UNIT 3Forces and

the Laws of Motion

Monday October 24th

2

FORCES & THE LAWS OF MOTION

TODAY’S AGENDA

Laws of MotionMini-Lesson: Everyday Forces (2nd Law

Problems)

UPCOMING…

Thurs: Newton’s 2nd Law LabFri: Quiz #2 2nd Law ProblemMon: Test ReviewTue: TEST #4

Monday, October 24

m1

m2

N

m1g

T

T

m2g

gmN0F 1y

amF 1x

amT 1

maF amTgm 22

amgmT 22

amgmam 221

21

2mmgm

a

Forces on m1

Forces on m2

23

m1a = T = m2g – m2a

Force Lab Notes

Everyday Forces

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

b) What acceleration would a 175 kg box have on this ramp?

A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2.

25

A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2.

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

FN

mgmgcos(25°)

Ff

mgsin(25°)

ΣFy = 0

FN = mgcos(25°) = 667 N

FNET = ma = mgsin(25°) - Ff

FNET = 270 N = 311- Ff

Ff = µFN = µ(667N) = 41N

µ = .0614

ΣFx ≠ 0

26

A 75.0-kg box slides down a 25.0˚ ramp with an acceleration of 3.60 m/s2.

b) What acceleration would a 175 kg box have on this ramp?

FN

mgmgcos(25°)

Ff

mgsin(25°)

FNET = ma

Ff = µFN

ΣFx ≠ 0

ma = mgsin(25°) - Ff

ma = mgsin(25°) – μmgcos(25˚)

mass does not matter, the acceleration is the same!!

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?

F

4.00 m

FN

Fg

Ffk

t = ?vi = 0

28

90.0N

FN

735.8 N

Ffk

= 30˚

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

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

77.9 N

45.0 N

29

90.0N

FN

735.8 N

Ffk

= 30˚

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

77.9 N

45.0 N

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

5. Solve for the normal force.

Net force !

FN = 45.0 + 735.8 N

FN = 781 N

= 781 N

30

90.0N

FN

735.8 N

Ffk

= 30˚

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

77.9 N

45.0 N

μkFN = Ff

= 781 N

(0.057) 781 N = 44.5 N Ff = 44.5 N

44.5 N

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

90.0N

FN

735.8 N

Ffk

= 30˚

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

77.9 N

45.0 N

= 781 N

44.5 N

t = 4.24 s

a = .445 m/s2

32

13

END