Newton’s Laws of Motion• 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

Newton’s Laws of Motion• 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 kgm/s2

• Diagrams of forces acting on bodies: free-body diagrams Draw all the external forces

acting on the hot dog cart

Newton’s Laws of Motion• Free-body diagram for the hot dog cart (neglecting

friction):

• Effect of all 3 forces acting on the cart same as effect of a single force equal to vector sum of individual forces – Total (net) force = vector sum of individual forces = – So– Since cart does not move up or down, sum of vertical

forces must be zero (same effect as no vertical forces):

“Normal” force ( to surface of contact) NF

1pFPulling force

Weight of cart WF

netF

FFFFF NWp

1net

1net pFF

(in this case)

Fp2y

Fp2x

Newton’s Laws of Motion• Suppose we add a 2nd pulling force:

– 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

• Set up coordinate system & determine vector components:

NF

1pF

WF

2pF

= hot dog cart x

y

2pF

NF

1pF

WF

Newton’s Laws of Motion• The net force acting on the hot dog cart can be

determined from the components of each individual force:

• How do forces affect motion?• First consider what happens when net force on a

body is zero:– Box at rest on floor:

– Box will remain at rest

xx FFnet

cos21net ppx FFF

yy FFnet

WNpy FFFF sin2net

2net2

netnet yx FFF

NF

WF 0 F

Newton’s Laws of Motion– Box sliding along freshly waxed floor:

– 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 1st 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 1st Law

NF

WFv = const.

0 F

v = constant a = 0

Newton’s Laws of Motion• 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:

• Newton’s 1st 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

0 F

0 xF 0 yF (Newton’s 1st Law)

Newton’s Laws of Motion• 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?

• Let’s examine 3 different pulling forces on the cart:

1pF 12 pF

a

a

2

12

1pF

a

2

1

Newton’s Laws of Motion– 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 2nd Law: or

– In (2 – D) component form:

• Remember that: – Newton’s 2nd Law is a vector equation– Newton’s 2nd Law refers to external forces (ma is not a

force, so don’t include it on free-body diagrams!)– Newton’s 2nd Law valid only in inertial reference frames,

like the 1st Law– A nonzero net force is a cause, acceleration is the effect

amF

xx maF yy maF

m

Fa

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

CQ3: Interactive Example Problem: Predict the Satellite’s Motion

Which animation correctly shows the motion of the satellite after the thruster force is applied?

PHYSLET #8.2.2, Prentice Hall (2001)

A) Animation 1

B) Animation 2

C) Animation 3

D) Animation 4

Weight• 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 2nd Law:

– Gravitational forces accelerate bodies with constant magnitude a = g = 9.8 m/s2, so (with a downward direction)

• Weight acts on a body all of the time• Magnitude of g can change, depending on the

location (gmoon = 1.62 m/s2, for example)

amF

mgW

Newton’s Third Law• Newton’s 3rd 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:

– 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 3rd Law are

called action – reaction pairs and they never act on the same body

AonBBonA FF

CQ4: A book is at rest on a horizontal table. What is the reaction force (as dictated by

Newton’s 3rd 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.

Problem-Solving Strategy for Newton’s Laws

1) Draw cartoon of physical situation & define your coordinate system

2) Draw free-body diagram of the object of interesta) 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?

Example Problem #4.31

Solution (details given in class):

(a) 78.4 N

(b) 105 N

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.

Example Problem #4.30

Geometry:

°

Solution (details given in class):

13 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?

Example Problem #4.36Find the acceleration of each block and the tension in the cable for the following frictionless system:

m1

m2+y

+x

Coupled system: mass m2 moves same distance in same time as mass m1 v1 = v2 a1 = a2 = a

Free–body diagrams: m1m2

N

m1g m2g

T

T

Apply Newton’s 2nd Law to block m1:

Fx = m1ax T = m1ax = m1a (1)

Fy = 0 m1g – N = 0 m1g = N (a = 0 in y–direction)

5.00 kg

10.0 kg

Example Problem (continued)

Apply Newton’s 2nd Law to block m2:

Fy = m2ay = m2a

m2g – T = m2a (2)

Combining equations (1) and (2):

m2g – m1a = m2a a = [m2 / (m1 + m2)]g = 6.53 m/s2

Using equation (1) and plugging in for a:

T = m1a = 32.7 N

(Note that this analysis will be useful for Experiment 5 in lab!)

CQ5: Interactive Example Problem: Rocket Blasts Off

What is the maximum height reached by the rocket?

ActivPhysics Problem #2.4, Pearson/Addison Wesley (1995–2007)

A) 0 m

B) 240 m

C) 960 m

D) 2880 m

E) 3840 m

Friction• Friction is a contact force between two surfaces

that always opposes motion

• The kind of friction that acts when a body slides over a surface is called the kinetic friction force (“kinetic” for motion)– The magnitude of is proportional to the magnitude of

the normal force N:

– Constant k = coefficient of kinetic friction (depends on

the two surfaces in contact)

W

NF

f

is always tof

N

kf

kf

Nf kk

Friction• 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!• 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:

sf

W

N(1) (no pushing)

W

N(pushing but no sliding)(2) F

fs

Friction

• fs has a variable magnitude:

– Constant s is the coefficient of static friction (s > k)

• In situation (1) above:• In situation (2) above:• In situation (3) above:• In situation (4) above:

Nf ss

Nff ss

Nff ss Nff kk

0f

W

N(just about to slide)(3) F

fs,max

fs,max = fs max. value

W

N(piano sliding)(4) F

fk

Friction• 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.

Example Problem #4.44

Solution (details given in class):

(a) 3.43 m/s2

(b) 3.14 m/s2

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?

Example Problem #4.69

N1

m1g

T

f1 f2

N2

m2g

T

Two boxes of fruit on a frictionless horizontal surface are connected by a light string (see figure below), where m1 = 10 kg and m2 = 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.

Solution (details given in class):

(a) a = 1.7 m/s2, T = 17 N

(b) a = 0.69 m/s2, T = 17 N

( f1 and f2 = 0 in part a)

Example Problem #4.53

Solution (details given in class):

3.30 m/s2

Incline with Friction Interactive

N

f m1g

TT

m2g

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.

+x+y