fisica1-youn-e-freedman-exercicios-resolvidos

11.79: a) Take torques about the point where wheel B is in contact with the track. With
respect to this point, the weight exerts a counterclockwise torque and the applied force
and the force of wheel A both exert clockwise torques. Balancing torques,
F
A
(2.00 m) + ( F)(1.60 m) = (950 N)(1.00 m). Using
F = µ
k
w = 494 N, FA
= 80 N, and FB
= w − FA
= 870 N. b) Again taking torques about the
point where wheel B is in contact with the tract, and using
F = 494 N as in part (a), (494 N) h = (950 N)(1.00N), so h = 1.92 m.
11.80: a) The torque exerted by the cable about the left end is TLsin θ . For any angle
θ , sin(180° − θ ) = sinθ,
so the tension T will be the same for either angle. The horizontal
component of the force that the pivot exerts on the boom will be
1
T cosθ
or Tcos (180° − θ ) = −T
cosθ
. b) From the result of part (a), T α sin
, and this
θ
becomes infinite as θ → 0 or → 180°
. Also, c), the tension is a minimum when sin θ is a
maximum, or θ = 90° , a vertical string. d) There are no other horizontal forces, so for the
boom to be in equilibrium, the pivot exerts zero horizontal force on the boom.
11.81: a) Taking torques about the contact point on the ground,
T ( 7.0 m)sinθ
= w(4.5
m)sinθ,
so T = (0.
64) w = 3664 N. The ground exerts a vertical
force on the pole, of magnitude w − T = 2052 N . b) The factor of sin θappears in both
terms of the equation representing the balancing of torques, and cancels.
11.82: a) Identifying x with ∆l
in Eq. (11.10), k = Y A .
2
2
b) 1 2) kx = Y Ax 2l
.
(
0
l 0
11.83: a) At the bottom of the path the wire exerts a force equal in magnitude to the
centripetal acceleration plus the weight,
2
2
3
F = m(((2.00
rev s)(2π
rad rev)) (0.50 m) + 9.80 m s ) = 1.07 × 10 N.
From Eq. (11.10), the elongation is
3
(1.07 × 10 N)(0.50 m)
= 5.5 mm.
11
− 4 2
(0.7 × 10 Pa)(0.014×
10 m )
b) Using the same equations, at the top the force is 830 N, and the elongation is
0.0042 m.