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B. Newton's Laws of Motion
Content Outline
AP Exam % Goals
B. Newton's laws of motion (including friction and centripetal force)
1. Static equilibrium (first law)
2. Dynamics of a single particle (second law)
3. Systems of two or more bodies (third law)
9
Unit I B Objectives
1. Distinguish contact forces and field forces. Name the four basic forces.
2. State the first law of motion; understand the concept of inertia.
3. State the second law of motion and relate it to the first law. Use the second law in solving
problems. Define the newton.
4. State the third law; understand that all forces act in pairs.
5. Use free-body diagrams to represent forces acting on an object.
6. Apply Newton's three laws of motion to real-world situations and to experiments we conduct.
7. Distinguish between weight and mass; use the second law to relate mass and weight.
8. Comprehend the nature of frictional forces and be able to use the coefficient of friction in
solving problems.
9. Use the concept of net force in solving F = ma.
10. State the requirements for equilibrium.
11. Define equilibrant and differentiate between resultant and equilibrant.
12. Resolve nonperpendicular vector components.
13. Specify axes, resolve vectors, and solve inclined plane problems.
14. Demonstrate an understanding of centripetal acceleration and Newton's second law applied to
circular motion; Solve problems of circular motion.
AP Objectives – See Course Objectives for AP Physics
Introduction
We began our study of mechanics by describing, both qualitatively and quantitatively, how
things move. We are now ready to examine the causes of motion, which involve the forces acting
on an object. Newton’s laws of motion provide the foundation for the study of dynamics.
dynamics- the study of the causes of motion; involves the connection between
forces and accelerations.
Kinds of Forces
force – a “push” or a “pull”
 Force is not a property of an object, like mass, but rather it is an interaction of
the object with an external agent.
 In the macroscopic world we encounter two classes of forces, namely contact
forces and field forces.
contact forces - the object producing the force actually “touches” the object
being acted upon.
 Contact forces are actually electromagnetic field forces acting at very small
distances.
field forces - forces that can exert their effect over a distance without touching
the object being acted upon.
4 field Forces: Gravitational, electromagnetic, weak nuclear, and strong nuclear.
 The stretching of springs (spring scale) is used conveniently to measure forces.
The length of the stretch is directly proportional to the force producing the
stretch: (Hooke's Law).
Units of Force:
SI- newton (N) = 1 kg m/s2
dyne = 1 g cm/s2
British- pounds (lb)
Conversion factor: 1 lb = 4.448 N
newton (N) - the amount of force required to accelerate a one kilogram mass
at a rate of one meter per second squared.
 Forces and the accelerations they cause are vector quantities, so we can use the
techniques of adding and resolving vectors to analyze the acceleration of
objects that have any number of forces exerted on them.
1. Static Equilibrium (Newton’s first law- the law of inertia)
AP 8
An object at rest will remain at rest, and an object in motion will continue in
motion with a constant velocity (that is, with a constant speed in a straight line),
unless it experiences a net external force.
a=0
F = 0
  = “sum of”. The mathematical expressions of Newton’s 1st law, a = 0 and F = 0, show
that if you add all the forces acting on an object and they come out to zero, then the object
will not speed up, slow down, nor change direction.
inertia - the tendency of an object not to change its motion.
 Mass is a measure of the quantity of matter and is used to measure inertia.
Anything that has mass has the property of inertia.
Inertial Reference Frames: An inertial reference frame is one in which the law of inertia holds.
i.e., the 1st law of Newton applies. Rotating and other kinds of accelerated reference frames are
examples of reference frames that are not inertial.
net force- the vector sum of all forces acting on an object (the resultant force).
Practice Problem 1 Block Party I
Two ropes are attached to a block of mass M resting on a frictionless, horizontal surface as
shown in the diagram below. The ropes are then pulled simultaneously so that concurrent forces,
F1 and F2, act on the block. F1 has a magnitude of 79.9 N directed at an angle of 53o relative to
the + x axis. F2 has a magnitude of 60.2 N directed at an angle of 143o relative to the +x axis.
a) Determine the net force (the resultant force) acting on the block before the two forces are applied.
b) Calculate the net force (the resultant force) acting on the block while the two forces are being applied.
c) Which way will the block move?
Looking down on the situation
+Y
FR

F2 = 60.2 N @ 143o
F1 = 79.9 N @ 53o
+X
Feq = 100 N @ 270o
Solution:
a. Since the block is initially at rest, the sum of the forces acting on it must be zero (F = 0).
The block does exert a force on the table due to its weight, but Newton’s 3rd law will show
us that the surface upon which it is resting exerts an equal and opposite force so the two
forces add to equal zero. The details about weight and Newton’s 3rd law are coming up
soon.
b.
Use geometry to obtain the needed angles
sin  = opp/hyp = F2/ FR
FR = F2 sin 
= (60.2 N)/(sin 37o)
= (60.2 N)/(.6018)
FR = 100 N @ 90 o
or
cos  = adj/hyp = F1/ FR
FR = F2 cos 
= (79.9 N)/(cos 37o)
= (79.9 N)/(.7986)
FR = 100 N @ 90 o
c. The block will move in the direction of the net force.
equilibrium- when two or more forces act concurrently (at the same time) on
an object, and their vector sum is zero.
 An object in equilibrium has no acceleration because the forces on it are
balanced. According to Newton's 1st law, this means that the object is “not
moving” or is moving at constant velocity.
 When 3 or more concurrent forces are in equilibrium they form a closed
triangle or polygon when drawn to scale and added together.
 Physical conditions (ropes, cables, gravity, friction, etc. dictate the magnitude
and directions of the components of these vectors.
equilibrant - the single additional force that, if applied at the same point as
other forces, will produce equilibrium.
 The equilibrant is equal in magnitude to the resultant, but opposite in
direction. To find the equilibrant force of two or more concurrent forces, first
find the resultant force then add 180o to the direction of the resultant (negative
of a vector).
 Sometimes, when a system is in a state of equilibrium, we may wish to find
one or more of the vectors involved in the equilibrium.
Practice Problem 2 Block Party II
Let’s go back to the previous problem to illustrate equilibrium. Before the two ropes were pulled,
the block was at rest (relative to the surface it was on) so, obviously, it was in a state of
equilibrium. After the two forces were applied by pulling on the ropes, the block moved in the
direction of the net force.
a) Describe the motion of the block if the two forces are continued.
b) Determine the magnitude and direction of a single force (the equilibrant), exerted through a
third rope attached to the block, that would be required to reestablish equilibrium.
c) Describe the motion of the block once equilibrium is reestablished
a) The block will accelerate in the direction of the net force.
b) 100 N @ 270 o
c) The block will move at a constant velocity, which is the final velocity
after it accelerated.
Practice Problem 3 Block Party III
Now let’s take our block of mass M and support it in midair by having students hold the first two
ropes so that it just hangs without moving. To simplify matters let’s keep the forces in the two
ropes the same (F1 = 79.9 N @ 53o F2 = 60.2 N @ 143o). The situation would look the same,
however, the +y axis is now up, not horizontal.
Also, let’s suppose that the block has a weight of 100 N, so we don’t need the third rope to
establish equilibrium because the three forces have a vector sum of zero. Notice the closed
triangle formed when the three vectors are drawn to scale and added together graphically.
a) Calculate the mass of the block.
b) Calculate the x-components of F1 and F2.
c) Sum the x-components of F1 and F2.
d) Calculate the y-components of F1 and F2.
e) Sum the y-components of F1 and F2.
f) Sum all vectors acting vertically.
+Y (up)
F2
F1
37o
53o
+X
Side view of the situation
W = -100 N
Solution:
a) W = FG = mg
m = W/g = 100 N / 9.8 m/s2
= 10.2 kg
Force
F1 = 79.9 N @ 53o
X-Component
Y-Component
F1 cos 53o
F1 sin 53o
= (79.9 N)(.6020)
= (79.9 N)(.7990)
= 48.1 N
= 63.8 N
d.
F2 = 60.2 N @ 143o
F2 cos 143o
F2 sin 143o
= (60.2 N)(- .7990)
= (60.2 N)(.6020)
= - 48.1 N
= 36.2 N
__________________________________________________________________________
c) and e)
Sum
0
+ 100 N
b)
f)
FY = F1Y + F2Y - W = 100 N - 100 N = 0
Practice Problem 4 Whoops, Rope Broke!
Suppose the rope exerting the 79.9 N force in the problem above breaks.
a) Calculate the net force acting on the block and b) the resulting acceleration of the block.
Using calculated values from PP 3
FY = F2Y - W
= 36.2 N – 100 N = - 73.8 N = 73.8 N @ 270o
FX = F2X = - 48.1 N = 48.1 N @ 180o
Now, just use the Pythagorean eq. and tan to find magnitude and direction of
resultant of 73.8 N @ 270o and 48.1 N @ 180o
+Y (up)
+X
Side view of the situation
W = -100 N
2. Dynamics of a Single Particle (Second Law)
The acceleration of an object is directly
proportional to the resultant force acting
on it and inversely proportional to its mass.
F = Fnet = ma
 Weight is the gravitational force acting on a mass m by some other mass M.
The weight of an object is what causes it to accelerate downward under the
influence of gravity.
Mass is a scalar because direction is not specified and is an inherent property
of a body.
Weight is a vector because force is a vector.
Thus, in accordance with Newton's 2nd law .
W = Fg = mg
(Eq)
 Free-body diagrams are very useful in solving dynamics problems. Each
object being considered in the diagram should be drawn. All the forces are
represented by labeled arrows indicating their direction and magnitude and
are labeled.

Exs: Free-body diagrams for the Block Party I situation
Before Pulling
While Pulling
FN
F1
F2
Fg
Force of Friction
 Since friction is often a factor in motion it should be considered in determining
the net force in Fnet = ma.
 Forces of friction are resistive in nature thus they are usually opposite the
direction of the motion but are sometimes in the direction of the change in
motion such as friction holding an object in a circular path.
friction- the force that opposes the motion of two surfaces that are in contact.
 The direction of the force is parallel to the surface and in a direction that is
opposite to the direction of motion.
Ffric
v
static friction (Fs)- the force that opposes the start of motion between
two surfaces that are not already in relative motion.
kinetic friction (Fk)- the force that opposes the relative motion
between two surfaces in contact with each other.
 The force of kinetic (sliding) friction is less than that of static friction.
Ffric  N
Ffric = force of friction  = coefficient of friction
N = normal force
 Ffric can be determined experimentally by measuring the applied force (FA)
required to slide an object at a constant velocity.
coefficient of friction (- a constant value describing the nature of the
surfaces in contact.
normal force (N)- the force pushing the surfaces together; is numerically equal
to the apparent weight of the object. (see Newton's 3rd law below)
 N always acts perpendicular to the surfaces in contact. The force of friction
depends only on the nature of the surfaces in contact and the normal force
pushing the surfaces together.
 On a surface with friction, the parallel force must equal the opposing force of
static friction before the object can begin to move.
 Once moving, the net force causing the object to accelerate is equal to the
parallel force minus the force of kinetic friction.
Practice Problem 5 Slide the Box!
A horizontal force of 100 N is applied to the right on a 20 kg box resting on the floor as shown
below, causing the box to slide to the right. The coefficient of friction of the box on the floor is
0.40.
Ffric
FA
a) Draw and label any other forces acting on the box.
b) Calculate the weight of the box and the normal force acting on the box.
c) Calculate the force of friction acting on the box.
d) Calculate the net force acting on the box.
e) Calculate the acceleration of the box.
a)
N
Ffric
FA
mg
b) W = FG = mg
= (20 kg) (9.8 m/s2)
= 196 N down
N = 196 N up
c) Ffric  N
= (0.40)(196 N)
= 78 N
d) Fnet = FA - Ffric
= 100 N – 78 N
= 22 N
e) a = Fnet / m
= 22 N / 20 kg
= 1.1 m/s2
3. Systems of Two or More Bodies (Third Law)
If two bodies interact, the force exerted on body one by body two is equal to and
opposite in direction to the force exerted on body two by body one.
F12 = -F21
 Action-reaction force pairs are always equal in magnitude and act on
different objects.
 Forces that cause acceleration described by Newton's 2nd law always act on the
same object and are not equal for linear motion.
 The action force acts upon the body whose motion we are studying and the
reaction force is on the other body producing the action.
 When working with tensions in a string, rope, or wire, the tension is a force
acting at every point in the wire producing action-reaction pairs. Thus, one
body may affect another through tension. Nevertheless, the reaction force acts
back on the body through the tension.
Practice Problem 6 Air-Track Lab
Suppose an air-track glider with a mass M of 1.000-kg is connected to a mass m of 0.015 kg as
shown below. Draw free-body diagrams for the glider and the hanging mass. What is the
acceleration of the glider?
1 kg
N
15 g
M
Mg
T
m
mg
T
Because the masses are joined by a light string, they move with the same
acceleration. The only force causing the acceleration of both masses is the
gravitational force acting on the 0.015 kg mass. Action-reaction force pairs
are propagated as tension T through the string.
a = Fnet / mtotal
= mg / (m + M)
= (0.015 kg) (9.8 m/s2)/(0.015 kg + 1.000 kg)
= 0.14 m/s2
[A] Complete the Newton’s Laws Worksheet provided.
Gravitational Force and Inclined Planes
P 11
 An object's weight, W, acts perpendicular to the surface of the earth because
the gravitational force acting on it is directed toward the center of the earth. W
is represented with a vertical line drawn downward.
 If the object rests on an inclined plane, the weight can be resolved into two
components perpendicular to each other. One component, F, is called the
perpendicular force because it acts perpendicular to the incline. The
perpendicular force is equal and opposite in direction to the normal force. The
2nd component, F, is called the parallel force and acts parallel to the incline.
Consider a block of mass m resting on an incline
The right triangle formed by the surface of the incline and the right triangle formed by W, F ,
and F are similar triangles because corresponding sides are mutually perpendicular. If the angle
of incline and W are both known, the vector diagram shown below can be drawn. The values of
the two components can then be calculated using trig. functions of the right triangle.
N = F= W cos  F = W sin 
W = mg(Eq)
N
F

F

F

F
W
Practice Problem 7 Physics of Sledding
W
The first big snow of the year inspired a group of PA physics students to conduct an
experiment on a super slick (frictionless) hill near the school. Ben sat on a disk sled being held in
place near the top of the 30o incline by two of his friends, Hannah and Marissa. The combined
mass of Ben and sled was 75 kg. Hannah and Marissa then released the sled allowing it to slide
down the incline. Use the diagram above to help answer the following questions
a) Calculate the force causing the acceleration of the sled down the hill.
F = W sin 
W = mg
2
F = (75 kg)(9.8 m/s ) sin 30o
F = (735 N) (.5000) = 367.5 N
b) Determine the acceleration of the sled assuming a frictionless surface and no air resistance.
On a frictionless surface, the parallel component of the weight provides the only force
causing the acceleration of the object down the incline and is equal to F = W sin .
a = F / m = W sin  / m
= 367.5 N / 75 kg
= 4.9 m/s2 down the incline
c) Determine the normal force between Ben + sled and the hill.
N = F= W cos 
F= W cos 30o
F= (735 N) (.8660) = 636.5 N
d) Find the actual acceleration given that friction is present. The coefficient of kinetic friction
between the sled and snow is .10.
Ff = N = F= W cos 
636.5 N)
 63.65 N
Add this force to the free-body diagram to help determine the net force.

F = Fnet = F- Ff
= 367.5 N - 63.65 N = 303.85 N
Fnet = ma
a = Fnet/m =
= (303.85 N)/(75 kg)
= 4.05 m/s2 down the incline
As the incline becomes steeper, F becomes greater and F becomes less.

Once the parallel component of the weight is calculated, use Newton's 2 nd law
to find the acceleration of the object.
a = F / m = W sin  / m since W/m = g a = g sin 
Please note that the last equation is a shortcut that will only work if friction is not present. If
friction is present (as it usually is in the real world) the opposing force that it supplies must be
subtracted from F in order to find the net force causing the acceleration down the incline.
Review Questions
e) Based on your answer in (d), how fast would Ben be going at the bottom of the hill if he
slides a total distance of 30 meters to the bottom of the hill?
vo = 0
v=?
xo = 0
x = 30 m
t = unk
a = 4.05 m/s2
v2 = vo2 + 2a (s - so)
= 0 + 2 (4.05)(30 - 0)
= 243
v = 15.6 m/s
h) How long would it take to reach the bottom of the hill.
v = vo + at
15.6 = 0 + 4.05 t
t = 3.85 s
or
x = xo + voxt + 1/2at2
30 = 0 + 0 + 1/2 (4.05)t2
t = 3.85 s
At the bottom of the hill Ben & sled slide onto level pavement and come to a screeching halt
in 1.5 s.
i) How far does he slide on the pavement?
When Ben sleds onto the pavement his motion changes from + acceleration to a acceleration. Since we are examining a different motion, one should consider the motion as
beginning when he hits the pavement. Thus, the final velocity from her acceleration down
the incline becomes the initial velocity of his "new" motion.
vo = 15.6 m/s
v=0
xo = 0
x= ?
t = 1.5 s
a = unk.
x = ½( v + vo ) t
does not involve the variable a
= ½ (0 + 15.6 m/s) 1.5 s
= 11.7 m
j) What is his acceleration on the pavement?
a = v/ t
= (0 - 15.6 m/s)/ 1.5 s
= - 10.4 m/s2
k) What is the frictional force between the sled and pavement?
Friction is now the only force acting on the sled since it is not on the hill so the net force
causing the negative acceleration is due to the frictional force.
Fnet = ma
= (75 kg) (- 10.4 m/s2)
= -780 N
l) What is the coefficient of kinetic friction between the sled and the pavement?
Ff = N
N = W = mg = 735 N
 Ff / N = 780 N / 735 N = 1.06
Practice Problem 8 Combining Concepts: An AP Free Response Question
Block 2
Block 1
m2
m1
M

a) On the figure below, draw and label all the forces on the block m1
(4 points)
N
T
f
m1g
For each correctly drawn and labeled force one point was awarded
In the absence of total weight showing both components m1g sin  and m1g cos
 received credit
One point was deducted from earned credit for one or more obvious incorrect
forces (no penalty for showing both weight and its components)
Express your answers to the following in terms of m1, m2, g, , and f
b) Determine the coefficient of kinetic friction between the inclined plane and
block 1.
(3 points)
Using the expression for frictional force:
f = N
1 point
For correct substitution of the normal force
N = m1g cos 
1 point
For correctly solving for the coefficient of friction in terms of the given
quantities.
f = N
=f/N
= f / m1g cos 
1 point
c) Determine the value of the suspended mass M that allows blocks 1 and 2 to
move with a constant velocity down the plane.
(5 points)
For an indication that the acceleration or the net force is zero
Using Newton's second law:
F = 0
1 point
Applying this to the three-block system, with the tensions treated as internal
forces (so they do not appear in the equation)
- m1g sin  + f - m2g sin  + 2f + Mg = 0
1 point
For correct weight components, with correct signs, in the above equation
1 point
For correct friction terms, with correct signs
Solving for Mg:
Mg = m1g sin  - f + m2g sin  - 2f
For correct answer in terms of the given quantities.
M = (m1 + m2) sin  - 3f/g
2 points
Only one point was awarded for a correct answer in terms of the quantities
other than the given ones
d) The string between blocks 1and 2 is now cut. Determine the acceleration of
block 1 while it is on the inclined plane.
(3 points)
Applying Newton's second law:
m1g sin  - f = Fnet = m1a
m1g sin  - f = m1a
For correctly including friction in the equation 1 point
For correctly including the weight component in the equation 1 point
For correct answer in terms of the given quantities. 1 point
a = g sin  - (f / m1)
AP Review Question # 6
[A] Complete the Incline Planes Worksheet provided.