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Ch 4 Force and Motion
Chapter 4
• Newton’s 1st Law, 2nd law and 3rd law
• More applications
• Weight and apparent weight
Force and Newton’s
Laws of Motion
(Continuation)
Example: two blocks
Example: Block on an incline
• A block rests on a surface
with adjustable
inclination. When the
angle increases to
60°the block starts
slipping down the slope.
Find the static friction
coefficient. If the kinetic
friction coefficient is 1.0,
what is the acceleration
of the block?
Two blocks are
connected by a rope
which wraps over a
idea pulley. The two
masses are m1=3kg
and m2=2kg. The
coefficient of
friction between m1
and table top is
µk=0.3. Find acc. and
y
r
a
r
FN
f
x
r
Fg
tension.
r
W
Weight
How much do you weigh on …
• Earth?
m = 70 kg
g(Earth)=9.8 m/s2
WE = mgE= 686 N
• The weight of a body is a force balancing the
gravity force when the body is at rest at the
ground (Earth or other planet).
v
FN
r
r
Fnet = ma
Fnet , y = ma y = 0
FN − W = 0
FN = W = mg
y
x
•The mass is the same, the weight changes.
•The mass is measured in kg, the weight in N.
m
r
W
• Mars?
m = 70 kg
g(Mars)=3.7 m/s2
WM = mgM= 259 N
v
FN
scale
m
scale
1
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Weight and Apparent Weight
r
Fapp
Exam 1 Fall 2010: Problem 11
Fext
• In a noninertial reference frame a spring
M
an apparent weight.
r scale measures
r
r
Fnet = ma
y
Fnet,y = may
r
FN
a
FN − W = ma
x
FN = W + ma = mg + ma
r
W
Fapp = FN = m( g + a)
y-axis
θ
M
Fext
θ
x-axis
• Near the surface of the Earth, a block of mass M = 2 kg
slides along the floor while an external force Fext is
applied at an upward angle θ = 26o. If the coefficient of
kinetic friction between the block and the floor is 0.488,
and the magnitude of the acceleration of the block is 1.89
m/s2, what is the magnitude of the external force?
Answer: 12 N
M (a x + µ k g )
Fext =
% Right: 59%
cos θ + µ sin θ
Fext cos θ − f k = Ma x
Fext sin θ + FN − Mg = 0
f k = µ k FN
k
=
(2kg )(1.89m / s 2 + (0.488)9.8m / s 2 )
≈ 12.0 N
cos(26o ) + (0.488) sin(26o )
Exam 1 Fall 2010: Problem 8
A
B
C
F
• Three blocks (A,B,C), each having mass MA = M, MB= 2M,
MC = M are connected by strings on a horizontal
frictionless surface as shown in the figure. Block C is
pulled to the right by a horizontal force of magnitude F
that causes the entire system to accelerate. What is the
magnitude of the net horizontal force acting on block B
due to the strings?
F
F
A
B
F
Answer: F/2 F = ( M A + M B + M C )a
C
% Right: 41% FB = ( M A + M B ) a
FA = M Aa
A
FnetB = FB − FA = M B a =
B
MB
2M
F=
F = 12 F
M A + M B + MC
M + 2M + M
2