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/s 2 dyne = 1 g cm/s 2
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 1 st 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 1 st
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,
F
1 and F
2
, act on the block. F
1
has a magnitude of 79.9 N directed at an angle of 53 o relative to the + x axis. F
2
has a magnitude of 60.2 N directed at an angle of 143 o
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
F
R
+Y

F
2
= 60.2 N @ 143 o F
1
= 79.9 N @ 53 o
+X
F eq
= 100 N @ 270 o
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 = F
F
R
= F
2 sin

2
/ F
R
= ( 60.2 N )/(sin 37 o )
= ( 60.2 N )/(.6018)
F
R
= 100 N @ 90 o
or cos

= adj / hyp = F
F
R
= F
2 cos

1
/ F
R
= ( 79.9 N )/(cos 37 o )
= ( 79.9 N )/(.7986)
F
R
= 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 1 st 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 180 o 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 (F
1
= 79.9 N @ 53 o
F
2
= 60.2 N @ 143 o
). 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 F
1
and F
2
. c) Sum the x-components of F
1
and F
2
. d) Calculate the y-components of F
1
and F
2
. e) Sum the y-components of F
1
and F
2
. f) Sum all vectors acting vertically.
+Y (up)
F
2
F
1
37 o 53 o
Side view of the situation
+X
W = -100 N
Solution: a) W = F
G
= mg
m = W/g = 100 N / 9.8 m/s 2
= 10.2 kg
Force X-Component Y-Component b) F
1
= 79.9 N @ 53 o F
1 cos 53 o F
1 sin 53 o
= ( 79.9 N)(.6020) = ( 79.9 N)(.7990)
= 48.1 N = 63.8 N d. F
2
= 60.2 N @ 143 o F
2 cos 143 o F
2 sin 143 o
= (60.2 N)(- .7990) = (60.2 N)(.6020)
= 48.1 N = 36.2 N
__________________________________________________________________________ c) and e) Sum 0 + 100 N

f)

F
Y
= F
1Y
+ F
2Y
- 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

F
Y
= F
2Y
- W
= 36.2 N – 100 N = - 73.8 N = 73.8 N @ 270 o

F
X
= F
2X
= - 48.1 N = 48.1 N @ 180 o
Now, just use the Pythagorean eq. and tan to find magnitude and direction of resultant of 73.8 N @ 270 o and 48.1 N @ 180 o
+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 = F net
= 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 2 nd law .
W = F g
= 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
F
N
F g
F
2
F
1
Force of Friction

Since friction is often a factor in motion it should be considered in determining the net force in F net
= 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.

F fric v
static friction (F s
)- the force that opposes the start of motion between two surfaces that are not already in relative motion.
kinetic friction (F k
)- 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.
F fric
 
N F fric
= force of friction

= coefficient of friction
N = normal force

F fric can be determined experimentally by measuring the applied force (F
A
) 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 3 rd 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.
F fric F
A 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
F fric F
A
mg b) W = F
G
= mg
= (20 kg) (9.8 m/s 2 )
= 196 N down N = 196 N up c) F fric
 
N
= (0.40)(196 N)
= 78 N d) F net
= F
A
F fric
= 100 N – 78 N
= 22 N e) a = F net
/ m
= 22 N / 20 kg
= 1.1 m/s 2

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.
F
12
= -F
21

Action reaction force pairs are always equal in magnitude and act on different objects.

Forces that cause acceleration described by Newton's 2 nd 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 m
T
T
mg

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 = F net
/ m total
= mg / (m + M)
= (0.015 kg) (9.8 m/s 2 )/(0.015 kg + 1.000 kg)
= 0.14 m/s 2
[A]
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
2 nd 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 
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 30 o 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

F  = (75 kg)(9.8 m/s 2 ) sin 30 o
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/s 2 down the incline c) Determine the normal force between Ben + sled and the hill.
N = F  = W cos

F  = W cos 30 o

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.
F f
=

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 = F net
= F  - F f
= 367.5 N - 63.65 N = 303.85 N
F net
= ma
a = F net
/m =
= (303.85 N)/(75 kg)
= 4.05 m/s 2 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? v o
= 0 v = ?
x o
= 0 x = 30 m t = unk a = 4.05 m/s 2 v 2 = v o
2 + 2a (s - s o
)
= 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 = v o
+ at or x = x o
+ v ox t + 1/2at 2
15.6 = 0 + 4.05 t 30 = 0 + 0 + 1/2 (4.05)t 2 t = 3.85 s 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. v o
= 15.6 m/s v = 0 x o
= 0 x = ?

t = 1.5 s a = unk.
 x = ½( v + v o
) 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/s 2 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.
F net
= ma
= (75 kg) (- 10.4 m/s 2 )
= -780 N l) What is the coefficient of kinetic friction between the sled and the pavement?
F f
=

N
N = W = mg = 735 N

F f
/ N = 780 N / 735 N = 1.06
Practice Problem 8 Combining Concepts: An AP Free Response Question
Block 1 m
1
Block 2 m
2
M
 a) On the figure below, draw and label all the forces on the block m
1
(4 points)

N
T
f m
1 g
For each correctly drawn and labeled force one point was awarded
In the absence of total weight showing both components m
1 g sin

and m

received credit
1 g cos
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 m
1, m
2, 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 = m
1 g cos

1 point
For correctly solving for the coefficient of friction in terms of the given quantities.
f =

N

= f / N
= f / m
1 g 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)
- m
1 g sin

+ f - m
2 g 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 = m
1 g sin

- f + m
2 g sin

- 2f
For correct answer in terms of the given quantities.
M = (m
1
+ m
2
) 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: m
1 g sin

- f = F net
= m
1 a m
1 g sin

- f = m
1 a
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 / m
1
)
AP Review Question # 6
[A] Complete the Incline Planes Worksheet provided.