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418
Chapter 4
EVALUATE:
The tension in the rope is not constant but increases from the bottom of the rope to the top. The
tension at the top of the rope must accelerate the rope as well the 5.00kg block. The tension at the top of the rope
is less than
F
; there must be a net upward force on the 6.00kg block.
4.55.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to the barbell and to the athlete. Use the motion of the barbell to calculate its
acceleration.
SET UP:
Let
be upward.
y
+
EXECUTE:
(a)
The freebody diagrams for the baseball and for the athlete are sketched in Figure 4.55.
(b)
The athlete&s weight is
. The upward acceleration of the barbell is found
from
2
(90.0 kg)(9.80 m/s )
882 N
mg
==
2
1
00
2
y
y
yy v
t a
t
−= +
.
2
0
22
2(
)
2(0.600 m)
0.469 m/s
(1.6 s)
y
yy
a
t
−
=
. The force needed to lift the barbell is given
by
lift
barbell
y
Fw
m
a
−=
. The barbell&s mass is
, so
.
2
(490 N) (9 80 m s )
50 0 kg
/.
/ = .
2
lift
barbell
490 N
(50.0 kg)(0.469 m/s )
490 N
23 N
513 N
m
a
=+
=
The athlete is not accelerating, so
floor
lift
athlete
0
FF
w
−
.
.
floor
lift
athlete
513 N 882 N
1395 N
w
=
+
=
EVALUATE:
Since the athlete pushes upward on the barbell with a force greater than its weight the barbell pushes
down on him and the normal force on the athlete is greater than the total weight, 1362 N, of the athlete plus
barbell.
Figure 4.55
4.56.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to the balloon and its passengers and cargo, both before and after objects are
dropped overboard.
SET UP:
When the acceleration is downward take
y
+
to be downward and when the acceleration is upward take
to be upward.
y
+
EXECUTE:
(a)
The freebody diagram for the descending balloon is given in Figure 4.56.
L
is the lift force.
(b)
y
y
Fm
a
∑=
gives
and
( /3)
Mg
L
M g
2
/3
L
Mg
=
.
(c)
Now
is upward, so
, where
m
is the mass remaining.
y
+
( /2)
Lm
gm
g
2/
3
L
Mg
=
, so
. Mass 5
must be dropped overboard.
4
/9
mM
=
M
EVALUATE:
In part (b) the lift force is greater than the total weight and in part (c) the lift force is less than the
total weight.
Figure 4.56
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View Full DocumentNewton&s Laws of Motion
419
4.57.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to the entire chain and to each link.
SET UP:
mass of one link. Let
m
=
y
+
be upward.
EXECUTE:
(a)
The freebody diagrams are sketched in Figure 4.57.
is the force the top and middle links
exert on each other.
is the force the middle and bottom links exert on each other.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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