fisica1-youn-e-freedman-exercicios-resolvidos

(4090 kg)(9.8 m s )
4
5.7: a) T cos θ W ,or T = W cos θ =
= 5.23×
10 N.
B
=
B
cos 40°
4
4
b) T = T sinθ
= (5.23×
10 N) sin 40°
= 3.36 × 10 N.
A
B
2
5.8: a) T w, T sin 30° + T sin 45°
= T = w,
and T cos30° − T cos45°
= 0.
Since
C
=
A
B
C
A
B
sin 45° = cos45°
, adding the last two equations gives T A
(cos30° + sin 30°
) = w,
and so
w
cos 30°
T = 0.732 . Then, T T 0.897w.
A
= w
1.366
B
=
A
=
cos 45°
b) Similar to part (a), T = w, − T cos60° + T sin 45°
w,
and
C A
B
=
w
TA = =
(sin 60 −cos 60°
)
T
A
sin 60° −TB
cos45°
= 0. Again adding the last two, 2.73w,
°
and
sin 60°
T T 3.35w.
B
=
B
=
cos 45°
2
5.9: The resistive force is w sinα = (1600 kg)(9.80 m s )(200 m 6000 m) = 523 N. .
5.10: The magnitude of the force must be equal to the component of the weight along the
2
incline, orW
sin θ = (180 kg)(9.80 m s )sin11.0°
= 337 N.
5.11: a) W = 60 N, T sinθ
= W , so T = ( 60 N) sin45°
, or T = 85 N.
b) F
1
= F2
= T cos θ,
F1
= F2
= 85 N cos 45°
= 60 N.
0.
110
5.12: If the rope makes an angle θ with the vertical, then sinθ =
1.
51
= 0. 073 (the
denominator is the sum of the length of the rope and the radius of the ball). The weight is
then the tension times the cosine of this angle, or
2
w mg (0.270 kg)(9.80m s )
T = =
=
= 2.65 N.
cosθ
cos(arcsin(.073)) 0.998
The force of the pole on the ball is the tension times sin θ , or ( 0.073) T = 0.193 N.
5.13: a) In the absence of friction, the force that the rope between the blocks exerts on
block B will be the component of the weight along the direction of the incline,
T = w sin α . b) The tension in the upper rope will be the sum of the tension in the lower
rope and the component of block A’s weight along the incline,
w sin α + w sin α = 2w
sin α.
c) In each case, the normal force is w cos α.
d) When
α = 0,
n = w,
when α = 90 ° , n = 0.