fisica1-youn-e-freedman-exercicios-resolvidos

5.83: Let the tension in the cord attached to block A be T
A
and the tension in the cord
attached to block C be T
C
. The equations of motion are then
T
C
− µ m g − T
mC
g − TC
= mCa.
a) Adding these three equations to eliminate the tensions gives
k
T
A
B
− m g = m a
A
A
= m
a ( mA
+ mB
+ mC
) = g(
mC
− mA
− µ
kmB),
solving for m
C
gives
mA( a + g)
+ mB(
a + µ
kg)
mC
=
,
g − a
and substitution of numerical values gives m = 12.9 kg.
b) T = m ( g + a)
= 47.2 N, T = m ( g − a)
= 101 N.
A
A
C
C
C
A
B
a
5.84: Considering positive accelerations to be to the right (up and to the right for the lefthand
block, down and to the right for the right-hand block), the forces along the inclines
and the accelerations are related by
T − ( 100 kg) g sin30°
= (100 kg) a,(50 kg) g sin53° − T = (50 kg) a,
where T is the tension
in the cord and a the mutual magnitude of acceleration. Adding these relations,
( 50 kg sin 53° −100 kg sin 30°
) g = (50 kg + 100 kg) a,
or a = −0.067
g.
a) Since a comes
out negative, the blocks will slide to the left; the 100-kg block will slide down. Of course,
if coordinates had been chosen so that positive accelerations were to the left, a would be
2
+ 0.067
g.
b) 0.067(9.80 m s ) = 0.658 m s .
c) Substituting the value of a (including the proper sign, depending on choice of
coordinates) into either of the above relations involving T yields 424 N.
2