Here is the free body diagram of the block, and the corresponding table of forces for
Example 15-4:
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Chapter 15 Newton’s Laws #2: Kinds of Forces, Creating Free Body Diagrams
BEWARE: By chance, in the examples
provided in this chapter, the normal force
UP
is upward. Never assume it to be upward.
The normal force is perpendicular to, and
a
away from, the surface exerting it. It
F
happens to be upward in the examples
N
because the object is in contact with the
top of a horizontal surface. If the surface
m
EAST
is a wall, the normal force is horizontal; if
F
S
it is a ceiling, downward; if an incline,
perpendicular to and away from the
incline. Never assume the normal force
F
g
to be upward.
Table of Forces
Symbol=? Name
Agent
Victim
Normal
The Horizontal
The
F
N
Force
Surface
Block
The
F = k x Spring Force The Spring
S
Block
F=mg
Gravitational The Earth’s
The
g
Force
Gravitational Field
Block
Example 15-5
From your vantage point, a crate of mass m is sliding rightward on a flat level
concrete floor. Nothing solid is in contact with the crate except for the floor. Draw
the free body diagram of the crate. Provide the corresponding table of forces.
From our experience with objects sliding on concrete floors, we know that the crate is slowing
down at the instant under consideration. It is slowing because of kinetic friction.
Kinetic Friction
A surface, upon which an object is sliding, exerts (in addition to the normal force) a
retarding force on that object. The retarding force is in the direction opposite that of the
velocity of the object. In the case of an object sliding on a dry surface of a solid body
(such as a floor) we call the retarding force a kinetic frictional force. Kinetic means
motion and we include the adjective kinetic to make it clear that we are dealing with an
object that is in motion.
The kinetic frictional formula given below is an empirical result. This means that it is
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Chapter 15 Newton’s Laws #2: Kinds of Forces, Creating Free Body Diagrams
derived directly from experimental results. It works only in the case of objects sliding on dry
surfaces. It does not apply, for instance, to the case of an object sliding on a greased surface.
We use the symbol F
k for the kinetic friction force. The kinetic frictional formula reads
f
F = µ F (15-3)
k
N
f
K
F
N is the magnitude of the normal force. Its presence in the formula indicates that the more
strongly the surface is pressing on the object, the greater the frictional force.
µ (mu-sub-K) is called the coefficient of kinetic friction. Its value depends on the materials of
K
which both the object and the surface are made as well as the smoothness of the two contact
surfaces. It has no units. It is just a number. The magnitude of the kinetic frictional force is
some fraction of the magnitude of the normal force; µ is that fraction. Values of µ for
K
K
various pairs of materials can be found in handbooks. They tend to fall between 0 and 1. The
actual value for a given pair of materials depends on the smoothness of the surface and is
typically quoted with but a single significant digit.
IMPORTANT: µ is a coefficient (with no units) used in calculating the frictional force. It is
K
not a force itself.
Here is the free body diagram and the table of forces for the case at hand. The crate is moving
rightward and slowing down—it has a leftward acceleration.
Table of Forces
a
Symbol
Name
Agent
Victim
The Concrete The
F
N
F
N
Normal Force Floor
Crate
Kinetic
The Concrete The
F = µ F
k
N
f
K
Friction Force Floor
Crate
m
The Earth’s
F
The
Gravitational
F
g = mg
Gravitational
k
Force
Crate
f
Field
F
g
Note: Not every object has a normal (“perpendicular-to-the-surface”) force acting on it. If the
object is not in contact with a surface, then there is no normal force acting on the object.
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Chapter 15 Newton’s Laws #2: Kinds of Forces, Creating Free Body Diagrams
Example 15-6
A person has pushed a brick along a tile floor toward an eastward-facing wall
trapping a spring of unstretched length L and force constant k between the wall and
o
the end of the brick that is facing the wall. That end of the brick is a distance d from
the wall. The person has released the brick, but the spring is unable to budge it—
the brick remains exactly where it was when the person released it. Draw the free
body diagram for the brick and provide the corresponding table of forces.
A frictional force is acting on an object at rest. Typically, an object at rest clings more strongly
to the surface with which it is in contact than the same object does when it is sliding across the
same surface. What we have here is a case of static friction.
Static Friction Force
A surface that is not frictionless can exert a static friction force on an object that is in
contact with that surface. The force of static friction is parallel to the surface. It is in the
direction opposite the direction of impending motion of the stationary object. The
direction of impending motion is the direction in which the object would accelerate if
there was no static friction.
In general, there is no formula for calculating static friction—to solve for the force of
static friction, you use Newton’s 2nd Law. The force of static friction is whatever it has to
be to make the net parallel-to-the-surface force zero.
We use the symbol F
sf to represent the magnitude of the static friction force.
SPECIAL CASE: Imagine trying to push a refrigerator across the floor.
Imagine that you push horizontally, and that you gradually increase the force
with which you are pushing. Initially, the harder you push, the bigger the force
of static friction. But it can’t grow forever. There is a maximum possible static
friction force magnitude for any such case. Once the magnitude of your force
exceeds that, the refrigerator will start sliding. The maximum possible force of
static friction is given by:
possible
max
F
= µ F (15-4)
sf
S
N
The unitless quantity µ is the coefficient of static friction specific to the type of
S
surface the object is sliding on and the nature of the surface of the object.
Values of µ tend to fall between 0 and 1. For a particular pair of surfaces, µ
S
S
is at least as large as, and typically larger than, µ .
K
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Chapter 15 Newton’s Laws #2: Kinds of Forces, Creating Free Body Diagrams
Clearly, this formula (
possible
max
F
= µ F ) is only applicable when the
sf
S
N
question is about the maximum possible force of static friction. You can use
this formula if the object is said to be on the verge of slipping, or if the
question is about how hard one must push to budge an object. It also comes
in handy when you want to know whether or not an object will stay put. In
such a case you would use Newton’s 2nd to find out the magnitude of the
force of static friction needed to keep the object from accelerating. Then you
would compare that magnitude with the maximum possible magnitude of the
force of static friction.
Here is the free body diagram of the brick and the table of forces for Example 15-6:
Up
a
=0
F
N
m
East
FS
F
sf
Table of Forces
F
g
Symbol=? Name
Agent
Victim
Normal
The
F
N
The Tile Floor
Force
Brick
Static
The
F
sf
Friction
The Tile Floor
Brick
Force
F = mg
Gravitational The Earth’s
The
g
Force
Gravitational Field
Brick
The
F = k x
S
Spring Force The Spring
Brick
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Chapter 16 Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
16 Newton’s Laws #3: Components, Friction, Ramps,
Pulleys, and Strings
When, in the case of a tilted coordinate system, you break up the gravitational force
vector into its component vectors, make sure the gravitational force vector itself forms
the hypotenuse of the right triangle in your vector component diagram. All too often,
folks draw one of the components of the gravitational force vector in such a manner that
it is bigger than the gravitational force vector it is supposed to be a component of. The
component of a vector is never bigger than the vector itself.
Having learned how to use free body diagrams, and then having learned how to create them, you
are in a pretty good position to solve a huge number of Newton’s 2nd Law problems. An
understanding of the considerations in this chapter will enable to you to solve an even larger
class of problems. Again, we use examples to convey the desired information.
Example 16-1
A professor is pushing on a desk with a force of magnitude F at an acute angle θ below
the horizontal. The desk is on a flat, horizontal tile floor and it is not moving. For the
desk, draw the free body diagram that facilitates the direct and straightforward
application of Newton’s 2nd Law of motion. Give the table of forces.
While not a required part of the solution, a sketch often makes it easier to come up with the
correct free body diagram. Just make sure you don’t combine the sketch and the free body
diagram. In this problem, a sketch helps clarify what is meant by “at an acute angle θ below the
horizontal.”
Pushing with a force that is directed at some acute angle below the horizontal is pushing
horizontally and downward at the same time.
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Chapter 16 Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
Here is the initial free body diagram and the corresponding table of forces.
a = 0
F
θ
N
FP
F
F
sf
g
Table of Forces
Symbol=? Name
Agent
Victim
Normal
F
N
The Floor
Desk
Force
Static
F
sf
Friction
The Floor
Desk
Force
F = mg
Gravitational The Earth’s
Desk
g
Force
Gravitational Field
Force of
F
Hands of Professor Desk
P
Professor
Note that there are no two mutually perpendicular lines to which all of the forces are parallel.
The best choice of mutually perpendicular lines would be a vertical and a horizontal line. Three
of the four forces lie along one or the other of such lines. But the force of the professor does not.
We cannot use this free body diagram directly. We are dealing with a case which requires a
second free body diagram.
Cases Requiring a Second Free Body Diagram in Which One of More of the Forces
that was in the First Free Body Diagram is Replaced With its Components
Establish a pair of mutually perpendicular lines such that most of the vectors lie along
one or the other of the two lines. After having done so, break up each of the other
vectors, the ones that lie along neither of the lines, (let’s call these the rogue vectors) into
components along the two lines. (Breaking up vectors into their components involves
drawing a vector component diagram.) Draw a second free body diagram, identical to the
first, except with rogue vectors replaced by their component vectors. In the new free
body diagram, draw the component vectors in the direction in which they actually point
and label them with their magnitudes (no minus signs).
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Chapter 16 Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
For the case at hand, our rogue force is the force of the professor. We break it up into
components as follows:
y
F
Px
θ
x
|F
Py |
FP
FPx = cosθ
FPy = sinθ
F
P
FP
F
= F cosθ
Px
P
F
= F sinθ
Py
P
Then we draw a second free body diagram, the same as the first, except with F replaced by its
P
component vectors:
a = 0
F
N
F
Px
F
g
F
sf
F
Py
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Chapter 16 Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
Example 16-2
A wooden block of mass m is sliding down a flat metal incline (a flat metal ramp) that
makes an acute angle θ with the horizontal. The block is slowing down. Draw the
directly-usable free body diagram of the block. Provide a table of forces.
We choose to start the solution to this problem with a sketch. The sketch facilitates the creation
of the free body diagram but in no way replaces it.
m
v
θ
Since the block is sliding in the down-the-incline direction, the frictional force must be in the up-
the-incline direction. Since the block’s velocity is in the down-the-incline direction and
decreasing, the acceleration must be in the up-the-incline direction.
Table of Forces
a
Symbol=?
Name
Agent
Victim
F
Normal
The
N
F
The Ramp
N
Force
Block
F
k f
Kinetic
The
F = µ F
kf
K
N
Friction
The Ramp
θ
Block
Force
F =mg
Gravitational The Earth’s
The
g
Force
Gravitational Field
Block
F
g
No matter what we choose for a pair of coordinate axes, we cannot make it so that all the vectors
in the free body diagram are parallel to one or the other of the two coordinate axes lines. At best,
the pair of lines, one line parallel to the frictional force and the other perpendicular to the ramp,
leaves one rogue vector, namely the gravitational force vector. Such a coordinate system is tilted
on the page.
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Chapter 16 Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
Cases Involving Tilted Coordinate Systems
For effective communication purposes, students drawing diagrams depicting phenomena
occurring near the surface of the earth are required to use either the convention that
downward is toward the bottom of the page (corresponding to a side view) or the
convention that downward is into the page (corresponding to a top view). If one wants to
depict a coordinate system for a case in which the direction “downward” is parallel to
neither coordinate axis line, the coordinate system must be drawn so that it appears tilted
on the page.
In the case of a tilted-coordinate system problem requiring a second free body diagram of the
same object, it is a good idea to define the coordinate system on the first free body diagram. Use
dashed lines so that the coordinate axes do not look like force vectors. Here we redraw the first
free body diagram. (When you get to this stage in a problem, just add the coordinate axes to
your existing diagram.)
y
a
F
N
F
kf
θ
F
g
x
Now we break up F into its x, y component vectors. This calls for a vector component diagram.
g
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Chapter 16 Newton’s Laws #3: Components, Friction, Ramps, Pulleys, and Strings
y
θ
Horizontal Line
θ
90° − θ
F
θ
g y
F
g
x
F
gx
Fgx = θ
F
sin
g y
=
θ
cos
Fg