The rectangular plate shown weighs 93 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust and no couples exerted on both A and B, determine (a) the tension in the cable, (b) the reactions at A and B.
The rectangular plate shown weighs 93 lb and is held in the position shown by hinges at A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust and no couples exerted on both A and B, determine (a) the tension in the cable, (b) the reactions at A and B.
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
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Transcribed Image Text:The rectangular plate shown weighs 93 lb and is held in the position shown by hinges at
A and B and by cable EF. Assuming that the hinge at B does not exert any axial thrust and
no couples exerted on both A and B, determine (a) the tension in the cable, (b) the
reactions at A and B.
Z
D
12 in.
30 in.
H
4 in.
A
E
8 in.
B
F
4 in. 25 in.
20 in.
x

Transcribed Image Text:• FBD analysis
1. There are in total
• Presenting all forces and couples using components:
1. Reacions on A: A =
2. Reactions on B: B =
3. Tension in cable EF:
o EF =
i+
• TEF=
TEFI +
4. Self weight of the plate W =
1. TAA =
2. [AB =
3. TAE =
4. TAW =
i+
1. MAA=
2. MAB=
3. ME=
4. MAG=
+1
+
i+
• Determination of arms for all forces, taking point A as rotating point:
i+
supports on the system, by assuming no couples on A and B, no axial trust on B, the reactions/forces on the FB are:
j+
j+
+!
i+
i+
| +
i+
j+
TEFI +
k (in);
k (in);
♦ j+
k;
• Determination of moments about A for each force:
• Equilibrium Equations:
1. ΣΜΑ = 0:
o Coefficient of i simplified to:
o Coefficient of j simplified to:
k (in);
k (in);
k (lb-in);
o Coefficient of k simplified to:
2. [F = 0:
o Coefficient of i simplified to:
o Coefficient of j simplified to:
o Coefficient of k simplified to:
TEFJ+
◆lb
TEFI +
k (lb-in);
◆k (lb);
◆ k (lb);
TEF +
TEF+
TEF+
TEF+
TEF +
TEF+
TEF K;
◆ k (lb-in);
TEFK (Ib-in);
◆
= 0, hence TEF=
By +
Bz = 0, hence Bz=
= 0, hence Ax =
+
◆ +
= 0, hence By =
lb;
+ +
lb;
= 0, hence Az =
lb;
lb;
= 0, hence Ay =
lb.
lb;
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