Newton’s Second Law

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# Newton’s Second Law - PowerPoint PPT Presentation

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

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### Newton’s Second Law

Chapter 6

The Second Law

Force = mass X acceleration

SF = ma

SF = 0 or SF = 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

Force of the material of the rock

Force of gravity

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

The Second Law

Situation Two: Moving Object: Constant Velocity

SF = 0

Fpedalling = Fair + Ffriction

Fpedalling

Fair

Ffriction

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

The Second Law

Situation Two: Moving Object: Accelerating

SF = ma

ma = Fpedalling – Fair - Ffriction

Fpedalling

Fair

Ffriction

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

The Second Law

Unit of Force = the Newton

SF=ma

SF = (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
• No motion
• Constant velocity

BOTH INDICATE NO ACCELERATION

SF=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
• SF=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/s

v2 = vo2 + 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 car

b. 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.

• Calculate the net force on the rocket
• Calculate the acceleration of the rocket
• Calculate the height and velocity of the rocket after 5 s
• Calculate the maximum height of the rocket even after the engine has burned out.
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 Velocity

a= 0

SF = 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

FN

mg

a is zero

Elevator Accelerating Upward

a = 4.9 m/s2

SF = FN – mg

ma = FN – mg

FN = ma + mg

FN = m(a + g)

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

FN = 1500 N

FN

mg

a is upward

Elevator Accelerating Downward

a = -4.9 m/s2

SF = FN – mg

ma = FN – mg

FN = ma + mg

FN = m(a + g)

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

FN = 500 N

FN

mg

a is down

At what acceleration will he feel weightless?

FN = 0

SF = FN – mg

ma = FN – mg

ma = 0 – mg

ma = -mg

a = -9.8 m/s2

Apparent weightlessness occurs if a > g

FN

mg

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

FN

Fg = W

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.

SF = FN+ Fp – mg

SF = 0 +100.0N – 98N = 2.0N

ma = 2N

a = 2N/10.0 kg = 0.2 m/s2

Fp = 100.0 N

Fg = mg = 98.0 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)

Fp = 40.0 N

30o

FN

mg

Friction
• Always opposes the direction of motion.
• Proportional to the Normal Force (more massive objects have more friction)

FN

Ffr

Ffr

Fa

FN

Fa

mg

mg

Friction

Static -opposes motion before it moves (ms)

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

Fmax = msFN

Kinetic - opposes motion while it moves (mk)

• Generally less than static friction

Ffr = mkFN

Friction and Rolling Wheels

Rolling uses static friction

• A new part of the wheel/tire is coming in contact with the road every instant

B

A

Braking uses kinetic friction

Point A gets drug across the surface

A

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

• How much force does she exert? (98 N)
• If she stops pushing, calculate the acceleration. (-1.96 m/s2)
• Calculate how far the box slides until it stops. (1.00 m)

A 100 kg box is on the back of a truck (ms = 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.

Inclines

What trigonometric function does this resemble?

Inclines

FN

q

mgcosq

mg

q

mgsinq

Inclines

FN

Ffr

mgsinq

mgcosq

q

• Calculate the magnitude of the static friction on the cabinet when the bed of the truck is tilted at 20.0o (170 N)
• Calculate the angle at which the cabinet will start to slide. (39o)
Given the following drawing:
• Calculate the acceleration of the skier. (snow has a mk of 0.10) (4.0 m/s2)
• Calculate her speed after 4.0 s? (16 m/s)
FGy = mgcos30o

FGx = mgsin30o

The pull down the hill is:

FGx = mgsin30o

The pull up the hill is:

Ffr= mkFN

Ffr= (0.10)(mgcos30o)

SF = pull down – pull up

SF = 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 mk

SF = pull down – pull up

ma = mgsin30o– (mk)(mgcos30o)

ma = mgsin30o– (mk)(mgcos30o)

a = gsin30o– (mk)(gcos30o)

Since the speed is constant, acceleration =0

0 = gsin30o– (mk)(gcos30o)

(mk)(gcos30o) = gsin30o

mk= gsin30o= sin30o = tan30o =0.577

gcos30o cos30o

Drag

D ≈ ¼Av2

D = drag force

A = Area

V = velocity

Fails 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
• Find the formula for terminal speed (a=0) for a freefalling body
• 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 (mk = 0.800). Calculate the stopping distance:

• On a level road. (57.0 m)
• Up a 10.0o incline (48.0 m)
• 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. (mk = 0.060)

• Calculate the acceleration of the sled. (0.833 m/s)
• Calculate T1 and T2 during the acceleration period. (140 N)
Formula Wrap-Up

SF=ma

Weight = mg (g = 9.8 m/s2)

Fmax = msFN

Ffr = mkFN