fisica1-youn-e-freedman-exercicios-resolvidos

5.77: See Exercise 5.40. a) The maximum tension and the weight are related by
Tmax
cos β = µ
k
( w − Tmax
sin β),
and solving for the weight w gives
⎛ cos β
max
sin ⎟ ⎞
w = T
⎜ + β .
⎝ µ
k ⎠
This will be a maximum when the quantity in parentheses is a maximum. Differentiating
with respect to β ,
d ⎛ cos β ⎞ sin β
⎜ + sin β
⎟ = − + cos β = 0,
dβ ⎝ µ
k ⎠ µ
k
or tan θ = µ
k
, where θ is the value of β that maximizes the weight. Substituting for µ
k
in terms of θ ,
⎛ cos θ ⎞
w = Tmax
⎜ + sinθ
sin θ cos θ
⎟
⎝
⎠
2 2
⎛ cos θ + sin θ ⎞
= Tmax
⎜
sin θ
⎟
⎝
⎠
Tmax
= .
sin θ
b) In the absence of friction, any non-zero horizontal component of force will be
enough to accelerate the crate, but slowly.
5.78: a) Taking components along the direction of the plane’s descent,
f = w sinα
and L = wcosα.
b) Dividing one of these relations by the other cancels the
weight, so tan α = f L.
c) The distance will be the initial altitude divided by the tangent
1300 N
of α . f = L tan α and L = w cosα,
therefore sin α = f w = g and so α = 5 .78°.
12,900 N
This makes the horizontal distance ( 2500 m) tan(5.78° ) = 24.7 km. d) If the drag is
reduced, the angle α is reduced, and the plane goes further.
5.79: If the plane is flying at a constant speed of 36 .1m s, then ∑ F = 0,
or
T − wsin α − f = 0. The rate of climb and the speed give the angle
α , α = arcsin(5 36.1) = 7.96°
. Then,
T = w sin α + f . T = (12,900 N) sin 7.96° + 1300 N = 3087 N. Note that in level flight
( α = 0), the thrust only needs to overcome the drag force to maintain the constant speed
of 36 .1m s.