fisica1-youn-e-freedman-exercicios-resolvidos

5.1: a) The tension in the rope must be equal to each suspended weight, 25.0 N. b) If the
mass of the light pulley may be neglected, the net force on the pulley is the vector sum of
the tension in the chain and the tensions in the two parts of the rope; for the pulley to be
in equilibrium, the tension in the chain is twice the tension in the rope, or 50.0 N.
5.2: In all cases, each string is supporting a weight w against gravity, and the tension in
each string is w. Two forces act on each mass: w down and T ( = w)
up.
5.3: a) The two sides of the rope each exert a force with vertical component T sin θ , and
the sum of these components is the hero’s weight. Solving for the tension T,
2
w (90.0 kg) (9.80 m s )
3
T = =
= 2.54×
10 N .
2 sin θ 2 sin 10.0°
b) When the tension is at its maximum value, solving the above equation for the angle
θ gives
2
arcsin ⎛ w ⎞ ⎛ (90.0 kg) (9.80 m s ⎞
θ = ⎜ ⎟ = arcsin
= 1.01°
.
2
⎜
2(2.50 10
4 N)
⎟
⎝ T ⎠ ⎝ × ⎠
5.4: The vertical component of the force due to the tension in each wire must be half of
the weight, and this in turn is the tension multiplied by the cosine of the angle each wire
w
makes with the vertical, so if the weight is 3 w
2
w = cos θ and θ = arccos = 48 .
, °
2 4
3
5.5: With the positive y-direction up and the positive x-direction to the right, the freebody
diagram of Fig. 5.4(b) will have the forces labeled n and T resolved into x- and y-
components, and setting the net force equal to zero,
F = T cosα
− nsinα
= 0
F
x
y
= n cosα
+ T sinα
− w = 0.
Solving the first for n = T cot α and substituting into the second gives
2
2
2
cos α
⎛ cos α sin α ⎞ T
T + T sinα
= T
⎜ +
⎟ = = w
sinα
⎝ sinα
sinα
⎠ sinα
and so n = T cot α = wsin
α cot α = wcosα,
as in Example 5.4.
2
3
5.6: w sin α = mg sin α = (1390 kg) (9.80 m s ) sin 17.5°
= 4.10×
10 N.