fisica1-youn-e-freedman-exercicios-resolvidos

11.76: (a) Writing an equation for the torque on the right-hand beam, using the hinge
as an axis and taking counterclockwise rotation as positive:
θ L θ L θ
Fwire Lsin
− Fc
cos − w sin = 0
2 2 2 2 2
whereθ is the angle between the beams, F
c
is the force exerted by the cross bar, and w is
the weight of one beam. The length drops out, and all other quantities except F
c
are
known, so
F
c
=
F
wire
θ
sin −
2
cos
1
2
1
2
θ
2
w sin
θ
2
= (2F
wire
θ
− w)
tan
2
Therefore
53°
F = 260 tan = 130 N
2
b) The cross bar is under compression, as can be seen by imagining the behavior of
the two beams if the cross bar were removed. It is the cross bar that holds them apart.
c) The upward pull of the wire on each beam is balanced by the downward pull of
gravity, due to the symmentry of the arrangement. The hinge therefore exerts no vertical
force. It must, however, balance the outward push of the cross bar: 130 N horizontally to
the left for the right-hand beam and 130 N to the right for the left-hand beam. Again, it’s
instructive to visualize what the beams would do if the hinge were removed.
11.77: a) The angle at which the bale would slip is that for which
f = µ
s
Ν = µ
swcos
β = w sin β,
or β = arctan( µ
s
) = 31.0°
. The angle at which the bale
would tip is that for which the center of gravity is over the lower contact point, or
0.25 m)
arctan = 26.6°
, or 27°
to two figures. The bale tips before it slips. b) The angle for
(
0.50 m)
tipping is unchanged, but the angle for slipping is arctan ( 0.40) = 21.8°
, or 22°
to two
figures. The bale now slips before it tips.
2
11.78: a) F = f = µ
k
Ν = µ
kmg
= (0.35)(30.0 kg)(9.80 m s ) = 103 N
b) With respect to the forward edge of the bale, the lever arm of the weight is
0.250 m
= 0.125 m and the lever arm h of the applied force is then h
2
=
µ
( 0.125 m)
mg
1 0.125
= (0.125 m) =
F
0. 35
k
m
= 0.36 m.