for an object to have over a short period of time.
Chapter Review, # 39
 A 4.00 kg block is pushed along the ceiling with a constant
applied force of 85.0 N that acts at an angle of 55.0o with the
horizontal. The block accelerates to the right at 6.00 m/s2.
Determine the coefficient of kinetic friction between the
block and the ceiling.
G.U.M.
Fapp
 Note that both FN and Fg

FK

FN


Fg
Fapp, x = Fapp cos q
Fapp, y = Fapp sin q


point down.
Fapp = 85 N
m = 4.00 kg
q = 55o
a=6 m/s2
K = ?
Accelerates in x,
equilibrium in y.
Method
K =
FK
FN
We will use the x - equation t o find FK , and
the y - equation t o find FN .
F
x
= max
Fapp, x  FK = max
FK = Fapp, x  max
FK = Fapp cos q  max
F
y
=0
Fapp, y  mg  FN = 0
Fapp, y  mg = FN
Fapp sin q  mg = FN
Implement and Evaluate the Solution
FK = 85 N cos(55 o )  4 kg(6 m/s 2 )
FK = 24.75 N
FN = 85 N sin(55 o )  4 kg(9.81 m/s 2 )
FN = 30.39 N
24.75
K =
= 0.814
30.39
This is less than 1, and is a reasonable number given
how much force friction appears to exert here.