Unit 2

Newton’s Laws

• Newton’s Laws of Motion1) Law of inertia: An object will remain in its

current state of motion unless acted upon by an external force.

2) ΣF = ma3) For every action (force) there is an equal and

opposite reaction (force). (Forces always come in pairs!)

ΣF = ma orFnet = ma

• Types of forces1) Field

a) Gravityb) Electromagnetic

2) Contacta) Tensionb) Normalc) Frictiond) Appliede) Resistance

• Gravity

• Fg = mg

• Also known as “weight”

• Tension

• T = mg for a suspended mass

• T < mg for a falling mass

• T > mg for a mass accelerating upward

• Normal

• Normal means perpendicular

• N = mg on a horizontal surface*• N = mgcosθ on a slope of angle θ*• *Applied forces that are not parallel to the

surface will affect the normal force

• Normal

• ΣF = ma

• mg + Fsinθ – N = ma

• But a = 0

• N = mg + Fsinθ

• Friction

• f = μN

• When two surfaces slide relative to each other it is kinetic friction μk

• When two surfaces are stationary relative to each other it is static friction μs

• μ is almost always less than 1• For a given pair of surfaces, μk < μs

• Applied

• Any external push or pull not already covered

• Resistance

• Usually air or water resistance

• A VERY common topic for differential equations

• A ball falls and is subject to a resistance force FR = –bv. Develop an equation for the velocity of the ball as a function of time. At t = 0, v = 0

mabvmg

dtdvmbvmg

bvmgdv

mdt

bvmgdv

mdt

Cbvmgb

tm

ln11

Cbvmgmbt

ln

Cbvmgmbt lnln

maF

• A ball falls and is subject to a resistance force FR = –bv. Develop an equation for the velocity of the ball as a function of time. At t = 0, v = 0

bvmgCe mbt /

Cbvmgmbt lnln

bvmgCmbt

ln

bvmgCe mbt /

0/0 bmgCe mb

mgC mbtmgemgbv /

mbteb

mgb

mgv /

mbteb

mgv /1

• Terminal velocity

• Constant velocity of a falling object

• What would be the terminal velocity in the previous example?

• A ball falls and is subject to a resistance force FR = –bv. Develop an equation for the velocity of the ball as a function of time. At t = 0, v = 0

mabvmg

maF

0a

0 Tbvmg

bmgvT

• Free Body Diagrams

• They show all external forces acting on an object

• A vector represents each force.– The head points in the direction of the force– The tail begins at the point of origin of the force– The length of the vector should be representative of

the magnitude of the force– When asked to draw a free body diagram on a test,

include only forces, not components– Label appropriately, using symbols from the problem

• Draw a free body diagram for mass m2 in the diagram

m2g

N

Tf

• Components of forces

• Always chose a coordinate system to minimize components

• Components of forces

F

θ

xy

• Components of forces

Fmg

f

N

mgcosθ

mgsinθ

xx maF

xmafFmg sin

yy maF

ymaNmg cos

θ

• Applications of Newton’s third law

• Determine the force of m2 on m1

amFF m 12

m1 m2

F a

amFm 21

ammF 21

21 mmFa

21

12 mmFmFF m

21

12 mmFmFFm

21

12 mm

FmFFm

amFm 21

21 mmFa

21

12 mm

FmFFm

21

21 mmFmFm

21

21 mm

FmFm

21

1?

21

2

mmFmF

mmFm

121

?

2 mmmm

21

1?

21

2 1mm

mmm

m

2112 mmmm

Tarzan’s Tension!

Tarzan’s Tension!• Determine the tension in the 12 m vine when 90.

kg Tarzan has swung to the point where he makes a 30.° angle with the vertical and is moving with a linear speed of 5.0 m/s.

mg

30°T

mgcosθ

mamgT cos

rvmmgT

2

cos

rvmmgT

2

cos

N 967120.59030cos1090

2

T

Jane’s Curves!• On the way to Nairobi one day to pick

up a few supplies, Jane was driving around a banked curve of radius r = 150 m. Some naughty monkeys had thrown banana peels onto the road making it essentially frictionless. If the road was banked at an angle of θ = 30.°, what speed would she need to drive to maintain her position (not slip up or down)?

Jane’s Curves!

maF

maN sinmg

N

mgθ

θ

rvmN

2

sin

rvmmg

2

sincos

2sincos vgr

sincosgrv

30sin30cos15010v

m/s 25v

Cheetah Dances!• On day, after he found Tarzan’s stash

of home-brewed Serengeti Special, and drank a couple bottles, Cheetah put on quite the show. If he completes two revolutions (4π radians) per second, determine the velocity of the empty bottle in his hand. The bottle is located 0.80 m from the center of his body, which is the axis of rotation.

m/s .1080.04122

rtdv

m/s .1080.044 rrv

Elephant Power!• Tarzan’s mother-in-law came for

an extended visit and when she was finally ready to leave, her Jeep was stuck. Tarzan called an elephant friend named Shep to help out. If the powerful pachyderm was pushing with a force of 2500 N when he had the vehicle moving at 20. m/s, what was his power output?

FvP

W000,05202500 P

• Vector Math

• Vectors are added head to tail

+ =a

a

b br

• Vector Math

• Vectors are subtracted by reversing the second and adding

– =a

ab b

r

• Vector Math

• Vectors are multiplied two ways: 1) Dot product2) Cross product

• Vector Math

• Fundamentally,• A dot product is

• A cross product is

cos baba

n sin baba

• Dot product

• Work is determined with the dot product:

• Dot products are scalar

dFW

• Dot product

• Evaluate if• and

• Cross product

• Angular momentum is calculated with a cross product:

• The direction of the cross product is given by the right hand rule.

• Cross products are vectors

• Evaluate if•