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PROBLEM 6.56
Determine the components of the reactions at A and E if a 750-N force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free-body: Entire frame:
The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial.
0: (750 N)(240 mm) (400 mm) 0
Ex
MAΣ= − − =
450 N 450 N
xx
A=−=A
0: 450 N 0 450 N 450 N
xx x x
FE EΣ= − = = =E
0: 750 N 0
y yy
F AEΣ= + − =
(1)
(a) Load applied at B.
Free body: Member CE:
240 mm ; 225 N
450 N 480 mm
yy
E= =E
From Eq. (1):
225 750 0; 525 N
yy
A+−= =A
Thus, reactions are
450 N
x
=
A
,
525 N
y=A
450N
x
=
E
,
225 N
y=E
(b) Load applied at D.
Free body: Member AC:
160 mm 150.0 N
450 N 480 mm
yy
A= =A
From Eq. (1):
750 N 0
150 N 750 N 0
yy
y
AE
E
+− =
+− =
600 N 600 N
yy
E= =E
Thus, reactions are
450 N
x
=A
,
150.0 N
y
=A
450 N
x=E
,
600 N
y=E
consent of McGraw-Hill Education.
SOLUTION Continued
From Eq. (1):
750 N 0
150 N 750 N 0
yy
y
AE
E
+− =
+− =
600 N 600 N
yy
E= =E
Thus, reactions are
300 N
x=A
,
150.0 N
y=A
300 N
x=E
,
600 N
y=E
consent of McGraw-Hill Education.
PROBLEM 6.57
Knowing that P = 90 lb and Q = 60 lb, determine the
components of all forces acting on member BCDE of the
assembly shown.
SOLUTION
Free body: Entire assembly:
Free body: Member BCDE:
( )( )
0: (12 in.) (60 lb)(4 in.) 90 lb 12 in. 0
C
MAΣ= − − =
110.0 lbA= +
110.0 lb=A
0: 60 lb 0
xx
FCΣ= + =
60 lb
x
C= −
60.0 lb
x=C
0: 110 lb 90 lb 0
yy
FCΣ= − − =
200 lb
y
C=
200 lb
y=C
( ) ( )( )
0: (10 in.) 200 lb (4 in.) 90 lb 8 in. 0
D
MBΣ= − − =
152.0 lbB= +
152.0 lb=B
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
SOLUTION Continued
60.0 lb
x
=D
0: 60 lb 0
xx
FD
Σ= − =
60.0 lb
x
D=
0: 200 lb 152 lb 90 lb 0
yy
FDΣ= + − − =
42.0 lb
y
D= +
42.0 lb
y
=D
PROBLEM 6.58
Determine the components of the reactions at A and
E, (a) if the 800-N load is applied as shown, (b) if
the 800-N load is moved along its line of action and
is applied at point D.
SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the 800-N force
on its line of action is immaterial.
0: (1200) (800 N)(900) 0
Ay
MEΣ= − =
600 N
y=E
0: 0 (1)
x xx
F AE
Σ= + =
0: 600 800 0
yy
FAΣ= + − =
200 N
y=A
(a) Free body: Member BE:
We note that E is directed along EB, since BE is a two-force member.
900
or 600 N
900 200 200
y
xx
E
EE
= =
2700 N
x
=E
From equation (1): 2700 N 0
x
A+=
2700 N
x
A=
2700 N
x
=A
SOLUTION Continued
(b) Free body: Member ABC:
We note that A is directed along AB, since ABC is a two-force member.
300
or 200 N
300 200 200
y
xx
A
AA
= =
300 N
x
=A
From equation (1): 300 N 0
x
E+=
300 N
x
E=
300 N
x
=
E
PROBLEM 6.59
Determine the components of the reactions at D and E if the
frame is loaded by a clockwise couple of magnitude 150 N∙m
applied (a) at A, (b) at B.
SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the couple is
immaterial.
0: (0.6 m) (150 N m) 0
Dy
MEΣ = − ⋅=
250 N
y=E
0: 0 (1)
x xx
F DEΣ= + =
0: 250 N 0
yy
FDΣ= + =
250 N
y=D
(a) Free body: Member BCE:
We note that E is directed along EC, since BCE is a two-force member.
1.2
or 250 N
1.2 0.40 0.40
y
xx
E
EE
= =
750 N
x=E
From equation (1): D 750 N 0
x
−=
750 N
x
D=
750 N
x=D
consent of McGraw-Hill Education.
SOLUTION Continued
(b) Free body: Member ACD:
0.6
or D 250 N
0.6 0.4 0.4
y
xx
D
D
= =
375 N
x
=
D
From equation (1): 375 N 0
x
E
+=
375 N
x
E= −
375 N
x
=E
consent of McGraw-Hill Education.
PROBLEM 6.60
The 48-lb load can be moved along the line of action shown and applied at
A, D, or E. Determine the components of the reactions at B and F if the
48-lb load is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
PROBLEM 6.56
Determine the components of the reactions at A and E if a 750-N force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free-body: Entire frame:
The following analysis is valid for both parts (a) and (b) since position of load on its line of action is immaterial.
0: (750 N)(240 mm) (400 mm) 0
Ex
MAΣ= − − =
450 N 450 N
xx
A=−=A
0: 450 N 0 450 N 450 N
xx x x
FE EΣ= − = = =E
0: 750 N 0
y yy
F AEΣ= + − =
(1)
(a) Load applied at B.
Free body: Member CE:
240 mm ; 225 N
450 N 480 mm
yy
E= =E
From Eq. (1):
225 750 0; 525 N
yy
A+−= =A
Thus, reactions are
450 N
x
=
A
,
525 N
y=A
450N
x
=
E
,
225 N
y=E
(b) Load applied at D.
Free body: Member AC:
160 mm 150.0 N
450 N 480 mm
yy
A= =A
From Eq. (1):
750 N 0
150 N 750 N 0
yy
y
AE
E
+− =
+− =
600 N 600 N
yy
E= =E
Thus, reactions are
450 N
x
=A
,
150.0 N
y
=A
450 N
x=E
,
600 N
y=E
consent of McGraw-Hill Education.
SOLUTION Continued
From Eq. (1):
750 N 0
150 N 750 N 0
yy
y
AE
E
+− =
+− =
600 N 600 N
yy
E= =E
Thus, reactions are
300 N
x=A
,
150.0 N
y=A
300 N
x=E
,
600 N
y=E
consent of McGraw-Hill Education.
PROBLEM 6.57
Knowing that P = 90 lb and Q = 60 lb, determine the
components of all forces acting on member BCDE of the
assembly shown.
SOLUTION
Free body: Entire assembly:
Free body: Member BCDE:
( )( )
0: (12 in.) (60 lb)(4 in.) 90 lb 12 in. 0
C
MAΣ= − − =
110.0 lbA= +
110.0 lb=A
0: 60 lb 0
xx
FCΣ= + =
60 lb
x
C= −
60.0 lb
x=C
0: 110 lb 90 lb 0
yy
FCΣ= − − =
200 lb
y
C=
200 lb
y=C
( ) ( )( )
0: (10 in.) 200 lb (4 in.) 90 lb 8 in. 0
D
MBΣ= − − =
152.0 lbB= +
152.0 lb=B
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
SOLUTION Continued
60.0 lb
x
=D
0: 60 lb 0
xx
FD
Σ= − =
60.0 lb
x
D=
0: 200 lb 152 lb 90 lb 0
yy
FDΣ= + − − =
42.0 lb
y
D= +
42.0 lb
y
=D
PROBLEM 6.58
Determine the components of the reactions at A and
E, (a) if the 800-N load is applied as shown, (b) if
the 800-N load is moved along its line of action and
is applied at point D.
SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the 800-N force
on its line of action is immaterial.
0: (1200) (800 N)(900) 0
Ay
MEΣ= − =
600 N
y=E
0: 0 (1)
x xx
F AE
Σ= + =
0: 600 800 0
yy
FAΣ= + − =
200 N
y=A
(a) Free body: Member BE:
We note that E is directed along EB, since BE is a two-force member.
900
or 600 N
900 200 200
y
xx
E
EE
= =
2700 N
x
=E
From equation (1): 2700 N 0
x
A+=
2700 N
x
A=
2700 N
x
=A
SOLUTION Continued
(b) Free body: Member ABC:
We note that A is directed along AB, since ABC is a two-force member.
300
or 200 N
300 200 200
y
xx
A
AA
= =
300 N
x
=A
From equation (1): 300 N 0
x
E+=
300 N
x
E=
300 N
x
=
E
PROBLEM 6.59
Determine the components of the reactions at D and E if the
frame is loaded by a clockwise couple of magnitude 150 N∙m
applied (a) at A, (b) at B.
SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the couple is
immaterial.
0: (0.6 m) (150 N m) 0
Dy
MEΣ = − ⋅=
250 N
y=E
0: 0 (1)
x xx
F DEΣ= + =
0: 250 N 0
yy
FDΣ= + =
250 N
y=D
(a) Free body: Member BCE:
We note that E is directed along EC, since BCE is a two-force member.
1.2
or 250 N
1.2 0.40 0.40
y
xx
E
EE
= =
750 N
x=E
From equation (1): D 750 N 0
x
−=
750 N
x
D=
750 N
x=D
consent of McGraw-Hill Education.
SOLUTION Continued
(b) Free body: Member ACD:
0.6
or D 250 N
0.6 0.4 0.4
y
xx
D
D
= =
375 N
x
=
D
From equation (1): 375 N 0
x
E
+=
375 N
x
E= −
375 N
x
=E
consent of McGraw-Hill Education.
PROBLEM 6.60
The 48-lb load can be moved along the line of action shown and applied at
A, D, or E. Determine the components of the reactions at B and F if the
48-lb load is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
