6 second law

8/11/2019 6 Second Law

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Newtons Second Law

Chapter 6

The Second Law

Force = mass X acceleration

F = 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

Equilibrium

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 Newton

F=ma

F = (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.


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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

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 cars 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 rocket

b. Calculate the acceleration of the rocket

c. Calculate the height and velocity of the

rocket after 5 s

d. Calculate the maximum height of the rocket

even after the engine has burned out.


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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)

Metric: Newton

English: 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 mg

FN

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

mg

FN

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

mg

FN

a is down


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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 > gmg

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 upwardwith a force of 100 N?

SF = FN+ Fp mgSF = 0 +100.0N 98N = 2.0N

ma = 2N

a = 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


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Friction

Always opposes the direction of motion.

Proportional to the Normal Force (more

massive objects have more friction)

mg

FN

mg

FN

FaFfr

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

Rolling Friction

Much lower than fs and fk 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 thesurface

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.


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Your little sister wants a ride on her sled. Should

you push or pull her?

Inclines

What trigonometric function does this resemble?

Inclines

q

mg

FN

Inclines

q mg

FN

q mgcosq

mgsinq

Inclines

q

FN

mgcosq

mgsinq

Ffr


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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 mk 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= 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 skiersmass)

To find the speed after 4 seconds:

v = vo + at

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


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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

A dog sled and rider have a mass of 200.0 kg. It

takes the dogs 15.0 m to reach their cruising

speed of 5.00 m/s. The ropes are connected

upwards to the two dogs at 10.0o. Calculate

the tension in the ropes at the start of the

race. (mk= 0.06)

(140 N)

Drag: Low SpeedsD -bv (works well for liquids)

D = drag force

b = constant

v = velocity

Derive the formula for the terminal speed of a

body falling through a fluid. Remember that

the sum of all forces will be zero.

Setup, but do not solve, a differential equationfor the velocity of the particle before reaching

terminal velocity.

Drag: High Speeds

D Av2 (works well for air)

D = drag force

A = Area

V = velocity

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)

c. Setup, but do not solve, a differential

equation for the velocity of the particle.


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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 from the ground,

one on each side connected to the dogs. (k

=

0.060)a. Calculate the acceleration of the sled. (0.833

m/s2)

b. Calculate T1 and T2 during the acceleration

period. (140 N)

Formula Wrap-Up

F=ma

Weight = mg (g = 9.8 m/s2)

Fmax = sFN

Ffr = kFN

4. 508 N

8. 4 m/s2 (0-3s) -6 m/s2 (3-6s)

8 N (at 1s) -12 N (at 7s)

10. a) T = 0 N b) F = T = 0N c) 250 N

12. a) 3.96 N b) 2.32 N

14.a) 540 N b) 89 N

16. 1035 N, 740 N, 590 N

20. 10, 154 N

26. 0.56 N, 57 g30. a ) 0.0036 N b) 0.0104 N

26. 0.56 N, 57 g

40. a) 15.7 N b) 2.9 m/s c) 4.4 m/s

42. 0.165

52. Stays at rest

6 second law

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