fisica1-youn-e-freedman-exercicios-resolvidos

11.18: a) Denote the length of the boom by L, and take torques about the pivot point.
The tension in the guy wire is found from
TL sin 60° = (5000 N) L cos 60.0° + (2600 N)(0.35 L)
cos 60.0°
,
so T = 3.14 kN. The vertical force exerted on the boom by the pivot is the sum of the
⎛ Fv
⎞
weights, 7.06 kN and the horizontal force is the tension, 3.14 kN. b) No; tan
⎜ ≠ 0.
F
⎟
⎝ H ⎠
11.19: To find the tension TL
in the left rope, take torques about the point where the
rope at the right is connected to the bar. Then,
T
L
( 3.00 m) sin150° = (240 N)(1.50 m) + (90 N)(0.50 m), so TL
= 270 N. The vertical
component of the force that the rope at the end exerts must be
( 330 N) − (270 N) sin 150°
= 195 N, and the horizontal component of the force is
− ( 270 N) cos150°
, so the tension is the rope at the right is T = 304
R
N. and θ = 39.9°
.
11.20: The cable is given as perpendicular to the beam, so the tension is found by
taking torques about the pivot point;
T ( 3.00 m) = (1.00 kN)(2.00 m) cos 25.0° + (5.00 kN)(4.50 m) cos 25.0°
, or T = 7.40 kN.
The vertical component of the force exerted on the beam by the pivot is the net weight
minus the upward component of T , 6.00 kN − T cos 25.0°
= 0.17 kN. The horizontal force
is T sin 25.0°
= 3.13 kN.
11.21: a) F1 ( 3.00 m) − F2
(3.00 m + l)
= (8.00 N)( −l).
This is given to have a
magnitude of 6 .40 N.m, so l = 0.80m.
b) The net torque is clockwise, either by
considering the figure or noting the torque found in part (a) was negative. c) About the
point of contact of F 2,
the torque due to F 1 is − F1
l,
and setting the magnitude of this
torque to 6 .40 N ⋅ m gives l = 0.80 m, and the direction is again clockwise.
11.22: From Eq. (11.10),
= l
0.200 m)
Y F = F
= (1333 m
− 2
−4
2
∆lΑ
(3.0 × 10 m)(50.0×
10 m )
F
0 ( −2
4
Then, F = 25.0 N corresponds to a Young’s modulus of 3.3× 10 Pa, and F = 500 N
corresponds to a Young’s modulus of 6.7 × 10
5
Pa.
).