Newton’s Second Law Chapter 6. The Second Law Force = mass X acceleration F = ma F = 0or F = ma...

Newton’s Second Law

Chapter 6

The Second Law

Force = mass X accelerationF = ma

F = 0 or F = ma -Still object -Accelerating object-Obj. at constant velocity

Sum of all the forces acting on a body

Vector quantity

The Second Law

Situation One:Non-moving Object• Still has forces

http://alfa.ist.utl.pt/~vguerra/Other/Rodin/thinker.jpg

Force of the material of the rock

Force of gravity

The Second Law

Situation Two: Moving Object: Constant Velocity

http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg

Fpedalling

Ffriction

Fair

F = 0

Fpedalling = Fair + Ffriction

The Second Law

Situation Two: Moving Object: Accelerating

http://2003.tour-de-france.cz/images/foto/05-07-2003/armstrong4.jpg

Fpedalling

Ffriction

Fair

F = ma

ma = Fpedalling – Fair - Ffriction

The Second Law

Unit of Force = the NewtonF=maF = (kg)(m/s2)1 N = 1 kg-m/s2 (MKS)

1 Newton can accelerate a 1 kg object from rest to 1 m/s in 1 s.

Equilibrium

1. No motion2. Constant velocity

BOTH INDICATE NO ACCELERATIONF=0

Three ropes are tied together for a wacky tug-of-war. One person pulls west with 100 N of force, another south with 200 N of force. Calculate the magnitude and direction of the third force.

100 N

200 N

?

A car with weight 15,000 N is being towed up a 20o slope (smooth) at a constant velocity. The tow rope is rated at 6000 N. Will it break?

Accelerating Systems

• F=ma• Must add up all forces on the object

What Force is needed to accelerate a 5 kg bowling ball from 0 to 20 m/s over a time period of 2 seconds?

Calculate the net force required to stop a 1500 kg car from a speed of 100 km/h within a distance of 55 m.

100 km/h = 28 m/sv2 = vo

2 + 2a(x-xo)

a = (v2 - vo2)/2(x-xo)

a = 02 – (28 m/s)2/2(55m) = -7.1 m/s2

A 1500 kg car is pulled by a tow truck. The tension in the rope is 2500 N and the 200 N frictional force opposes the motion. The car starts from rest.a. Calculate the net force on the carb. Calculate the car’s speed after 5.0 s

A 500.0 gram model rocket (weight = 4.90 N) is launched straight up from rest by an engine that burns for 5 seconds at 20.0 N.

a. Calculate the net force on the rocketb.Calculate the acceleration of the rocketc. Calculate the height and velocity of the rocket

after 5 sd.Calculate the maximum height of the rocket

even after the engine has burned out.

Calculate the sum of the two forces acting on the boat shown below. (53.3 N, +11.0o)

Mass vs. Weight

Mass– The amount of matter in an object/INTRINSIC

PROPERTY– Independent of gravity– Measured in kilograms

Weight– Force that results from gravity pulling on an object– Weight = mg (g = 9.8 m/s2)

Mass vs. Weight

• Weight = mg is really a re-write of F=ma.– Weight is a force– g is the acceleration (a) of gravity

• Metric unit of weight is a Newton • English unit is a pound

A 60.0 kg person weighs 1554 N on Jupiter. What is the acceleration of gravity on Jupiter?

Elevator at Constant Velocitya= 0F = FN – mg

ma = FN – mg

0 = FN – mg

FN = mg

Suppose Chewbacca has a mass of 102 kg:

FN = mg = (102kg)(9.8m/s2)

FN = 1000 N mg

FN

a is zero

Elevator Accelerating Upward

a = 4.9 m/s2

F = FN – mg

ma = FN – mg

FN = ma + mg

FN = m(a + g)

FN=(102kg)(4.9m/s2+9.8 m/s2)

FN = 1500 Nmg

FN

a is upward

Elevator Accelerating Downward

a = -4.9 m/s2

F = FN – mg

ma = FN – mg

FN = ma + mg

FN = m(a + g)

FN=(102kg)(-4.9m/s2+9.8 m/s2)

FN = 500 Nmg

FN

a is down

At what acceleration will he feel weightless?

FN = 0

F = FN – mg

ma = FN – mg

ma = 0 – mgma = -mga = -9.8 m/s2

Apparent weightlessness occurs if a > g

mg

FN

A 10.o kg present is sitting on a table. Calculate the weight and the normal force.

Fg = W

FN

Suppose someone leans on the box, adding an additional 40.0 N of force. Calculate the normal force.

Now your friend lifts up with a string (but does not lift the box off the table). Calculate the normal force.

What happen when the person pulls upward with a force of 100 N?

F = FN+ Fp – mgF = 0 +100.0N – 98N = 2.0N

ma = 2Na = 2N/10.0 kg = 0.2 m/s2

Fg = mg = 98.0 N

Fp = 100.0 N

Free Body Diagrams: Ex. 3

A person pulls on the box (10.0 kg) at an angle as shown below. Calculate the acceleration of the box and the normal force. (78.0 N)

mg

FN

Fp = 40.0 N

30o

Friction

• Always opposes the direction of motion.• Proportional to the Normal Force (more

massive objects have more friction)

mg

FN

mg

FN

FaFfr

F fr

F a

Friction

Static -opposes motion before it moves (s)

• Generally greater than kinetic friction• Fmax = Force needed to get an objct moving

Fmax = sFN

Kinetic - opposes motion while it moves (k)

• Generally less than static frictionFfr = kFN

Friction and Rolling Wheels

Rolling uses static friction– A new part of the wheel/tire is coming in contact

with the road every instant

A

B

Braking uses kinetic friction

A

Point A gets drug across the surface

A 50.0 kg wooden box is pushed across a wooden floor (k=0.20) at a steady speed of 2.0 m/s.

a. How much force does she exert? (98 N)b.If she stops pushing, calculate the

acceleration. (-1.96 m/s2)c. Calculate how far the box slides until it stops.

(1.00 m)

A 100 kg box is on the back of a truck (s = 0.40). The box is 50 cm X 50 cm X 50 cm.

a. Calculate the maximum acceleration of the truck before the box starts to slip.

Your little sister wants a ride on her sled. Should you push or pull her?

Inclines

What trigonometric function does this resemble?

Inclines

mg

FN

Inclines

mg

FN

mgcos

mgsin

Inclines

FN

mgcos

mgsin

Ffr

A 50.0 kg file cabinet is in the back of a dump truck (s = 0.800).

a. Calculate the magnitude of the static friction on the cabinet when the bed of the truck is tilted at 20.0o (170 N)

b.Calculate the angle at which the cabinet will start to slide. (39o)

Given the following drawing:a. Calculate the acceleration of the skier. (snow

has a k of 0.10) (4.0 m/s2)

b.Calculate her speed after 4.0 s? (16 m/s)

First we need to resolve the force of gravity into x and y components:

FGy = mgcos30o

FGx = mgsin30o

The pull down the hill is:FGx = mgsin30o

The pull up the hill is:Ffr= kFN

Ffr= (0.10)(mgcos30o)

F = pull down – pull upF = mgsin30o– (0.10)(mgcos30o)ma = mgsin30o– (0.10)(mgcos30o)ma = mgsin30o– (0.10)(mgcos30o)a = gsin30o– (0.10)(gcos30o)a = 4.0 m/s2

(note that this is independent of the skier’s mass)

To find the speed after 4 seconds:

v = vo + at

v = 0 + (4.0 m/s2)(4.0 s) = 16 m/s

Suppose the snow is slushy and the skier moves at a constant speed. Calculate k

F = pull down – pull upma = mgsin30o– (k)(mgcos30o)

ma = mgsin30o– (k)(mgcos30o)

a = gsin30o– (k)(gcos30o)

Since the speed is constant, acceleration =00 = gsin30o– (k)(gcos30o)

(k)(gcos30o) = gsin30o

k= gsin30o = sin30o = tan30o =0.577

gcos30o cos30o

Drag

D ≈ ¼Av2

D = drag forceA = AreaV = velocityFails for• Very small particles (dust)• Very fast (airplanes)• Water and dense fluids

Finding Acceleration with Drag

Derive the formula for the acceleration of a freefalling object including the drag force.

Terminal Speed

a. Find the formula for terminal speed (a=0) for a freefalling body

b. Calculate the terminal velocity of a person who is 1.8 m tall, 0.40 m wide, and 75 kg. (64 m/s)

A 1500 kg car is travelling at 30 m/s when the driver slams on the brakes (k = 0.800). Calculate the stopping distance:

a. On a level road. (57.0 m)b.Up a 10.0o incline (48.0 m)c. Down a 10.0o incline (75.0 m)

A dogsled has a mass of 200 kg. The sled reaches cruising speed, 5.0 m/s in 15 m. Two ropes are attached to the sled at 10.0o, one on each side connected to the dogs. (k = 0.060)

a. Calculate the acceleration of the sled. (0.833 m/s)

b.Calculate T1 and T2 during the acceleration period. (140 N)

Formula Wrap-Up

F=maWeight = mg (g = 9.8 m/s2)Fmax = sFN

Ffr = kFN