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UNIT 3
Forces and
the Laws of Motion
Monday October 24th
FORCES & THE LAWS OF MOTION
2
TODAY’S AGENDA
Monday, October 24
 Laws of Motion
 Mini-Lesson: Everyday Forces (2nd Law Problems)
UPCOMING…




Thurs: Newton’s 2nd Law Lab
Fri:
Quiz #2 2nd Law Problem
Mon: Test Review
Tue:
TEST #4
 Fy  0  N  m1g
Force Lab Notes
 Fx  m1a
T  m1a
Forces on m1
N
 F  ma
T
T  m 2 g  m 2a
m1
T
m1g
m2
m2g
Forces on m2
m 2 g  T  m 2a
m1a = T = m2g – m2a
m1a  m 2g  m 2a
a
m 2g
m1  m 2
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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.
b) What acceleration would a 175 kg box
have on this ramp?
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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.
ΣFy = 0
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°)
mgsin(25°)

µ = .0614
mg

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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°)
mg
mass does not matter,
the acceleration is the same!!
mgsin(25°)
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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?
FN
vi = 0

F
t=?
4.00 m
Ffk
Fg
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FN
90.0N

77.9 N
Ffk
 = 30˚
45.0 N
735.8 N
1. Draw a free-body diagram to find the net force.
2. Convert all force vectors into x- and y- components.
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FN = 781 N
90.0N

77.9 N
Ffk
 = 30˚
45.0 N
735.8 N
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.
5. Solve for the normal force.
FN = 45.0 + 735.8 N
FN = 781 N
30
FN = 781 N
90.0N

77.9 N
Ffk
 = 30˚
45.0 N
44.5 N
735.8 N
6. Given the μk = 0.057, find the frictional force.
μkFN = Ff
(0.057) 781 N = 44.5 N
Ff = 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
FN = 781 N
90.0N

77.9 N
Ffk
 = 30˚
45.0 N
a = .445 m/s2
44.5 N
735.8 N
8. Which constant acceleration equation has a, vi, x, and t?
t = 4.24 s
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END
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