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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
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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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: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
• 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;
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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