fisica1-youn-e-freedman-exercicios-resolvidos

11.61: a) Taking torques about the hinged end of the pole
( 200 N)(2.50 m) (600 N) × (5.00 m) − T (5.00 m) = 0 . Therefore the y-component of
the tension is
T
x
=
attached is
T
y
(1000 N)
2
+
y
= 700 N . The x-component of the tension is then
− (700 N)
(
714
2
= 714 N . The height above the pole that the wire must be
700
5.00 m) = 4.90 m . b) The y-component of the tension remains 700 N and
4.90 m
the x-component becomes ( 714 N)
4.40 m
(795 N)
2
2
+ (700 N) = 1059 N, an increase of 59 N.
= 795 N , leading to a total tension of
11.62: A and B are straightforward, the tensions being the weights suspended;
2
2
ΤA = (0.0360 kg)(9.80 m s ) = 0.353 N, TB
= (0.0240 kg + 0.0360 kg)(9.80 m s ) = 0.588 N.
To find T C
and T D
, a trick making use of the right angle where the strings join is available;
use a coordinate system with axes parallel to the strings. Then,
T
C
= TB
cos 36.9°
= 0.470 N, TD
= TB
cos 53.1°
= 0.353 N, To find T
E
, take torques about
the point where string F is attached;
T (1.000 m) = T sin 36.9°
(0.800 m) + T sin 53.1°
(0.200 m)
so T
E
D
2
+ (0120 . kg)(9.
80 m s )(0.
500 m)
= 0.833 N ⋅ m,
E
= 0.833 N. TF
may be found similarly, or from the fact that T
E
+ TF
C
must be the
2
total weight of the ornament. (0.180 kg)(9.80 m s ) 1.76 N, from which T = 0.931 N.
=
F
11.63: a) The force will be vertical, and must support the weight of the sign, and is 300
N. Similarly, the torque must be that which balances the torque due to the sign’s weight
about the pivot, ( 300 N)(0.75 m) = 225 N ⋅ m . b) The torque due to the wire must balance
the torque due to the weight, again taking torques about the pivot. The minimum tension
occurs when the wire is perpendicular to the lever arm, from one corner of the sign to the
2
2
other. Thus, T (1.50 m) + (0.80 m) = 225 N ⋅ m, or T = 132 N. The angle that the wire
0.80
makes with the horizontal is 90° − arctan ( ) = 62.0°
.
1.50
Thus, the vertical component of
the force that the pivot exerts is (300 N) –(132 N) sin 62 .0° = 183 N and the horizontal
force is ( 132 N)cos62. 0°
= 62 N , for a magnitude of 193 N and an angle of 71 ° above
the horizontal.