• We begin our discussion of dynamics (study of how forces produce motion in objects)
• We’ll use kinematic quantities (displacement, velocity, acceleration) along with force & mass to analyze principles of dynamics
– What is the relationship between motion and the forces that cause the motion?
• Newton’s Laws of Motion (3 of them) describe the behavior of dynamical motion
– Form the foundation of classical mechanics (or
Newtonian mechanics) – the physics of “everyday life”
– They are deterministic in nature

• Concept of force: Push or pull experienced as a result of interaction between objects or between object and its environment
– Contact forces (involves direct contact between objects)
– Long-range forces (forces that act at a distance, e.g. gravity)
– Weight (force of gravitational attraction that Earth exerts on an object)
• Force is a vector quantity (has magnitude & direction)
• SI unit of force: the Newton (N)
– 1 N = 1 kg  m/s 2
• Diagrams of forces acting on bodies: free-body diagrams
 Draw all the external forces acting on the hot dog cart

• Free-body diagram for the hot dog cart (neglecting friction):
“Normal” force (  to
 surface of contact) F
N
Pulling force
F p 1
Weight of cart

F
W
• Effect of all 3 forces acting on the cart same as effect of a single force equal to vector sum of individual forces
– So

F net


F p 1


F
W


F
N
 

F

F
– Since cart does not move up or down, sum of vertical net
F net

F p 1
(in this case)

• Suppose we add a 2 nd pulling force:

F p 2

F
N

F p 1

F
W
– Easier to add forces if we use the components of each force
– Any force can be replaced by its component vectors, acting at the same point y
• Set up coordinate system components: q
= hot dog cart
F p
F p 2 x q

F p 1
F
N

F
W x

• The net force acting on the hot dog cart can be determined from the components of each individual force:
F net x
 
F x
F net y
 
F y
F net x
 
F p 1

F net

F p 2
 cos q
F net y
    net x
 net y
2

F p 2 sin q 
F
N

F
W
• How do forces affect motion?
• First consider what happens when net force on a body is zero:
– Box at rest on floor:

F
N

F
W


F

0
– Box will remain at rest

– Box sliding along freshly waxed floor: v = constant
 a = 0

F


0

F
N
 v = const.
F
W
– v will remain constant ( a = 0) if there is no friction between box and floor (approximately true for a slick floor)
– No force is needed to keep box sliding once it has started moving (it would slow down and stop only if friction, another force, were present)
• Newton’s 1 st Law of Motion: A body with zero net force acting on it moves with constant velocity
(which may be zero) and zero acceleration
– It’s the net force that matters in Newton’s 1 st Law

• Inertia is a property that indicates the tendency of a body to keep moving once it’s set in motion, or the tendency of a body at rest to remain at rest
– For example, your inertia is what causes you to feel like you are being “pushed” against the side of your car when you exit quickly from the highway onto an exit ramp
• When the net force on a body is zero, we say that the body is in equilibrium:

F


0

F x

0

F y

0
(Newton’s 1 st Law)
• Newton’s 1 st Law is valid only in an inertial reference frame, i.e. not a reference frame that is accelerating with respect to the earth
– For example, an accelerating car does not form an inertial reference frame

• Now consider what happens when net force on a body is not zero:
– From experiments, we learn that the presence of a net force acting on a body causes the body to accelerate
• What is the relationship between net force and acceleration?

F
• Let’s examine 3 different pulling forces on the cart:
 a
2

F p 1 p 1
2
 a
1
2 a

1
2

F p 1

– Magnitude of acceleration is directly proportional to the magnitude of the net force acting on the body
– Constant of proportionality is the mass m of the body
• Newton’s 2 nd 

F
– In (2 – D) component form:

F x

• Remember that: ma x

 m
 a
F
 y ma y
 a

 m

F
– Newton’s 2 nd Law is a vector equation
– Newton’s 2 nd Law refers to external forces ( ma is not a force, so don’t include it on free-body diagrams!)
– Newton’s 2 nd Law valid only in inertial reference frames, like the 1 st Law
– A nonzero net force is a cause, acceleration is the effect

CQ1: A 50-kg skydiver and a 100-kg skydiver open their parachutes and reach a constant velocity. The net force on the larger skydiver is:
A) equal to the net force on the smaller skydiver.
B) twice as great as the net force on the smaller skydiver.
C) four times as great as the net force on the smaller skydiver.
D) half as great as the net force on the smaller skydiver.

CQ2: If F is the force of air resistance on an object with mass m moving at a constant velocity, which of the following best describes the acceleration of the object when the force of air resistance is reduced by a factor of 4?
A) F / m
B) ½
F / m
C) ¼
F / m
D) ¾
F / m

Which animation correctly shows the motion of the satellite after the thruster force is applied?
A) Animation 1
B) Animation 2
C) Animation 3
D) Animation 4
PHYSLET #8.2.2, Prentice Hall (2001)

• We’ve already seen (from the hot dog cart example) that there is a force called weight that is exerted on bodies due to the gravitational pull of the earth
• What is the magnitude of this force?
– From Newton’s 2 nd Law:


F

 m a
– Gravitational forces accelerate bodies with constant magnitude a = g = 9.8 m/s 2 W
 mg direction)
• Weight acts on a body all of the time
• Magnitude of g can change, depending on the location ( g moon
= 1.62 m/s 2 , for example)

• Newton’s 3 rd Law: When a force from object
A is exerted on object B, B will exert a force on A that is equal in magnitude but opposite in direction to the force that A exerts on B:

F
A on B
 

F
B on A
– If I push on a wall, the wall pushes back on me with a force that is equal to mine in magnitude but opposite in direction
– Forces thus always come in pairs
– Force pairs resulting from Newton’s 3 rd Law are called action – reaction pairs and they never act on the same body

CQ4: A book is at rest on a horizontal table.
What is the reaction force (as dictated by
Newton’s 3 rd law) to the weight of the book?
A) The force that the table exerts upward on the book.
B) The force that the book exerts downward on the table.
C) The force of gravity on the book.
D) The force of gravity that the book exerts on Earth.

ProblemSolving Strategy for Newton’s Laws
1) Draw cartoon of physical situation & define your coordinate system
2) Draw free-body diagram of the object of interest a) Draw force vectors for each external force acting on the object b) NEVER include ma in a free-body diagram
3) Apply Newton’s Laws of motion as appropriate
4) Repeat steps 1 – 3 for multiple objects if necessary
5) Check your results – do they make sense?

A setup similar to the one shown at right is often used in hospitals to support and apply a traction force to an injured leg. (a) Determine the force of tension in the rope supporting the leg. (b)
What is the traction force exerted on the leg? Assume the traction force is horizontal.
Solution (details given in class):
(a) 78.4 N
(b) 105 N

An object of mass 2.0 kg starts from rest and slides down an inclined plane 80 cm long in 0.50 s. What net force is acting on the object along the incline?
Solution (details given in class):
13 N
Geometry: q q q
90 °  q q

Find the acceleration of each block and the tension in the cable for the following frictionless system:
5.00 kg m
1
+ x
Coupled system: mass m
2 moves same distance in same time as mass m
1
 v
1
= v
2
 a
1
= a
2
= a m
2
+ y
Free–body diagrams:
10.0 kg
N m
1 m
1 g
Apply Newton’s 2 nd Law to block m
1
:
T m
2
T m
2 g
S
F x
= m
1 a x

T = m
1 a x
= m
1 a (1)
S
F y
= 0
 m
1 g
–
N = 0
 m
1 g = N ( a = 0 in y
–direction)

Apply Newton’s 2 nd Law to block m
2
:
S
F y
= m
2 a y
= m
2 a
 m
2 g
–
T = m
2 a (2)
Combining equations (1) and (2):
 m
2 g
– m
1 a = m
2 a
 a = [ m
2
/ ( m
1
+ m
2
)] g = 6.53 m/s 2
Using equation (1) and plugging in for a :

T = m
1 a = 32.7 N
(Note that this analysis will be useful for Experiment 5 in lab!)

What is the maximum height reached by the rocket?
A) 0 m
B) 240 m
C) 960 m
D) 2880 m
E) 3840 m
ActivPhysics Problem #2.4, Pearson/Addison Wesley (1995 –2007)

• Friction is a contact force between two surfaces that always opposes motion
F
N f
 is always
 to

N f W
• The kind of friction that acts when a body slides f
 over a surface is called the kinetic friction force
(“kinetic” for motion) f
 the normal force N : f k
 m k
N
– Constant m k
= coefficient of kinetic friction (depends on the two surfaces in contact)

• When pushing a car, have you ever noticed that it’s harder to start car moving than to keep it moving?
• The magnitude of the frictional force varies!
f

• There is a static frictional force , with variable magnitude, that is almost always larger (at it’s maximum value) than the kinetic frictional force
• For the case of pushing a piano across the floor:
N
(1) (no pushing)
W
(2) F
N
(pushing but no sliding) f s
W

(3) F
N
W
(just about to slide) f s ,max
= f s max. value f s, max
F
N
(4)
(piano sliding)
• f s f k
W has a variable magnitude: f s
 m s
N
– Constant m s is the coefficient of static friction ( m s
• In situation (1) above: f

0
> m k
)
• In situation (2) above:
• In situation (3) above:
• In situation (4) above: f f f


 f s f s f k


 m s
N m s m k
N
N

• Variable magnitude of the friction force as a function of pushing force F summarized in the graph at right
• Friction responsible for motion of wheeled vehicles
(Note that f is in the same direction as the motion of the car, but opposite to the motion of the tires!)

CQ6: If the rear wheels of a truck drive the truck forward, then the frictional force on the rear tires due to the road is:
A) kinetic and in the backward direction.
B) kinetic and in the forward direction.
C) static and in the backward direction.
D) static and in the forward direction.

A crate of mass 45.0 kg is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck’s flatbed is 0.350, and the coefficient of kinetic friction is 0.320. (a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck’s flatbed?
(b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed.
What is the acceleration of the crate relative to the ground?
Solution (details given in class):
(a) 3.43 m/s 2
(b) 3.14 m/s 2

Two boxes of fruit on a frictionless horizontal surface are connected by a light string (see figure below), where m
1 and m
2
= 10 kg
= 20 kg . A force of 50 N is applied to the 20-kg box. (a)
Determine the acceleration of each box and the tension in the string. (b) Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10.
N
2
N
1 T T f
1 m
1 g f
2
( f
1 and f
2
= 0 in part a) m
2 g
Solution (details given in class):
(a) a = 1.7 m/s 2 , T = 17 N
(b) a = 0.69 m/s 2 , T = 17 N

Find the acceleration reached by each of the two objects shown in the figure at right if the coefficient of kinetic friction between the 7.00-kg object and the plane is 0.250.
N
T
+ y
+ x f m
1 g
T m
2 g
Solution (details given in class):
3.30 m/s 2
Incline with Friction Interactive