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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
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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.
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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.
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
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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
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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
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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
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END