fisica1-youn-e-freedman-exercicios-resolvidos

4.52: a) m = mass of one link
The downward forces of magnitude 2ma and ma for the top and middle links are the
reaction forces to the upward force needed to accelerate the links below.
2
b) (i) The weight of each link is mg = (0.300 kg)(9.80 m /s ) = 2.94 N . Using the freebody
diagram for the whole chain:
Fnet
12 N − 3(2.94 N) 3.18 N
2
2
a = =
= = 3.53 m /s or 3.5 m / s
3 m 0.900 kg 0.900 kg
(ii) The second link also accelerates at
2
3 .53 m /s , so:
F = F − ma − 2mg
= ma
F
net
top
= 2ma
+ 2mg
= 2(0.300 kg)(3.53 m / s
2
top
+
= 2 .12 N + 5.88 N = 8.0 N
)
2(2.94 N)
4.53: Differentiating twice, the acceleration of the helicopter as a function of time is
r
3 ˆ
2
a = (0.120 m / s ) ti
− (0.12 m / s ) kˆ,
and at t = 5.
0s , the acceleration is
r
2
2
a = (0.60 m / s )ˆ i − (0.12 m / s ) kˆ.
The force is then
2
[(0.60 m / s )ˆ i − (0.12 m / s ) kˆ
]
5
r w r ( 2.75×
10 N)
2
F = ma
= a =
2
g (9.80 m / s )
4
3
= (1.7 × 10 N)ˆ i − (3.4 × 10 N) kˆ.