chapter 4 forces and the laws of motion

CHAPTER 4 FORCES and the LAWS of

MOTION

THE NORMAL FORCE

a force that acts on a surface in a direction perpendicular to the

surface

Fn

THE NORMAL FORCE

the television is in equilibrium, so the normal force from the table must be equal in magnitude and opposite in direction to

the gravitational force exerted on the television.

Fn

THE NORMAL FORCEFn is ALWAYS perpendicular to the surface,

but NOT ALWAYS

opposite to the FgFn = mg ∙ cos

The Net ForceIs the vector sum of all forces acting on the body

• 4 forces acting on the car

• Add them as vectors

• The resultant is then the NET

FORCE• If the NET FORCE

is 0, then you know that the car

is travelling at constant velocity

(zero acceleration)

Free Body Diagrams (FBDs)

are diagrams used to show the relative magnitude and direction of all forces

acting upon an object in a given situation.

Free Body Diagrams (FBDs)

• 4 forces acting on this object, but there MIGHT be more or less forces

• Represent the object as a SQUARE or a CIRCLE• You MUST show all the forces acting on the object

Free Body Diagrams (FBDs)A puck is sliding on a

sheet of ice to the right

How many forces are acting on the puck?

Label all of them!

Fapp

Fgravity

Ffriction

FA

Free Body Diagrams (FBDs)A puck is sliding on a

sheet of ice to the right

Fapp

Fgravity

Ffriction

FA

If the object is moving, you MUST show

the direction of acceleration

Free Body Diagrams (FBDs)A puck is sliding on a

sheet of ice to the right

Fapp

Fgravity

Ffriction

FA

If the object is moving, you MUST show

the direction of acceleration

a

Interpreting FBD (Quantitative = with numbers)

NET FORCE??

Fapp

Fgravity

Ffriction

FAa

Interpreting FBD (Quantitative = with numbers)

NET FORCE??

25 N

40 N

10 N

40 N a Fnet = 0 N in

vertical direction

Fnet = + 15 N in

horizontal direction

Interpreting FBD (Quantitative = with numbers)

100 N

47 N

100 N

47 N a Fnet = 0 N in

vertical direction

Fnet = 0 N in

horizontal direction

If the Net Force is 0 (Fnet = 0 N), the object is in EQUILIBRIUM

4.1 The Newton’s FIRST Law:Tells us what happens when all existing forces acting on a body

are balancedIf the Net Force is 0 (Fnet = 0 N), the

object is in EQUILIBRIUM

100N

47 N

100 N

47 N If the forces acting on a body are

balanced, then the

acceleration of the body will be 0 m/s2

4.2 The Newton’s Second Law:

Tells us what happens when an unbalanced force acts on a

bodyIf the Net Force is NOT 0 (Fnet ≠ 0 N), the

object is NOT in EQUILIBRIUM

200N

47 N

100 N

47 N aIf an

UNBALENCED force acts on a body,

the body will ACCELARAT

E

4.2 The Newton’s Second Law:

Tells us what happens when an unbalanced force acts on a

bodyThe rate at which the body will

ACCELERATE depends DIRECTLY on the UNBALANCED FORCE (Fnet) and INVERSELY on the

mass of the body. ∝a ∝ Fnet

a ∝ 1/m

4.2 The Newton’s Second Law:

Tells us what happens when an unbalanced force acts on a

bodyThe rate at which the body will

ACCELERATE depends DIRECTLY on the UNBALANCED FORCE (Fnet) and INVERSELY on the

mass of the body. ∝

Fnet = a · m

a = Fnet / m

First Law vs. Second Law

A 45.0 N block is being pushed along a floor, where the coefficient of kinetic friction is 0.333. If a horizontal force of 25.0 N is applied, at what rate will

the block accelerate? 1. FBD2. Known?3. Unknown

?4. Formula(s

)

Fg = 45.0 N

µk = 0.333

Fapp = 25.0 N

Fk = ???

Fapp = 25.0 N

Fg = 45.0 N

Fk = ???

FN = ???

m = Fg/g

= 45.0/9.81

= 4.59 kg

Fnet = a · m

Fg = m · g

Fk = µk · FN

A 45.0 N block is being pushed along a floor, where the coefficient of kinetic friction is 0.333. If a horizontal force of 25.0 N is applied, at what rate will

the block accelerate? 1. FBD2. Known?3. Unknown

?4. Formula(s

)

ClassworkPage: 112Problems: all problems

22

QUIZOn Monday: 15 minutes

Problems: from 4.1 and 4.2

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