fisica1-youn-e-freedman-exercicios-resolvidos

5.125: Take all accelerations to be positive downward. The equations of motion are
straightforward, but the kinematic relations between the accelerations, and the resultant
algebra, are not immediately obvious. If the acceleration of pulley B is a
B,
then
a B
= −a 3
, and a
B
is the average of the accelerations of masses 1 and 2, or
a1 + a2
= 2aB = −2a3.
There can be no net force on the massless pulley B, so T C
= 2T
A
.
The five equations to be solved are then
m g − T = m a
1
m g − T
2
m3g
− TC
= m3a3
a 1
+ a 2
+ 2a
3
= 0
2TA
− TC
= 0.
These are five equations in five unknowns, and may be solved by standard means. A
symbolic-manipulation program is of great use here.
a) The accelerations a
1
and a
2
may be eliminated by using
a = −(
a + a ) = −(2g
− TA ((1 m ) + (1 ))).
A
A
1
2
1
= m a
2
3 1 2
1
m2
The tension T
A
may be eliminated by using
T = 1 2) T = (1 2) m ( g −
A
(
C
3
a3
Combining and solving for a
3
gives
− 4m1m
2
+ m2m3
+ m1m
3
a3
= g
.
4m m + m m + m m
1
b) The acceleration of the pulley B has the same magnitude as a
3
and is in the
opposite direction.
TA
TC
m3
c) a = g − = g − = g − ( g −
3).
1
m 2m
2m
a
1
2
Substituting the above expression for a
3
gives
4m1m
2
− 3m2m3
+ m1m
3
a1
= g
.
4m1m
2
+ m2m3
+ m1m
3
d) A similar analysis (or, interchanging the labels 1 and 2) gives
4m1m
2
− 3m1m
3
+ m2m3
a2
= g
.
4m1m
2
+ m2m3
+ m1m
3
e) & f) Once the accelerations are known, the tensions may be found by substitution
into the appropriate equation of motion, giving
4m1m2m3
8m1m2
m3
T A
= g
, TC
= g
.
4m m + m m + m m 4m m + m m + m m
1
2
2 3
3
2m
1
3
g) If m 1
= m 2
= m and m = , all of the accelerations are zero, T C
= 2mg
and
T A
= mg. All masses and pulleys are in equilibrium, and the tensions are equal to the
weights they support, which is what is expected.
2
3
1
2
1
1
3
).
2
1
2
3
1
3