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➤ Consider FBD of pulley H: F₁, FH 54" Y TBF = 0.2666053799P = 500N TGF = 0.43137655663P = 500 N By applying the equilibrium conditions, we can solve for FS,FH and TH in terms of P. → EF₂ = -FSFH Cos 54 +3TH) cos 78.701 = 0 + EF = FS,FH sin 54 +37HJ sin 78.701 - P = 0 FSFH = 0.2666053799P = (0.250m) (4000N/m) THI= 0.2666045474P = 500 N 3TH1 TBF 78.701" P Because the computed value of FS.FH is positive, it means that spring FH is in tension. To solve for the maximum tensile force the spring can support, the largest possible length of stretch 0.250m is multiplied to the spring constant. 36 > Consider FBD of pulley F: Since we already have the value of FS,FH in terms of P, we can solve for TBF and TGF in terms of P by applying the equilibrium equations. EF=-TBF sin 36+ FFH COS 54 = 0 12F₂=-TBF Cos 36 - FS,FH sin 54 + TGF TGF 54" P= 3750.8620 N P= 1875.4369 N F₁, FH = 0 P = 1875.4310 N P= 1159.080115 N

➤ Consider FBD of pulley H: F₁, FH 54" Y TBF = 0.2666053799P = 500N TGF = 0.43137655663P = 500 N By applying the equilibrium conditions, we can solve for FS,FH and TH in terms of P. → EF₂ = -FSFH Cos 54 +3TH) cos 78.701 = 0 + EF = FS,FH sin 54 +37HJ sin 78.701 - P = 0 FSFH = 0.2666053799P = (0.250m) (4000N/m) THI= 0.2666045474P = 500 N 3TH1 TBF 78.701" P Because the computed value of FS.FH is positive, it means that spring FH is in tension. To solve for the maximum tensile force the spring can support, the largest possible length of stretch 0.250m is multiplied to the spring constant. 36 > Consider FBD of pulley F: Since we already have the value of FS,FH in terms of P, we can solve for TBF and TGF in terms of P by applying the equilibrium equations. EF=-TBF sin 36+ FFH COS 54 = 0 12F₂=-TBF Cos 36 - FS,FH sin 54 + TGF TGF 54" P= 3750.8620 N P= 1875.4369 N F₁, FH = 0 P = 1875.4310 N P= 1159.080115 N

International Edition---engineering Mechanics: Statics, 4th Edition
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter4: Coplanar Equilibrium Analysis
Section: Chapter Questions
Problem 4.125P: The figure shows a three-pin arch. Determine the horizontal component of the pin reaction at A...
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➤ Consider FBD of pulley H: F₂, FH 54° TBF TGF 3TH1 By applying the equilibrium conditions, we can solve for FS,FH and TH in terms of P. → EF₂ = -F₁,FH Cos 54 +37H cos 78.701 = 0 4EF₂ = FS.FH sin 54 +37HJ sin 78.701 - P = 0 FS.FR = 0.2666053799P = (0.250m)(4000N/m) THI = 0.2666045474P = 500 N 78.701" y P Because the computed value of FSFH is positive, it means that spring FH is in tension. To solve for the maximum tensile force the spring can support, the largest possible length of stretch 0.250m is multiplied to the spring constant. 36" > Consider FBD of pulley F: Since we already have the value of FS,FH in terms of P, we can solve for TBF and TGF in terms of P by applying the equilibrium equations. TGF 54" P= 3750.8620 N P = 1875.4369 N F₁, FH →2F₂=-TBF sin 36 + FFH COS 54 = 0 EF=-TBF cos 36 - FS,FH sin 54 + TGF = 0 0.2666053799P = 500N 0.43137655663P = 500 N P= 1875.4310 N P = 1159.080115 N
Transcribed Image Text:➤ Consider FBD of pulley H: F₂, FH 54° TBF TGF 3TH1 By applying the equilibrium conditions, we can solve for FS,FH and TH in terms of P. → EF₂ = -F₁,FH Cos 54 +37H cos 78.701 = 0 4EF₂ = FS.FH sin 54 +37HJ sin 78.701 - P = 0 FS.FR = 0.2666053799P = (0.250m)(4000N/m) THI = 0.2666045474P = 500 N 78.701" y P Because the computed value of FSFH is positive, it means that spring FH is in tension. To solve for the maximum tensile force the spring can support, the largest possible length of stretch 0.250m is multiplied to the spring constant. 36" > Consider FBD of pulley F: Since we already have the value of FS,FH in terms of P, we can solve for TBF and TGF in terms of P by applying the equilibrium equations. TGF 54" P= 3750.8620 N P = 1875.4369 N F₁, FH →2F₂=-TBF sin 36 + FFH COS 54 = 0 EF=-TBF cos 36 - FS,FH sin 54 + TGF = 0 0.2666053799P = 500N 0.43137655663P = 500 N P= 1875.4310 N P = 1159.080115 N
PROBLEM 6 A pulley-and-cable assembly supports block W and requires a downward force at P for equilibrium. Determine the largest tension at P if the tension in the cables were not to exceed 500 N, the springs were not to stretch by more than 250 mm nor compress by 200 mm, and the tension in the bar were not to exceed 900 N nor the compression greater than 800 N. Also, calculate the deformation in the springs and force in the bar when P is at maximum. (Use k = 4 kN/m.) A 15 25° B kum 54 www 42 E W H 78.701
Transcribed Image Text:PROBLEM 6 A pulley-and-cable assembly supports block W and requires a downward force at P for equilibrium. Determine the largest tension at P if the tension in the cables were not to exceed 500 N, the springs were not to stretch by more than 250 mm nor compress by 200 mm, and the tension in the bar were not to exceed 900 N nor the compression greater than 800 N. Also, calculate the deformation in the springs and force in the bar when P is at maximum. (Use k = 4 kN/m.) A 15 25° B kum 54 www 42 E W H 78.701
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cos 78-701= ~ FS, FH COS 54 + 3 THJ cos" Tuy 0-5877 FFH+0.587787 Ti15 FS, FM Sin 54 +- 3 TH5 Sin 78-701 = p 0:8090 FS+ 2 9418 THS P FH 11.376552663, 0-80.90 E 0.90 FS, FH + 0·8091T, = 0 S.FH +0.8090. 0.Fs. Fs, FH + 2.9418 THS - P 3.7509 THE P THE = 0-266026 P from 1), F.S.E. H E FS = S,FH 1.0001361240-2660 0.2660622. 1). 56026/p þ
Transcribed Image Text:cos 78-701= ~ FS, FH COS 54 + 3 THJ cos" Tuy 0-5877 FFH+0.587787 Ti15 FS, FM Sin 54 +- 3 TH5 Sin 78-701 = p 0:8090 FS+ 2 9418 THS P FH 11.376552663, 0-80.90 E 0.90 FS, FH + 0·8091T, = 0 S.FH +0.8090. 0.Fs. Fs, FH + 2.9418 THS - P 3.7509 THE P THE = 0-266026 P from 1), F.S.E. H E FS = S,FH 1.0001361240-2660 0.2660622. 1). 56026/p þ
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