Newton’s 1 st Law of Motion. Newton’s 1 st Law Newton’s 1 st Law: An object at _______stays at _______and an object in motion stays in motion, unless.

Newton’s 1st Law of Motion

Newton’s 1st Law• Newton’s 1st Law: An object at

_______stays at _______and an object in motion stays in motion, unless acted upon by an ___________(i.e. net or outside) force

• This motion is in a ___________line, at a _____________speed

• Often called an objects inertia

What Does Inertia Have To Do With Rugby?

• Jonah Lomu:– New Zealand All Blacks– Mass: 120 kg (translates

to 273 lbs)– Height: 6’ 5’’ – Speed: Runs 100 meters

in 10.8 seconds(Carl Lewis: 100 meters in

9.88 seconds)

What Does Inertia Have To Do With Rugby?

• Jonah Lomu:– Very Big (lots of mass – 120 kg)– Lots of Inertia– Because of Large Mass (i.e., large amount of

inertia) he is very resistant to changes in motion • HARD TO STOP WHEN MOVING• HARD TO START MOVING WHEN AT REST

What Does Inertia Have To Do With Rugby?

• So when Jonah Lomu is in motion and encounters someone less massive (i.e., with less INERTIA)…• HE IS MORE RESISTANT TO THEIR

ATTEMPTS TO CHANGE HIS STATE OF MOTION (i.e., bring him to a rest)

What Does Inertia Have To Do With Rugby?

• So when Jonah Lomu is at rest and encounters someone less massive (i.e., with less INERTIA)…• He is more resistant to their attempts to change his motion.

What Does Inertia Have To Do With Rugby?

• Because of his mass, it is harder to change his direction when in motion.

• WHY?– INERTIA!

• Straight Line Motion• Inertia resists changes

in motion (in this case the DIRECTION of the motion)

•The GREATER the MASS, the HARDER it is to CHANGE DIRECTION

Challenge!• Why does my water bottle roll under

my seat when leaving school!

• Why do you shift left when you turn your car right?

Newton’s 2nd Law of Moton

Newton’s 2nd Law• The acceleration of an object is directly

proportional to the net external force acting on the object and inversely proportional to the mass of the object

or:____________________

Forces

• A push or pull• The cause of an _________________• Cause of a change in an object’s state of

motion– Cause objects to speed up or slow down– Cause a change of direction

• Unit of force: ________________________

Types of Forces• Common forces:

– Applied – (FA or F) – Generic name for any force that is applied

– Tension - (Ft or T) - The force that a “string” _________on an object

– Normal Force - (FN or N) - Force that a surface applies to an object (the direction is to the surface)

– Weight – (Fg or w) – The force due to ______________(mg)

– Friction - (Ff or f ) - To be defined later

FN, Normal Force (N)• A force that a surface applies to an

object• “Normal” means perpendicular• The direction of the normal force is

perpendicular to the surface

surfacesurface

Free Body Diagrams• Used to analyze the forces affecting

the motion of a single object• Shows only the forces acting on an

object

Free Body Diagrams•One object only•Shows only forces acting on the

object•Forces represented as arrows•Fg down, FN perpendicular to

surface

Examples

• Force of Friction: Ff , Unit: N – Newton

• μ: ____________of Friction (Unitless)

• Static vs Kinetic•Static Friction –Two surfaces at _________relative to one another (not sliding)

• Always parallel to the surface

• Kinetic Friction – Two surfaces ___________relative to one another (Sliding)

• Always ______________to the surface and opposes the motion

Friction

Common μ’sMaterials μOak on oak, dry 0.30Steel on steel, dry 0.41

greasy 0.12Steel on ice 0.01Rubber on asphalt, Dry 1.07

wet 0.95Rubber on ice 0.005

Sample Problems

• It takes 50 N to pull a 6.0 kg object along a desk at constant speed. What is the coefficient of friction?

• The coefficient of friction between two materials is 0.35. A 5.0 kg object made of one material is being pulled along a table made of another material. What is the force of friction?

Newton’s 3rd Law• Every action force has an ______________and

equal reaction force.– The two forces are called force pairs.– They act on ________________objects.

– How do we walk by pushing backwards?

– Why do cannons recoil when fired?

Challenge• Is the force of gravity that the Earth exerts

on the Moon greater than, equal to, or less than the force the Moon exerts on the Earth?

• How does a hammer push a nail into wood in light of Newton’s 3rd Law?

Centripetal Force• ac, __________________acceleration (m/s2)

– “Center seeking”– Necessary to keep the object traveling in a circle

• Fc, centripetal force (N)

– Newton’s 2nd Law

• Newton’s 2nd Law for Circles

• NOTE: CENTRIFUGAL FORCE IS AN IMAGINARY FORCE TO EXPLAIN INERTIA! DOES NOT EXIST!

Centripetal Force• The centripetal acceleration is caused by

another force such as:– Friction– Tension– Normal Force– Gravity

Centripetal Force• Steps to solve centripetal force problems:

– Draw a free body diagram – Identify the force causing the centripetal force– Set the causal force = centripetal force eq.– Solve for unknown

A 0.50 kg box is attached to string on a frictionless horizontal table. The box revolves in a circle of radius 2.8 m. If the box completes 1 revolution every 2.0 seconds, what is the tension in the string?

r

FN

FgFT

A 1200 kg car approaches a circular curve with a radius of 45.0 m. If the coefficient of friction between the tires and road is 1.20, what is the maximum speed at which the car can negotiate the curve?

r

A 0.025 kg rubber stopper connected to a string is swung in a horizontal circle of radius 1.20 m. If the stopper completes 5 revolutions in 2 seconds. Calculate the period of revolution of the stopper, the magnitude of the velocity of the stopper, the magnitude of the stopper’s centripetal acceleration and the tension in the string.

In a popular amusement park ride, a cylinder of radius 3.0 m is set in rotation at a speed of 5.6 m/s.

The floor drops away, leaving the riders suspended against the wall in a vertical position What minimum coefficient of friction between a 100.0 kg rider’s clothing and the wall of the cylinder is needed to keep the rider from slipping?

Vertical Circle I

Vertical circle 2

Vertical circle 3

r=13 m

A 800 kg car drives down a hill. If it is going 11 m/s at the bottom of the hill, what is the force the road is exerting on the car (FN)?

Fg

FN

cN FmgF

mgr

vmFN 2

8.9*8001311800

2

NF

NFN 15286

r=13 m

Fg

cN FFmg

cN FmgF

13118008.9*800

2

NF

FN

A 800 kg car goes up a hill. If it is going 11 m/s at the top of the hill, what is the force the road is exerting on the car (FN)?

N394