5.8.
I
DENTIFY
:
Apply Newton’s 1st law to the wrecking ball. Each cable exerts a force on the ball,
directed along the cable.
S
ET
U
P
:
The force diagram for the wrecking ball is sketched in Figure 5.8.
Figure 5.8
E
XECUTE
:
(a)
y
y
F
ma
=
∑
cos40
0
B
T
mg
° −
=
2
4
(4090 kg)(9.80 m/s
)
5.23
10
N
cos40
cos40
B
mg
T
=
=
=
×
°
°
(b)
x
x
F
ma
=
∑
sin40
0
B
A
T
T
° −
=
4
sin40
3.36
10
N
A
B
T
T
=
° =
×
E
VALUATE
:
If the angle
40
is replaces by
(cable
B
is vertical), then
and
°
0
°
B
T
mg
=
0.
A
T
=
5.14.I
DENTIFY
:
Apply
m
=
∑
F
a
G
G
to each block.
0
a
=
.
S
ET
U
P
:
Take
perpendicular to the incline and
y
+
x
+
parallel to the incline.
E
XECUTE
:
The free-body diagrams for each block,
A
and
B
, are given in Figure 5.14.
(a)
For
B
,
x
x
F
ma
=
∑
gives
1
sin
0
T
w
α
−
=
and
1
sin
T
w
α
=
.
(b)
For block
A
,
x
x
F
ma
=
∑
gives
1
2
sin
0
T
T
w
α
−
−
=
and
2
2
sin
T
w
α
=
.
(c)
y
y
F
ma
=
∑
for each block gives
cos
A
B
n
n
w
α
=
=
.
(d)
For
0
α
→
,
and
. For
,
1
2
0
T
T
=
→
A
B
n
n
w
=
→
90
α
→
°
1
T
w
=
,
and
.
2
2
T
=
w
0
A
B
n
n
=
=
E
VALUATE
:
The two tensions are different but the two normal forces are the same.
Figure 5.14a, b
5.21.I
DENTIFY
:
Apply
m
∑
F =
a
G
G
to each block. Each block has the same magnitude of acceleration
a
.

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*