CHAPTER 4 The Laws of Motion Newton’s First Law: Newton’s First Law: An object at rest remains at rest and an object in motion continues in motion with.

CHAPTER 4CHAPTER 4The Laws of MotionThe Laws of Motion

Newton’s First Law: Newton’s First Law: An object at rest remains at rest and an object in motion continues in motion with constant velocity (constant speed in straight line) unless acted on by a net external force.

““in motion”in motion” or““at rest”at rest” – with respect to the chosen frame of reference““net force”net force” – vector sum of all the external forces acting on the

object – FNet,x and FNet,y calculated separately

Forces:Forces: Contact Forces*Applied Forces (push or pull)

*Normal Force (supporting force)

*Frictional Force (opposes motion)

Field Forces*Gravitational·Magnetic·Electrostatic

*The typical four forces analyzed in our study of classical mechanics

Newton’s Second Law:Newton’s Second Law: The acceleration of an object is directly proportional to the net force acting on itFNet = ma

Mass – The measurement of inertia (“inertial mass”)Inertia – The tendency of an object to resist any attempt

to change its motionBook Example:

1. Strike golf ball w/golf club2. Strike bowling ball w/golf club

Which has greatest inertia?Which has greatest mass?

Dimensional AnalysisDimensional AnalysisF = ma = kg x m/s2

= newton= N

1 newton = 1 kg · m/s2

Weight and the Gravitational ForceWeight and the Gravitational Force

MassMass – an amount of matter (“gravitational mass”)“Your mass on the Moon equals your mass on Earth.”

WeightWeight – the magnitude of the force of gravity acting on an amount of matter

F = maFg = mgw = mg

NOTE: Your text treats weight (w) as a scalar rather than as a vector.

ExampleExampleYour mass is 80kg. What is your weight?

w = 80kg · 9.8m/s2

w = 780 kg·m/s2

w = 780 N

Newton’s Third Law:Newton’s Third Law: If two objects interact, the force exerted on object 1 by object 2 is equal in magnitude but opposite in direction to the force exerted on object 2 by object 1

Example:Example: (Contact Force)Book

Table

Book pushes down on table with force of 9.8.NTable pushes up on book with force of 9.8.N

Net Force on book =9.8N – 9.8N = 0NHence, book does not accelerate up or down.

Example:Example: (Field Force)

Earth FMoonMoonFEarth

Earth pulls on Moon equal to the force the Moon pulls on Earth.

Problem Solving StrategyProblem Solving StrategyRemember: We are working now with only 4 forces.

• Applied Force Fa• Normal Force FN• Frictional Force Ff• Gravitational Force Fg

Draw a SketchDraw a Sketch FN

Fa

Fg

Ff

Determine the Magnitude of Forces in “x” and in “y” DirectionFN often equals Fg (object does not accelerate up off surface or accelerate downward through surface)FNet,y = FN – Fg = 0 NFNet,x = Fa – Ff = maFf < Fa

Label forces on SketchSolve Problem

Example 1:Example 1: Sliding “Box” Problem (Horizontal Fa)“Box” = hockey puck

= shopping cart= tire= dead cat= etc.

A 55 kg shopping cart is pulled horizontally with a force of 25N. The frictional force opposing the motion is 15N. How fast does the cart accelerate?

Fa=25N

FN=540N

Fg=540NFf=15N

Fa = 25NFf = 15NFNet,x = 25N – 15N = ma

= 10.N = 55kg·aa = .18m/s2

Fg = mg = 55kg · 9.8m/s2

= 540NFN = Fg = 540N

Example 2:Example 2: Sliding “Box” Problem (Pulled at an Angle)A dead cat with a mass of 7.5kg is pulled off the road by a passing motorist. The motorist pulls the cat by its tail which is at an angle of 37° to the horizontal. A force of 25N is applied. The force of friction opposing motion is 18N. How fast does the cat accelerate?

m = 7.5kgFa = 25N Fa,x = Fa cos37 = 20.N

Fa,y = Fa sin 37 = 15NFf = 18NFNet,x = Fa,x – Ff = ma

20.N – 18N = 7.5kg · a

a = .27m/s2

FN + Fa,y = Fg (up forces equal down forces)

Fg = mg = 74NFN = 74N – 15NFN = 59N

=74N

=15N=20N

59N=

18N=Fa,x

Fa,yFN

Fg

Ff

F a=25N

FrictionFrictionFriction opposes motion.Kinetic Friction opposes motion of a moving object.Static Friction opposes motion of a stationary object.

Ff = FN

static = coefficient of static frictionkinetic = coefficient of kinetic friction

s > k Why?Static condition: peaks and valleys of the two surfaces overlap each other.Kinetic Condition: surfaces slide over each other touching only at their peaks

s > k Ff,s > Ff,k

Applied Physics Example: Anti-lock Brakes

Example 3:Example 3: Sliding “Box” (Pulled at Angle: advanced)A box is pulled at a 37° angle with increasingly applied force. The box which has a mass of 15kg begins to move when the applied force reaches 50.N. What is the coefficient of static friction between the box and the surface?

FN

Fg

Ff

F a

Fa,x

Fa,y37 °

Fa =

Fa,x =

Fa,y =

Fg =

FN + Fa,y = Fg

FN = 120N

Ff,s = At the point where box started to move

Ff,s = s FN = Fa,x

= s· 120N = 40Ns = .33

50.N

Fa cos 37 = 40.N

Fa sin 37 = 30.N

mg = 150N

Fa,x

Forces on an Inclined PlaneForces on an Inclined Plane

Fg is always directed straight down.

We then choose a Frame of Reference where the x-axis is parallel to the incline and the y-axis perpendicular to the incline.

Fg,x = Fg sinFg,y = Fg cosFN = Fg,y (in opposite direction)Fa and Ff will be along our new x-axis

= 30°

Fg

Fgy

Fgx

xy

Example ProblemExample Problem (Inclined Plane)A 25.0kg box is being pulled up a 30° incline with a force of 245N. The coefficient of kinetic friction between the box and the surface is .567. Calculate the acceleration of the box.

Draw a SketchDraw a Sketch

Fg · sin = 245N · sin30= 123N (to left along x-axis)

Label Forces on your sketch Label Forces on your sketch Solve the ProblemSolve the Problem

kFN = .567 · 212N = 120N

(to left along x-axis)

Determine the Magnitude of the forces in x and y directionsDetermine the Magnitude of the forces in x and y directions

= 30°

Fg

Fgy

Fgx

xy

m = Fg =

Fg,x =

Fg,y =

mg = 25.0kg · 9.80m/s2 = 245N (down)

25.0 kg

Fg · cos = 245N · cos30= 212N (down along y-axis)

Fa =FN =

Ff =

245N (to right along x-axis)Fg cos = 212N (up along y-axis)

Solve the ProblemSolve the Problem

Fa – Ff – Fg,x

= 245N – 120.N – 123N

= 2N

FNet,x = max

2N = 25.0kg · ax

ax = .08 m/s2

NOTE: The box may be moving up the incline at any velocity. However, at the specified conditions it will

be accelerating.

FNet,x =

Example Problem (Connected Objects – Flat Surface)Example Problem (Connected Objects – Flat Surface)

Two similar objects are pulled across a horizontal surface at constant velocity. The required Fa is 350.N. The mass of the leading object is 125kg while the mass of the trailing object is 55kg. The values for k are the same for each object. Calculate k and calculate the Force of “Tension” in the connecting rope.

NOTE: FT = Force of Tension is not a new type of force. It is just a specific type of applied force.

• Label the forces.• Calculate the magnitude of the forces.• Solve the problem(s).

FT Fa

Fg,1 =

Fg,2 =

FN,1 =FN,2 = Ff,1 = Ff,2 = FNet,x =

Fa = k = .20FT = FT = 110N

FT

Ff,1

m1 = 125kg

Fg,1

FN,1

Fa=350.N

Ff,2

m2 = 55kg

Fg,2

FN,2

m1g = 125kg · 9.80m/s2

= 1230N (down)m2g = 55kg · 9.80m/s2

= 540N (down)1230N (up)540N (up)k · 1230N (left)k · 540N (left) 0N Constant Velocity a = 0m/s2 = maFf,1 + Ff,2 = k · 1230N + k · 540N

Fa = 350.N = k (1230N + 540N)

k · 540N

Example ProblemExample Problem (Elevators)

m1

m2

Two weights are connected across a frictionless pulley by

weightless string. Mass of object 1 is 25.0kg. The mass of object 2

is 18.0kg. Determine the acceleration of the two objects.

m1g = 25.0kg · 9.80m/s2

= 245N (down on right)m2g = 18.0kg · 9.80m/s2

= 176N (down on left) 245N – 176N= 69N (down on right)

a = 1.60 m/s2

m1 accelerates downm2 accelerates up

69N = ma = (m1 + m2) · a69N = (25.0kg + 18.0kg) · a

Fg,1 =

Fg,2 =

Fg,net =