P H Y S I C S   

Newton’s Laws Overview

Review of Newton’s Laws of Motion

Objects in motion stay in motion* and objects at rest stay at rest

if there is zero net force (balanced)

ΣF = m·a(the forces will be unbalanced)

Every force has an

equal and opposite force

* straight line/constant speed

1st

2nd

3rd

Inertia Depends on mass

More mass more resistance Less mass less resistance

Demo: NFL Hits

Equilibrium

Equilibrium: Net force is zero (ΣF = 0) ΣFx = 0

ΣFy = 0

FN

Fg

FAirFEngine

Equilibrium

ΣF = 0 Newton’s First Law applies An object in equilibrium can be:

in motion (straight line/constant speed) at rest

Ff Fengine

FN

Fg

Fair

Terminal Velocity

Once the forces of air resistance and gravity become balanced equilibrium is reached

No more

acceleration

Newton’s Second Law

If there is a net force the object will accelerate

ΣF = m·a

Units:

ΣF net force (N)

m mass (kg)

a acceleration (m/s2)

Newton’s Second Law Equations:

F = ma a = F/m m = F / a

F = net forcem = Massa= Acceleration

Use one of the equations you just wrote down…

Acceleration

Increase acceleration by: Increasing force Decreasing mass

Weight vs Mass

Weight Force Fg

Fg = m ·g

Mass: Amount of matter (does not change)

Weight: Pull of gravity (changes)

Weight Force (Fg)

g = 9.8 m/s2 g = 1.6 m/s2 g = 26 m/s2

m = 50 kgFg = 490 N

( 110 lb)

m = 50 kgFg = 80 N( 18 lb)

m = 50 kgFg = 1300 N

( 292 lb)

In-Class Problem #1

A 2000 kg car has a push force of 5000 N from its engine. If it experiences a friction force of 3000 N determine it’s (a) acceleration, (b) weight and (c) the normal force acting on it.

a = 1 m/s2

Fg = 19,600 N

FN = 19,600 N

Review of Newton’s Laws of Motion

First Law

a = 0 m/s2

in motion*at rest

stays

in motion*

stays

at rest

Second Law

Accelerates

depends inversely on mass

depends on net force

* Straight line/constant speed

ΣF = 0 ΣF ≠ 0

Force that resists motion due to imperfections in surfaces

FRICTION MOTION

Friction

Two Types

1. Static (rest): Keeps object from moving

2. Kinetic (moving): Slows moving object

Friction Force Equation

Coefficient of Friction (): Ratio between friction force and normal force:

friction force (N) coefficient of friction

normal force (N)

μk

(kinetic)

μs

(static)

Coefficient of Friction Table

In-Class Problem #2

A 30 kg desk is at rest on the floor. It takes 200 N of force to start it in motion. Determine the static coefficient of friction between the desk and the floor.

μs = 0.68

In-Class Problem #3

Once the desk in the previous problem is set in motion the 200 N force continues to be applied. Determine the acceleration of the desk if the coefficient of kinetic friction is 0.52.

a = 1.57 m/s2