Physics 2011

Lecture 4: Newton’s Laws

S.Norr

Sir Isaac Newton• Born: 1642 Died: 1727• Philosophiae Naturalis Principia Mathematica

(Mathematical Principles of Natural Philosophy) (1687)

In Principia Mathematica:

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial

reference frame.

Law 2: For any object, FFNET = FF = maa

Law 3: Forces occur in pairs: FFA-B = - FFB-A

(For every action there is an equal and opposite reaction.)

Netwon’s First Law:

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

An Object in motion tends to stay in motion unless acted upon by an external force

The application of an external force results in an Acceleration of the object

What makes a good Inertial Reference Frame?

• Acceleration must be negligible. Is Duluth, MN a good IRF?

• Consider the UCM equations:

Calculating the Centripetal Acceleration at Duluth:

• Duluth is on the surface of Earth, rotating with a period of 1 Day, with a radius of less than ½ of Earth’s Diameter:

• T = 1 Day * 24 Hrs * 3600 Sec/Hr = 86400 Sec• R = approx. 6 x 106 meters

Duluth is a fairly good IRF

BUT….Earth orbits the sun…

• T = 1 year = 365 * 24* 3600 = 31.6 Msec

• R = 150 x 109 meters

• a = 0.006 m/s2 - So, again a very small acceleration DULUTH is an IRF

Newton’s Second Law

• The net resultant Force (the graphical sum of all forces) acting on a Body is equivalent to the product of its Mass and the Acceleration it is experiencing.

Fnet = m*aIn other words, any imbalance of forces on an object causes the object to accelerate.

That acceleration is directly proportional to the net force and inversely proportional to its mass.

Defining Forces:• Forces can be a push or a pull

• The SI unit of Force is the Newton (1 kg*m/s2)• Forces act on a Body with a magnitude at some

direction: A VECTOR! (hold for applause)

Combining Forces:

• Forces add and subtract just like vectors, as one might expect:FNET(x,y,z) = F1(x,y,z) + F2(x,y,z) = [F1x + F2x] Î + etc

Adding Force Vectors Graphically

Superposition of Forces• Superposition states that total reaction from

multiple disturbances is the algebraic sum of the individual reactions from each disturbance.

• Thus:– Any force vector can be replaced by its component

vectors, all acting at the same point– Any number of forces acting at the same point can be

replaced by a single Resultant Force equal to the vector sum of the individual forces.

Component Vectors

Components and 2nd Law

• Components of FF = maa : FX = maX

FY = maY

FZ = maZ

• Suppose we know m and FX , we can solve for aX

and apply the things we learned about kinematics over the last few weeks:

tavv

ta21

tvxx

xx0x

2xx00

+=

++=

Example: Pushing a Box on Ice.

• A skater is pushing a heavy box (mass m = 100 kg) across a sheet of ice (horizontal & frictionless). He applies a force of 50 N in the ii direction. If the box starts at rest, what is its speed v after being pushed a distance d = 10m ?

d

FF

v

m a

ii

Calculations:• Start with F = ma.

Or a = F / m.

– Recall that v2 - v02 = 2a(x - x0 ) from Chap.2

– So v2 = 2Fd / m ; where:

F = 50 N, d = 10 m, m = 100 kg

Thus, V = +/- 3.2 m/s2 and we’ll discard the –tive solution

d

FF

v

m a

ii

Review: Newton's Laws

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

Law 2: For any object, FFNET = maa

Where FFNET = FF

Law 3: Forces occur in action-reactionaction-reaction pairs, FFA ,B = - FFB ,A.

Where FFA ,B is the force acting on object A due to its interaction with object B and vice-versa.

The Free Body Diagram

• Newton’s 2nd Law says that for an object FF = maa.

• Key phrase here is for an objectfor an object..

• So before we can apply FF = maa to any given object we isolate the forces acting on this object:

The Free Body Diagram...

• Consider the following case– What are the forces acting on the plank ?

P = plank

F = floor

W = wall

E = earth

FFW,P

FFP,W

FFP,F FFP,E

FFF,P

FFE,P

The Free Body Diagram...

• Consider the following case– What are the forces acting on the plank ?

Isolate the plank from

the rest of the world.

FFW,P

FFP,W

FFP,F FFP,E

FFF,P

FFE,P

The Free Body Diagram...

• The forces acting on the plank should reveal themselves...

FFP,W

FFP,F FFP,E

Aside...

• In this example the plank is not moving...– It is certainly not accelerating!– So FFNET = maa becomes FFNET = 0

– This is the basic idea behind statics, which we will discuss in a few weeks.

FFP,W + FFP,F + FFP,E = 0

FFP,W

FFP,F FFP,E

Example

• Example dynamics problem:

A box of mass m = 2 kg slides on a horizontal frictionless floor. A force Fx = 10 N pushes on it in the xx direction. What is the acceleration of the box?

FF = Fx ii aa = ?

m

y y

x x

Example...

• Draw a picture showing all of the forces

FFFFB,F

FFF,BFFB,E

FFE,B

y y

x x

Example...• Draw a picture showing all of the forces.• Isolate the forces acting on the block.

FFFFB,F

FFF,BFFB,E = mgg

FFE,B

y y

x x

Example...• Draw a picture showing all of the forces.• Isolate the forces acting on the block.• Draw a free body diagram.

FFFFB,F

mgg

y y

x x

Example...• Draw a picture showing all of the forces.• Isolate the forces acting on the block.• Draw a free body diagram.• Solve Newton’s equations for each component.

– FX = maX

– FB,F - mg = maY

FFFFB,F

mgg

y y

x x

Example...• FX = maX

– So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.

• FB,F - mg = maY

– But aY = 0– So FB,F = mg.

• The vertical component of the forceof the floor on the object (FB,F ) isoften called the Normal Force Normal Force (N).

• Since aY = 0 , N = mg in this case.

FX

N

mg

y y

x x

Example Recap

FX

N = mg

mg

aX = FX / m y y

x x

Normal Force• A block of mass m rests on the floor of an

elevator that is accelerating upward. What is the relationship between the force due to gravity and the normal force on the block?

m

(a)(a) N > mgN > mg

(b)(b) N = mgN = mg

(c)(c) N < mgN < mga

Solution

m

N

mg

All forces are acting in the y direction, so use:

Ftotal = ma

N - mg = ma

N = ma + mg

therefore N > mg

a

Tools: Ropes & Strings• Can be used to pull from a distance.• TensionTension (T) at a certain position in a rope is the magnitude of the

force acting across a cross-section of the rope at that position.– The force you would feel if you cut the rope and grabbed the ends.– An action-reaction pair.

cut

TT

T

Tools: Ropes & Strings...

• Consider a horizontal segment of rope having mass m:–Draw a free-body diagram (ignore gravity).

• Using Newton’s 2nd law (in xx direction): FNET = T2 - T1 = ma

• So if m = 0 (i.e. the rope is light) then T1 =T2

T1 T2

m

a x x

Tools: Ropes & Strings...• An ideal (massless) rope has constant tension along the

rope.

• If a rope has mass, the tension can vary along the rope– For example, a heavy rope

hanging from the ceiling...

• We will deal mostly with ideal massless ropes.

T = Tg

T = 0

T T

Tools: Ropes & Strings...

• The direction of the force provided by a rope is along the direction of the rope:

mg

T

m

Since ay = 0 (box not moving),

T = mg

Force and acceleration• A fish is being yanked upward out of the water using a fishing line that breaks when the tension reaches 180

N. The string snaps when the acceleration of the fish is observed to be is 12.2 m/s2. What is the mass of the fish?

m = ?a = 12.2 m/s2

snap ! (a) 14.8 kg

(b) 18.4 kg

(c) 8.2 kg

Solution:

• Draw a Free Body Diagram!!T

mg

m = ?a = 12.2 m/s2

Use Newton’s 2nd lawin the upward direction:

FTOT = ma

T - mg = ma

T = ma + mg = m(g+a)

mT

g a

kg28

sm21289

N180m

2.

..