slide1
Download
Skip this Video
Download Presentation
UNIT 4 Forces and the Laws of Motion

Loading in 2 Seconds...

play fullscreen
1 / 11

UNIT 4 Forces and the Laws of Motion - PowerPoint PPT Presentation


  • 190 Views
  • Uploaded on

UNIT 4 Forces and the Laws of Motion. N. T. m 1. T. m 1 g. m 2. m 2 g. Force Lab Notes. Forces on m 1. m 1 a = T = m 2 g – m 2 a. Forces on m 2. 23. Everyday Forces. A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s 2.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' UNIT 4 Forces and the Laws of Motion' - garfield-jonathon


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1
UNIT 4Forces and

the Laws of Motion

slide2
N

T

m1

T

m1g

m2

m2g

Force Lab Notes

Forces on m1

m1a = T = m2g – m2a

Forces on m2

23

slide3
Everyday Forces

A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of

3.60 m/s2.

a) Find the μk between the box and the ramp.

What acceleration would a 175 kg box

have on this ramp?

25

slide4
A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of

3.60 m/s2.

ΣFy = 0

a) Find the μ between the box and the ramp.

FN = mgcos(25°) = 667 N

ΣFx ≠ 0

FNET = ma = mgsin(25°) - Ff

FN

FNET = 270 N = 311- Ff

Ff

Ff = µFN = µ(667N) = 41N

mgcos(25°)

µ = .0614

mg

mgsin(25°)

26

slide5
A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of

3.60 m/s2.

ΣFx ≠ 0

b) What acceleration would a 175 kg box

have on this ramp?

Ff = µFN

FN

FNET = ma

ma = mgsin(25°) - Ff

Ff

ma = mgsin(25°) – μmgcos(25˚)

mgcos(25°)

mass does not matter,

the acceleration is the same!!

mg

mgsin(25°)

27

slide6
Everyday Forces

A 75.0-kg box is pushed with a 90.0N exerted downward at a 30˚

below the horizontal. If the coefficient of kinetic friction

between box and the floor is 0.057, how long does it take to move

the box 4.00m, starting from rest?

A 75.0-kg box is pushed with a 90.0N exerted downward at a 30˚

below the horizontal. If the coefficient of kinetic friction

between box and the floor is 0.057, how long does it take to move

the box 4.00m, starting from rest?

FN

t = ?

vi = 0

F

4.00 m

Ffk

Fg

28

slide7
90.0N

77.9 N

45.0 N

FN

 = 30˚

735.8 N

1. Draw a free-body diagram to find the net force.

2. Convert all force vectors into x- and y- components.

Ffk

29

slide8
= 781 N

90.0N

77.9 N

45.0 N

FN

735.8 N

 = 30˚

3. Is this an equilibrium or net force type of problem?

Net force !

4. The sum of all forces in the y-axis equals zero.

Ffk

FN = 45.0 + 735.8 N

5. Solve for the normal force.

FN = 781 N

30

slide9
= 781 N

90.0N

77.9 N

45.0 N

44.5 N

FN

735.8 N

 = 30˚

6. Given the μk = 0.057, find the frictional force.

μkFN = Ff

(0.057) 781 N = 44.5 N

Ff = 44.5 N

Ffk

7. Given this is a net force problem, net force equals m times a.

77.9 N – 44.5 N = (75 kg) a

a = .445 m/s2

31

slide10
= 781 N

90.0N

77.9 N

a = .445 m/s2

45.0 N

44.5 N

FN

735.8 N

 = 30˚

8. Which constant acceleration equation has a, vi, x, and t?

t = 4.24 s

Ffk

32

ad