Assignment 10 Key solutions
Chapter 6, Problem 86.
Determine the components of the reactions at
A
and
E
when a
counterclockwise couple of magnitude 192 lb
⋅
in. is applied to
the frame (
a
) at
B
, (
b
) at
D
.
Chapter 6, Solution 86.
(
a
)
FBD
AC
:
(
b
)
FBD
CE
:
Note:
CE
is a twoforce member
(
29
(
29
1
1
0:
8 in.
2 in.
192 lb in.
0
2
2
A
CE
CE
M
F
F
Σ
=


+
⋅
=
19.2
2 lb,
CE
F
=
19.20 lb
x
=
E
t
19.20 lb
y
=
E
t
0:
19.2 lb = 0,
x
x
F
A
Σ
=

19.20 lb
x
=
A
t
0:
19.2 lb = 0,
y
y
F
A
Σ
=

19.20 lb
y
=
A
t
Note:
AC
is a twoforce member
(
29
1
4
0:
3 in.
192 lb in. = 0
17
17
E
AE
AE
M
F
F
Σ
=
+
+
⋅
12.8 17 lb,
AE
F
= 
4
,
17
x
AE
A
F
=
51.2 lb
x
=
A
t
1
,
17
y
AE
A
F
=
12.80 lb
y
=
A
t
0:
51.2 lb = 0,
x
x
F
E
Σ
=

51.2 lb
x
=
E
t
0:
12.80 lb = 0,
y
y
F
E
Σ
=

12.80 lb
y
=
E
t
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Chapter 6, Problem 89.
The 120N load can be moved along the line of action shown
and can be applied at
A
,
D
, or
E
. Determine the components of
the reactions at
B
and
F
when the 120N load is applied (
a
) at
A
,
(
b
) at
D
, (
c
) at
E
.
Chapter 6, Solution 89.
(
a
)
FBD
ACF
:
(
b
)
FBD
BCE
:
Note:
BC
is a twoforce member
(
29
(
29
(
29
1
0:
0.1 m
120 N
0.3 m
0,
26
F
BC
M
F
Σ
=

=
40
26 N,
BC
F
=
200 N
x
=
B
t
40.0 N
y
=
B
t
(
29
5
0:
40
26 N
0,
26
x
x
F
F
Σ
=

=
200 N
x
=
F
t
(
29
1
0:
120 N
40
26 N
0,
26
y
y
F
F
Σ
=


=
160.0 N
y
=
F
t
Note
ACF
is a twoforce member
(
29
(
29
(
29
1
0:
0.4 m
120 N
0.3 m
0,
5
B
CF
M
F
Σ
=

=
160 5 N,
CF
F
=
320 N
x
=
F
t
160.0 N
y
=
F
t
(
29
2
0:
160 5 N
0,
5
x
x
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 Spring '11
 Hsia
 Statics, Trigraph, lb, FBD Frame

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