The section of rope between Member-AD and Sheave-F is _____. options: 1 a two-force member 2 a multi-force member 3 a zero-force member 4 a torsional member
The section of rope between Member-AD and Sheave-F is _____. options: 1 a two-force member 2 a multi-force member 3 a zero-force member 4 a torsional member
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
The section of rope between Member-AD and Sheave-F is _____.
options:
|
1
|
a two-force member |
|
2
|
a multi-force member |
|
3
|
a zero-force member |
|
4
|
a torsional member |
![i E E
206 Chapter 4 Structures
3 m
SAMPLE PROBLEM 4/6
The frame supports the 400-kg load in the manner shown. Neglect the
weights of the members compared with the forces induced by the load and com-
pute the horizontal and vertical components of all forces acting on each of the
1.5 m
0.5 m
0.5 m
members.
1.5 m
Solution. We observe first that the three supporting members which consti-
O tute the frame form a rigid assembly that can be analyzed as a single unit. We
also observe that the arrangement of the external supports makes the frame sta-
tically determinate.
From the free-body diagram of the entire frame we determine the external
1.5 m
400 kg
Helpful Hints
reactions. Thus,
1 We see that the frame corresponds to
the category illustrated in Fig. 4/14a,
2 Without this observation, the prob-
lem solution would be much longer,
because the three equilibrium equa-
tions for member BF would contain
four unknowns: B, By, E, and E.
Note that the direction of the line
joining the two points of force appli-
cation, and not the shape of the mem-
ber, determines the direction of the
forces acting on a two-force member.
[0 = "W3)
5.5(0.4)(9.81) - 5D = 0
D = 4.32 kN
A- 4.32 = 0
A, = 4.32 kN
[o = "3)
A, - 3.92 = 0
A, = 3.92 kN
[o = 13]
Next we dismember the frame and draw a separate free-body diagram of
each member. The diagrams are arranged in their approximate relative positions
to aid in keeping track of the common forces of interaction. The external reac-
tions just obtained are entered onto the diagram for AD. Other known forces are
the 3.92-kN forces exerted by the shaft of the pulley on the member BF, as ob-
tained from the free-body diagram of the pulley. The cable tension of 3.92 kN is
also shown acting on AD at its attachment point.
Next, the components of all unknown forces are shown on the diagrams.
2 Here we observe that CE is a two-force member. The force components on CE
have equal and opposite reactions, which are shown on BF at E and on AD at C.
We may not recognize the actual sense of the components at B at first glance, so
they may be arbitrarily but consistently assigned.
The solution may proceed by use of a moment equation about B or E for
member BF, followed by the two force equations. Thus,
3.92(5) – E,(3) = 0
[0 = "W3)
[0 = 3]
B, + 3.92 – 13.08/2 = 0
E, = 13.08 kN
Ans.
%3D
0.4(9.81)
B, = 2.62 kN
Ans.
N可268=
B, + 3.92 – 13.08 = 0
B = 9.15 kN
Ans.
[0 = "3]
Positive numerical values of the unknowns mean that we assumed their direc-
tions correctly on the free-body diagrams. The value of C, = E, = 13.08 kN ob-
tained by inspection of the free-body diagram of CE is now entered onto the
diagram for AD, along with the values of B, and By just determined. The equa-
tions of equilibrium may now be applied to member AD as a check, since all the
forces acting on it have already been computed. The equations give
3.92 kN
A, = 3.92 kN
NY 76'E
= "v
3.92 kN
3.92 kN Y3.92 kN
[0 = °W3)
4.32(3.5) + 4.32(1.5) – 3.92(2) – 9.15(1.5) = 0
4.32 - 13.08 + 9.15 + 3.92 + 4.32 = 0
[0 = "3)
-13.08/2 + 2.62 + 3.92 = 0
[0 = 3)
4.32 kN](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6cf78908-a1f9-4330-9ece-f4021df10613%2F59d2d1e6-d1e6-4068-9ccf-46089397c191%2Fh8yzl4c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:i E E
206 Chapter 4 Structures
3 m
SAMPLE PROBLEM 4/6
The frame supports the 400-kg load in the manner shown. Neglect the
weights of the members compared with the forces induced by the load and com-
pute the horizontal and vertical components of all forces acting on each of the
1.5 m
0.5 m
0.5 m
members.
1.5 m
Solution. We observe first that the three supporting members which consti-
O tute the frame form a rigid assembly that can be analyzed as a single unit. We
also observe that the arrangement of the external supports makes the frame sta-
tically determinate.
From the free-body diagram of the entire frame we determine the external
1.5 m
400 kg
Helpful Hints
reactions. Thus,
1 We see that the frame corresponds to
the category illustrated in Fig. 4/14a,
2 Without this observation, the prob-
lem solution would be much longer,
because the three equilibrium equa-
tions for member BF would contain
four unknowns: B, By, E, and E.
Note that the direction of the line
joining the two points of force appli-
cation, and not the shape of the mem-
ber, determines the direction of the
forces acting on a two-force member.
[0 = "W3)
5.5(0.4)(9.81) - 5D = 0
D = 4.32 kN
A- 4.32 = 0
A, = 4.32 kN
[o = "3)
A, - 3.92 = 0
A, = 3.92 kN
[o = 13]
Next we dismember the frame and draw a separate free-body diagram of
each member. The diagrams are arranged in their approximate relative positions
to aid in keeping track of the common forces of interaction. The external reac-
tions just obtained are entered onto the diagram for AD. Other known forces are
the 3.92-kN forces exerted by the shaft of the pulley on the member BF, as ob-
tained from the free-body diagram of the pulley. The cable tension of 3.92 kN is
also shown acting on AD at its attachment point.
Next, the components of all unknown forces are shown on the diagrams.
2 Here we observe that CE is a two-force member. The force components on CE
have equal and opposite reactions, which are shown on BF at E and on AD at C.
We may not recognize the actual sense of the components at B at first glance, so
they may be arbitrarily but consistently assigned.
The solution may proceed by use of a moment equation about B or E for
member BF, followed by the two force equations. Thus,
3.92(5) – E,(3) = 0
[0 = "W3)
[0 = 3]
B, + 3.92 – 13.08/2 = 0
E, = 13.08 kN
Ans.
%3D
0.4(9.81)
B, = 2.62 kN
Ans.
N可268=
B, + 3.92 – 13.08 = 0
B = 9.15 kN
Ans.
[0 = "3]
Positive numerical values of the unknowns mean that we assumed their direc-
tions correctly on the free-body diagrams. The value of C, = E, = 13.08 kN ob-
tained by inspection of the free-body diagram of CE is now entered onto the
diagram for AD, along with the values of B, and By just determined. The equa-
tions of equilibrium may now be applied to member AD as a check, since all the
forces acting on it have already been computed. The equations give
3.92 kN
A, = 3.92 kN
NY 76'E
= "v
3.92 kN
3.92 kN Y3.92 kN
[0 = °W3)
4.32(3.5) + 4.32(1.5) – 3.92(2) – 9.15(1.5) = 0
4.32 - 13.08 + 9.15 + 3.92 + 4.32 = 0
[0 = "3)
-13.08/2 + 2.62 + 3.92 = 0
[0 = 3)
4.32 kN
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