fisica1-youn-e-freedman-exercicios-resolvidos

5.72: The key idea in solving this problem is to recognize that if the system is
accelerating, the tension that block A exerts on the rope is different from the tension that
block B exerts on the rope. (Otherwise the net force on the rope would be zero, and the
rope couldn’t accelerate.) Also, treat the rope as if it is just another object. Taking the
“clockwise” direction to be positive, the Second Law equations for the three different
parts of the system are:
Block A (The only horizontal forces on A are tension to the right, and friction to the left):
− µ
kmA g + TA
= mAa.
Block B (The only vertical forces on B are gravity down, and tension up):
mB g − TB
= mBa.
Rope (The forces on the rope along the direction of its motion are the tensions at either
end and the weight of the portion of the rope that hangs vertically):
d
mR (
L
) g + TB
− TA
= mRa.
To solve for a and eliminate the tensions, add the left hand sides and right hand sides of
−
d
m + m −µ
m
km A
+
B
+
R L
= (
A
+
B
+
R
=
( mA
+ mB
+ mR
)
B R L k A
the three equations: µ g m g m ( d ) g m m m ) a,
or a g .
m (
d
B + mR
)
L
(a) When µ = , = g . As the system moves, d will increase, approaching
k
0 a
( mA
+ mB
+ mR
)
L as a limit, and thus the acceleration will approach a maximum value of
mB
+ mR
= g
a
( m A + mB
+ mR
.
)
d
(b) For the blocks to just begin moving, a > 0,
so solve 0 = [ m
B
+ mR( ) − µ
smA]
L
for d.
Note that we must use static friction to find d for when the block will begin to move.
L
1 .0 m
Solving for d, d = µ m − m ), or d = (.25(2 kg) − .4 kg) = .63 m. .
(
mR
s A B
.160 kg
1 .0 m
. 04 kg, =
.04 kg
(.25(2 kg) − .4 kg) =
(c) When m R
= d
2.50 m. This is not a physically
possible situation since d > L.
The blocks won’t move, no matter what portion of the
rope hangs over the edge.
5.73: For a rope of length L, and weight w, assume that a length rL is on the table, so that
a length ( 1− r) L is hanging. The tension in the rope at the edge of the table is then
( 1− r)
w,
and the friction force on the part of the rope on the table is f
s
= µ
srw.
This must
be the same as the tension in the rope at the edge of the table, so
µ rw = 1−
r)
w and r = 1 (1 + ). Note that this result is independent of L and w for a
s
( µ
s
uniform rope. The fraction that hangs over the edge is − r = µ (1 + ) ; note that if
µ
s
= 0, r = 1and1−
r = 0.
1
s
µ
s