Applications of Newton’s Laws
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Sample Problems
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Homework
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Sample Problem 1
In the figure below, a block B of mass M = 15.0 kg hangs
by a cord from a knot K of mass mK , which hangs from
the ceiling by means of two other cords. The cords have
negligible mass, and the magnitude of the gravitational
force on the knot is negligible compared to the gravitational force on the block. What are the tensions in the
three cords?
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Sample Problem 1 (cont’d)
Block
⇒
X
Fy = T 3 − F g = 0
T3 = Fg = M g = (15.0 kg) 9.8 m/s
Knot
⇒
X
2
= 147 N
Fx = T2 cos 47◦ − T1 cos 28◦ = 0
cos 28◦
T2 =
T1 = 1.29T1
cos 47◦
X
Fy = T1 sin 28◦ + T2 sin 47◦ − T3 = 0
T1 sin 28◦ + (1.29T1) sin 47◦ − T3 = 0
T3
T1 =
sin 28◦ + 1.29 sin 47◦
147 N
T1 =
= 104 N
◦
◦
sin 28 + 1.29 sin 47
T2 = 1.29T1 = 1.29 (104 N ) = 134 N
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Sample Problem 2
In the figure below, a cord holds stationary a block of
mass m = 15 kg, on a frictionless plane that is inclined at
an angle θ = 27◦. What are the magnitudes of the tension
in the cord and the normal force on the block?
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Sample Problem 2 (cont’d)
X
Fx = T − Fg sin θ = 0
T = Fg sin θ = mg sin θ
2
T = (15 kg) 9.8 m/s sin 27◦ = 67 N
X
Fy = N − Fg cos θ = 0
N = Fg cos θ = mg cos θ
2
N = (15 kg) 9.8 m/s cos 27◦ = 130 N
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Sample Problem 3
In the figure below, a passenger of mass m = 72.2 kg
stands on a platform scale in an elevator. (a) Find a general expression for the scale reading, whatever the vertical motion of the cab. (b) What does the scale read if
the elevator is stationary or moving upward at a constant
speed of 0.50 m/s? (c) What does the scale read if the
elevator accelerates upward at 3.20 m/s2 and downward
at 3.20 m/s2?
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Sample Problem 3 (cont’d)
(a) Find a general expression for the scale reading, whatever the vertical motion of the cab.
X
Fy = N − Fg = may
N = Fg + may = mg + may = m (g + ay )
(b) What does the scale read if the elevator is stationary
or moving upward at a constant speed of 0.50 m/s?
2
N = (72.2 kg) 9.8 m/s + 0 = 708 N
(c) What does the scale read if the elevator accelerates
upward at 3.20 m/s2 and downward at 3.20 m/s2?
2
U p ⇒ N = (72.2 kg) 9.8 m/s + 3.20 m/s
2
2
Down ⇒ N = (72.2 kg) 9.8 m/s − 3.20 m/s
7
= 939 N
2
= 477 N
Sample Problem 4
Consider Atwood’s machine shown in the figure. Assume m2 > m1 and the masses of the cord and the pulley
are negligible. Find an expression for the acceleration.
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Sample Problem 4 (cont’d)
X
Fm 1 = T − m 1 g = m 1 a
X
Fm 2 = m 2 g − T = m 2 a
Adding these two eqns. we get
m2 g − m 1 g = m 1 a + m 2 a
m2 − m 1
a=
g
m1 + m2
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Homework Set 8 - Due Mon. Sept. 27
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Read Section 4.7
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Do Problems 4.22, 4.25, 4.26, 4.30, 4.31 & 4.34
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