PROBLEM 6.77
For the frame and loading shown, determine the force
acting on member ABC (a) at B, (b) at C.
SOLUTION
FBD ABC:
Note: BD is two-force member
(a)
MC
0: (0.09 m)(200 N) (2.4 m)
3
FBD
5
0
125.0 N
36.9
C 125.0 N
36.9
FBD
(b)
Fx
0: 200 N
Fy
0:
3
FBD
5
4
(125 N) C x
5
Cy
0
Cy
0
Cx
3
(125 N)
5
100 N
75 N
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PROBLEM 6.79
For the frame and loading shown, determine the components of all forces
acting on member ABC.
SOLUTION
Free body: Entire frame:
Fx
0: Ax 18 kN 0
Ax
ME
0:
A x 18.00 kN
18 kN
(18 kN)(4 m)
Ay (3.6 m) 0
Ay
Fy
0:
20 kN F
F
20 kN
Ay
20.0 kN
0
20 kN
F
20 kN
Free body: Member ABC
Note: BE is a two-force member, thus B is directed along line BE.
MC
0: B(4 m) (18 kN)(6 m) (20 kN)(3.6 m) 0
B 9 kN
Fx
0: Cx 18 kN 9 kN 0
Cx
Fy
B 9.00 kN
0: C y
9 kN
Cx
9.00 kN
20 kN 0
Cy
20 kN
Cy
20.0 kN
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PROBLEM 6.83
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.
MA
0: E y (1200) (800 N)(900) 0
Fx
0: Ax
Ex
Fy
0: Ay
600 800 0
Ey
600 N
Ay
200 N
0 (1)
(a) Free body: Member BE:
We note that E is directed along EB, since BE is a two-force member.
Ex
900
Ey
200
or E x
900
600 N
200
Ex
From equation (1): Ax
2700 N
Ax
2700 N
0
2700 N
Ax
2700 N
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PROBLEM 6.83 (Continued)
(b) Free body: Member ABC:
We note that A is directed along AB, since ABC is a two-force member.
Ax
300
Ay
200
or Ax
300
200 N
200
Ax
300 N
From equation (1): 300 N
Ex
300 N
Ex
Ex
300 N
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