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34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N
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34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Mar 28, 2015

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Beau Waldman
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Page 1: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

34 N

270

13 N

510

1/ 1 cm:5N 2/ 1 cm:2N

Page 2: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

1 cm:4 N

200

6.5 cm 26N

10N

1 cm:10 N

4.2 cm 42N

42N

460

1 cm:8 N1 cm:3 N

Fx = 40 N Fy = 0

Fx = 0

Fy = -6 N

Page 3: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.
Page 4: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Newton’s 1st Law of Motion:

If the ____________________ force, Fnet, acting

on an object equals ________, then that object has____________________ . Its acceleration a = _____ .It is said to be in __________________ .

zero

constant velocity

net (resultant)

equilibrium0

Because forces are ______________ and have____________ , the x and y _______________ of the forces often must be added up _______________when there are forces along different _________.

Fnet = 0 really means:________ = 0

________ = 0

vectorsdirection components

separately

net Fx

net Fy

axes

Page 5: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Ex: A 20-N box on a tableat rest

20-N

FN = 20N

w = 20 N

Ex: A 20-N box at rest on a table being pushed down with a 5 N force.

"At rest" __________

20-N

Free body diagrams:

w = FN

5-N

w = 20 N

5 N

FN = 25 N

"At rest" _______________w + push = FN

Both of these boxes are in ___________________ .equilibrium

Page 6: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Ex: Same box sliding to the right on a horizontalfrictionless table at a constant velocity 2.0 m/s

20-N

FN = 20N

Draw only the __________ , not the ____________

Notice:1/ The free body diagram is the _________ as thefirst example. Both are in ___________________ .2/ No force is ___________ to keep box moving.The box will continue its motion ________________.

equilibrium

v = 2.0 m/s

forces velocities

needed

forever

w = 20N

same

"constant velocity" __________w = FN

Page 7: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Ex: The forces acting on a 20-N box are shown at right:

FN = 20N

w = 20 N

pull= 10N

Ff = 10 N

This box is in ___________________. It is either:

1/ _____________ (constant velocity = _____);

2/ or _____________ at a constant velocity.

Unless other information is given, it is _______________ to tell if it is at rest or moving.

at rest

moving

equilibrium

0

impossible

Page 8: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

3.0 N

4.0 N

Ex. The three forces below are the only onesacting on an 2.0-kg object. What is the acceleration of the object?

5.0 N

This object could be _______________ or moving with ____________________________ .

0

at restconstant velocity

The net force Fnet = _____ so a must = ____ .0

Add them up:

4.0 N

3.0 N

5.0 N

Page 9: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Restatement of 1st Law: If an object is ____________ or moving with _____________________, it is_________________.

Then the net force (Fnet ) acting on it _________ .

lift = 1.3 MN

w= 1.3 MN

thrust = 300 kN

Ff (drag)= 300 kN

weight ≈ 1.3 x 106 Nthrust ≈ 75 x 103 N/engine x 4 engines

at restconstant velocity in equilibrium

= 0

Ex: A B-2 bomber flying at constant velocity:

Page 11: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Ex. The two forces shown below act on an object. What third force is needed to produceequilibrium?

F1

First way to solve:• Find resultant force F• The answer is -F:

F

-F

-F ___________________ the other two forces.

F2

balances

Page 12: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Second way to solve:1. Add the two forces using head to tail method.2. Add a force to "bring the total force to zero."

original

Notice:• Both ways give ________________________• -F is NOT the ________________ of the other vectors. It is the force that ________________

the other vectors, making Fnet =________. It is

sometimes called the ________________ because it is what is needed to produce ___________________ .

the same result

resultant"balances out"

equilibrantequilibrium

-F

F1

F2

F1F2

head to tail:

0

Page 13: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

wire

beam

build

ing

weight

Ex. Draw the vector that represents the force that the beam exerts on point A.

A A

T

w

beamforce

The beam force is the ________________ . It must be added to the other forces so that ___________

equilibrantFnet = 0

Page 14: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Ex. Increasing angle between 2 wires supporting a picture:

Case 1: vertical wires

Case 2: greater angle

Case 3: even greater angle

10 N

w = 10 N

T = ____in eachwire

w = 10 N

w = 10 N

10 N

10 N

1 2

1 12 2

12

1 2

T = ____in eachwire

T = ____in eachwire

As increases, the tension _______________ .

51 2

7

10

increases

Page 15: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

A net force is ________________ to maintain motion. An object will maintain its motion in the ________________ of a net force. This idea was discovered by _____________. An object in motion tends to ____________________ , and anobject at rest tends to _________________ . The _______________________ that “makes” this true is called its ___________ .

more __________ more___________in

ert

iamass

Ex. What has more inertia, a truck or a baseball?

property of matter

inertia

absence

not needed

Galileostay in motion

stay at rest

massinertia

truck

Page 16: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Inertia is the tendency of an object to _________________________ . More inertia (mass) means it is __________________ for an object to…1/ …_____________________ when it is at rest;2/ … __________ when it is already moving; and,3/ … ____________________ from a straight path.In other words, ______________ is why an objectneeds __________________ to maintain its motion.This is why Newton's _____________ is calledthe ________________ .

resist changes in motion

more difficultstart moving

stopmove sideways

Law of Inertia

1st Law

no net forceinertia

Fnet a

equilibrium

= 0 = 0

Lawof

Inertia

Page 17: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.

Open your 3-ring binder to theWorksheet Table of Contents.

Record the title of the worksheet:

Newton’s First Law WS

Page 18: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.
Page 19: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.
Page 20: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.
Page 21: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.
Page 22: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.
Page 23: 34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F.