UNIT 3 Forces and the Laws of Motion

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# UNIT 3 Forces and the Laws of Motion - PowerPoint PPT Presentation

UNIT 3 Forces and the Laws of Motion. Mon day October 24 th. FORCES & THE LAWS OF MOTION. TODAY’S AGENDA. Mon day , October 24. Laws of Motion Mini-Lesson: Everyday Forces ( 2 nd Law Problems ). UPCOMING…. Thurs : Newton’s 2 nd Law Lab Fri: Quiz #2 2 nd Law Problem

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## UNIT 3 Forces and the Laws of Motion

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UNIT 3Forces and

the Laws of Motion

### Monday October 24th

FORCES & THE LAWS OF MOTION

TODAY’S AGENDA

Monday, October 24

• Laws of Motion
• Mini-Lesson: Everyday Forces (2ndLaw Problems)

UPCOMING…

• Thurs: Newton’s 2nd Law Lab
• Fri: Quiz #2 2ndLaw Problem
• Mon: Test Review
• Tue: TEST #4

N

T

m1

T

m1g

m2

m2g

Force Lab Notes

Forces on m1

m1a = T = m2g – m2a

Forces on m2

23

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

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

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

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

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

= 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

= 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

= 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