fisica1-youn-e-freedman-exercicios-resolvidos

11.49: The horizontal component of the force exerted on the bar by the hinge must
balance the applied force F r , and so has magnitude 120.0 N and is to the left. Taking
torques about point A , (120.0 N)(4.00 m) + FV
(3.00 m), so the vertical component is
− 160 N , with the minus sign indicating a downward component, exerting a torque in a
direction opposite that of the horizontal component. The force exerted by the bar on the
hinge is equal in magnitude and opposite in direction to the force exerted by the hinge on
the bar.
11.50: a) The tension in the string is w = 50 2
N, and the horizontal force on the bar
must balance the horizontal component of the force that the string exerts on the bar, and
is equal to ( 50 N) sin 37° = 30 N, to the left in the figure. The vertical force must be
⎛ 50 N ⎞
2
2
(50 N) cos 37°
+ 10 N = 50 N, up. b) arctan
⎜ = 59°
.c) (30 N) + (50 N) = 58 N.
30 N
⎟
⎝ ⎠
d) Taking torques about (and measuring the distance from) the left end,
( 50 N) x = (40 N)(5.0 m) , so x = 4.0 m , where only the vertical components of the
forces exert torques.
11.51: a) Take torques about her hind feet. Her fore feet are 0.72 m from her hind feet,
(190 N) (0.28 m)
and so her fore feet together exert a force of = 73.9 N, so each foot exerts a
(0.72 m)
force of 36.9 N, keeping an extra figure. Each hind foot then exerts a force of 58.1 N.
b) Again taking torques about the hind feet, the force exerted by the fore feet is
(190 N) (0.28 m) + (25 N) (0.09 m)
= 105.1 N,
0.72 m
so each fore foot exerts a force of 52.6 N and each hind
foot exerts a force of 54.9 N.
11.52: a) Finding torques about the hinge, and using L as the length of the bridge and
w and w for the weights of the truck and the raised section of the bridge,
T
B
( 3
1
L) cos 30° + w ( L) cos °
TL sin 70° = wT
B
30
4
2
, so
T
=
3 1
( m + m )
4
T
2
9.80 m s
sin 70°
)cos 30°
= 2.57 × 10
2
B
( 5
N.
T Vertical: w + w − T sin 40°
5
b) Horizontal: cos( 70° − 30°
) = 197 . × 10 N.
= 2.46×
10
5
N.
T
B