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Ch 5 lecture.notebook 1 January 08, 2015 kok. 307:11 PM kok. 307:11 PM Chapter Five: Forces in Two Dimensions Vector review: vectors have magnitude and direction add vectors tail to tip, any order draw a vector from tail of first vector to the tip of the last vector = resultant

Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

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Page 1: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

1

January 08, 2015

kok. 30­7:11 PM

kok. 30­7:11 PM

Chapter Five:  Forces in Two Dimensions

Vector review:

    vectors have magnitude and direction

    add vectors tail to tip, any order

draw a vector from tail of first vector to the tip of the last vector

= resultant

Page 2: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

2

January 08, 2015

kok. 30­7:11 PM

Vectors at right angle to each other : use pythagorean theorem

Bob walks 3 miles west then turns and walks 4 miles north, how far is Bob from his original starting point?

R2 =  A2  +  B2

kok. 30­7:11 PM

Any triangle:  can find resultant with Law of Cosines

R2 = A2  +  B2  ­ 2AB cos 

Bob walks 10 miles due east, then turns 25 degrees to the north and walks 4 more miles.  How far is Bob from his original starting point?

Page 3: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

3

January 08, 2015

kok. 30­7:11 PM

Practice drawing vectors on a coordinate plane

15 degrees north of west 40 degrees east of north30 degrees west of south

kok. 30­7:11 PM

Find the magnitude of the sum of two forces, one 20.0 N and the other 7.0 N when the angle between them is 30.0 degrees.

Page 4: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

4

January 08, 2015

kok. 30­7:11 PM

Components of vectors

+ x

­ y

+ x

+ y

­ x

+ y

­ x

­ y

SOHCAHTOA

sin      =    opposite

     hypotenuse

cos      =    adjacent

     hypotenuse

tan      =    opposite

       adjacent

kok. 30­7:11 PM

7.0 m

35 degrees north of east

vector resolution:  breaking a vector into its x­ and y­ components

x­component :  parallel to x­axis

y­component : parallel to y­axisR=

Rx

Ry

in this case, Ry = side opposite the angle

so use sin 0 =   opposite

hypotenuse

Rx = side adjacent to the angle

use cos 0 =  adjacent

hypotenuse

Page 5: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

5

January 08, 2015

kok. 30­7:11 PM

find the x and y components of the vector

kok. 30­7:11 PM

50 degres west of north

23.5 N

Find the x and y components

Page 6: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

6

January 08, 2015

kok. 30­7:11 PM

43 degrees west of south

Find the x and y components

14.5 m/s

kok. 30­7:11 PM

73.7 degrees south of east

173 m

Page 7: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

7

January 08, 2015

kok. 30­7:11 PM

Algebraic Addition of Vectors

travel 5.0 m east, then 7.5 m north

kok. 30­7:11 PM

but what about more than 2 vectors?

5.0 m east, 7.4 m north, 1.5 m east, 2.2 m south

Draw resultant

find x and y components of the right triangle of the resultant

Page 8: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

8

January 08, 2015

Dec 3­1:38 PM

What about more complicated vectors?

5.0 m  Northeast, then 3.0 m due east, then 7.0 m  15 north of west

Find x and y components of each, THEN add all x's to = Rx, and add all y's to = Ry

But.........also need a direction.

tan 0 = opposite

adjacent

Dec 3­1:38 PM

A pack of four arctic wolves are exerting four different forces on the carcass of a 500 kg dead polar bear.  Wolf 1 is pulling with a force of 20. N due North.  Wolf 2 is pulling with 45 N, 30 north of west.  Wolf 3 is pulling with 25 N due West, and Wolf four is pulling with 35 N, 10 west of south.

Page 9: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

9

January 08, 2015

Dec 3­1:38 PM

5.2 Friction

force that opposes motion

requires contact between 2 surfaces

2 types:

kinetic ­ at least one surface is in motion

static ­ no motion between the surfaces

Depends on:

materials the surfaces are made of

normal force

Dec 3­1:38 PM

graph on p 127

as Fn   ,  kinetic friction force also

(lines are sloping upward)

sandpaper has steeper slope than polished table

slope = coefficient of kinetic friction

 

F f, kinetic = normal force

F f, static =  

coefficient of static friction

Page 10: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

10

January 08, 2015

Dec 3­1:38 PM

if an object is sitting on the floor and not moving, draw all of the forces in both directions acting upon it

What if someone pushed on the object to the right, but it still didn't move?  Draw all of the forces acting upon it.

Dec 3­1:38 PM

Now someone pushes it with enough force that it travels at a constant velocity to the right.  Draw all of the forces acting upon it.

Page 11: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

11

January 08, 2015

Dec 3­1:38 PM

someone pushes on the box so that it is accelerating to the right.  Draw all of the forces acting upon it.

Dec 3­1:38 PM

If you exert a 35.0 N force to the right on a 14.0 kg box of books so that it travels at a constant velocity of 0.75 m/s, what is the force of friction acting upon the box of books?

Page 12: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

12

January 08, 2015

Dec 3­1:38 PM

What is the friction force on a 10.0 N sled that is moving at 3.0 m/s if the coefficient of static friction is 0.30 and the coefficient of kinetic friction is 0.12?

Dec 3­1:38 PM

What is the friction force when a refrigerator with a mass of 150.0 kg is being pushed at a constant velocity?   0.43 0.77

Page 13: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

13

January 08, 2015

Dec 3­1:38 PM

If you use a horizontal force of 30.0 N to slide a 12.0 kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?

Dec 3­1:38 PM

What is the minimum force with which you would need to push to get a dresser moved from its spot?  The dresser has a mass of 450. kg and  0.62 0.89

Page 14: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

14

January 08, 2015

Dec 3­1:38 PM

Unbalanced friction forcesIf there is an acceleration     there is an Fnet

previous problems were BALANCED friction forces 

= constant velocity 

= no acceleration 

= no Fnet

Dec 3­1:38 PM

Fg

Fn

Pushing on the box to the right with a force of 25.0 N and the friction force is 15.0 N.

What will be the acceleration of the box?

What is the coefficient of kinetic friction?

Ff Fpush14.0 kg

.

Page 15: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

15

January 08, 2015

Dec 3­1:38 PM

A  32.0 kg fireman slides down a pole at constant velocity. What is the friction force of his hands on the pole?

If  he loosens his grip and decreases the friction force to 310 N, what is his acceleration downward?

Dec 3­1:38 PM

At a wedding reception, a small, 25 kg boy is being drug by his heels across the dance floor by his cousin.  If the coefficient of friction between they boy's pants and the floor is 0.15, what is the frictional force acting upon him as he slides?

If the cousin is pulling with 32 N of force eastward, what is the small boy's acceleration?

Page 16: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

16

January 08, 2015

Dec 3­1:38 PM

Dec 3­1:38 PM

5.3 Force and motion in Two Dimensions

an object is in equilibrium when the net force on it is ZERO

ex:  

Fg

Fair resistance object is NOT acceleration because the net force = zero

will be at rest OR at constant velocity

Page 17: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

17

January 08, 2015

Dec 3­1:38 PM

IF there is a net force, then the object is accelerating

Fg

Fair resistance = 5 N

=  ­ 12N

Net force = 5N ­ 12N

= ­ 7N

the equilibrant is the force that will put an object in equilibrium

so.....if the net force is 7 N downward, the equilibrant is 7 N upward

   because then all of the forces added together will equal zero

5N ­ 12N  + 7N  = 0N

Dec 3­1:38 PM

Practice:Find the equilibrant for:

3 N north and 4 N east

12 N left plus 15 N right

20.0 N north, 5 N west, 15 N at 54 degrees north of east

Page 18: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

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January 08, 2015

Dec 3­1:38 PM

Motion along an inclined plane

motion of boxmo

tion of 

box

you need to tip your coordinate system to match the slope of the hill

Dec 3­1:38 PM

Fg is straight down towards center of earth

Fgy

FgxFg

Fn = Fgy

Page 19: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

19

January 08, 2015

Dec 3­1:38 PM

Dec 3­1:38 PM

Page 20: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

20

January 08, 2015

Dec 3­1:38 PM

Dec 3­1:38 PM

Page 21: Ch 5 lecture.notebook · Ch 5 lecture.notebook 17 January 08, 2015 Dec 31:38 PM IF there is a net force, then the object is accelerating Fg Fair resistance = 5 N = 12N Net force =

Ch 5 lecture.notebook

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January 08, 2015

Dec 3­1:38 PM