fisica1-youn-e-freedman-exercicios-resolvidos

5.94: The banked angle of the track has the same form as that found in Example 5.24,
2
v
0
tan β = ,
gR
where v
0
is the ideal speed, 20 m s in this case. For speeds larger than v
0
, a
frictional force is needed to keep the car from skidding. In this case, the inward force will
consist of a part due to the normal force n and the friction force
f ; n sin β + f cos β = marad.
The normal and friction forces both have vertical
components; since there is no vertical acceleration, n cos β − f sin β = mg.
Using
2 (1.5v
2
)
R R
f n
v 0
= µ
s
and a = = = 2.25 g tan β,
these two relations become
rad
Dividing to cancel n gives
nsin
β + µ ncos
β = 2.25 mg tan β,
s
ncos
β − µ nsin
β = mg.
s
sin β + µ
s
cos β
= 2.25 tan β.
cos β − µ sin β
Solving for µ
s
and simplifying yields
1.25 sinβ
cos β
µ
s
=
.
2
1+
1.25sin β
2
β gives µ = 0.34
s
.
(20 m s)
Using = arctan( ) = 18. 79°
2
(9.80 m s )(120 m)
s
5.95: a) The same analysis as in Problem 5.90 applies, but with the speed v an unknown.
The equations of motion become
2
nsin
β + µ ncos
β = mv R,
Dividing to cancel n gives
s
ncos
β − µ nsin
β = mg.
s
2
sin β + µ
s
cos β v
= .
cos β − µ sin β Rg
s
Solving for v and substituting numerical values gives v = 20.9 m s (note that the value
for the coefficient of static friction must be used).
b) The same analysis applies, but the friction force must be directed up the bank; this
has the same algebraic effect as replacing f with − f , or replacing µ
s
with − µ
s
(although coefficients of friction may certainly never be negative). The result is
2 sin β − µ
s
cos β
v = ( gR)
,
cos β + µ sin β
and substitution of numerical values gives v = 8.5 m s.
s