fisica1-youn-e-freedman-exercicios-resolvidos

11.84: a)
b) The ratio of the added force to the elongation, found from taking the slope of the
graph, doing a least-squares fit to the linear part of the data, or from a casual glance at the
F
4
data gives = 2.00 × 10 N m.
From Eq. (11.10),
∆l
= F l0
4
(3.50 m)
11
Y = (2.00 × 10 N m)
= 1.8×
10 Pa.
3 2
∆l
A
( π (0.35×
10
− m) )
c) The total force at the proportional limit is 20 .0 N + 60 N = 80 N, and the stress at
this limit is
(80 N)
−
π (0.35×
10 m)
3 2 = 2.1×
10
8
Pa.
11.85: a) For the same stress, the tension in wire B must be two times in wire A, and so
the weight must be suspended at a distance ( 2 3)(1.05 m) = 0.70 m from wire A..
b) The product Y A for wire B is ( 4 3)
that of wire B, so for the same strain, the
tension in wire B must be ( 4 3)
that in wire A, and the weight must be 0.45 m from wire
B.
11.86: a) Solving Eq. (11.10) for ∆ l and using the weight for F,
Fl0 ( 1900 N)(15.0 m)
−4
∆l
= =
= 1.8×
10 m.
11
− 4 2
YA (2.0 × 10 Pa)(8.00×
10 m )
b) From Example 5.21, the force that each car exerts on the cable is
2 w 2
F = mω l = w l and so
0
g
0 ,
Fl wω l
∆l
= =
YA gYA
2 2
0 0
( −4
2
2
1900 N)(0.84 rad s) (15.0 m)
=
= 1.9×
10
2
11
− 4 2
(9.80 m s )(2.0×
10 Pa)(8.00×
10 m )
m.