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EQUILIBRIUM OF RIGID BODIES KESETIMBANGAN BENDA TEGAR PowerPoint Presentation
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EQUILIBRIUM OF RIGID BODIES KESETIMBANGAN BENDA TEGAR

EQUILIBRIUM OF RIGID BODIES KESETIMBANGAN BENDA TEGAR

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EQUILIBRIUM OF RIGID BODIES KESETIMBANGAN BENDA TEGAR

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  1. EQUILIBRIUM OF RIGID BODIES KESETIMBANGAN BENDA TEGAR roghibin's blog

  2. Equilibrium Of POINT ( Kesetimbangan Titik) Base on object of Equilibrium Equilibrium of Rigid Bodies ( Kesetm. Benda Tegar) roghibin's blog

  3. Equilibrium Of POINT( Kesetimbangan Titik) T1 T2 • Syarat Setimbang: • Σ Fx = 0 • Σ Fy = 0 Point W roghibin's blog

  4. T1 T2 Point W Equilibrium Of POINT( Kesetimbangan Titik) T1 sinα Σ Fx = 0 T1 cos α- T2 cos β = 0 Σ Fy = 0 T1 sinα + T2 sinβ – W = 0 T2 sinβ β T1 cos α α T2 cos β roghibin's blog

  5. Ditermine tension of each string T1 , T2 and T3 ! Example 60o 30o T3 T2 T1 5 Kg roghibin's blog

  6. Answer T1 = W = m.g = 50 N T3 sin60o T2 T3 Σ Fx = 0 T2 sin30o T2 cos 30o - T3 cos60o = 0 T2 ½ = T3 ½ T3 = T2 60o 30o T2 cos 30o T3 cos60o Σ Fy = 0 T2 sin30o + T3 sin60o – W = 0 T2 ½ + T3 ½ = 50 T2 + T3 = 100 T2+T2.3=100 4 T2=100 T2= 25 N T1 T3= 25 N roghibin's blog

  7. Shortcut Formula T2 T1 α3 α1 α2 Notes : Di kwadran 2 berlaku : Sin ( 180 – α ) = sin α T3 roghibin's blog

  8. Ditermine tension of each string T1 , T2 and T3 ! Example 60o 30o T3 T2 T1 5 Kg roghibin's blog

  9. Answer 60o 30o T3 90o T2 150o 120o T1 5 Kg T1 = W = 50 N roghibin's blog

  10. Moment Of Force F l sinα α poros α l F F sinα roghibin's blog

  11. Moment of Force is Vector + As clockwise _ Anti clockwise roghibin's blog

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  16. EquilibriumOf Rigid Bodies Prerequisite Of Equilibrium of Rigid Bodies roghibin's blog

  17. Example No: 1 A B O 2m X = ? 60 kg 100 kg To becomes equilibrium condition, so where are B object must be placed from O ?  X = ? roghibin's blog

  18. N O 2m X = ? WB=600N wA = 1000N WB.x – WA.2=0 600x – 1000.2 =0 600x = 2000 X = 2000/600 X = 3,3 m N – WA – WB =0 N – 1000 -600 =0 N = 1600 N roghibin's blog

  19. Dua orang A dan B ingin membawa beban 1200 N dengan menggunakan batang homogen yang masanya dapat diabaikan. Panjang batang 4 meter. Dimanakah beban harus diletakkan ( diukur dari B ) agar B menderita gaya 2 kali dari A. B A roghibin's blog

  20. NB NA 4m NA+NB-w=0 NA+NB=1200 NA+2NA=1200 3NA = 1200 NA= 400 N NB= 800 N X=? (4-x) 1200N NA.(4-X)-NB.X=0 400(4-X)-800X=0 1600-400X-800X=0 1600-1200X=0 1600=1200X X= 1,33 m NB = 2 NA B A roghibin's blog

  21. From the following picture, how far C must be placed from B so that equilibrium system ! mA = 80 kg mB = 30 Kg mC = 20 kg AO = 1,5 m OB = 1,2 m B C A X =? o roghibin's blog

  22. B C A X =? N N – 800 – 300 – 200 = 0 N = 1300 N Poros O 1,5m 1,2m o 300.1,2 + 200(1,2+x)-800.1,5 = 0 360+240+200x=1200 300N 200N 600 + 200 x = 1200 800N 200 x = 600 X = 600/200 X = 3 meter roghibin's blog

  23. WEIGHT POINT ( X1, Y1) ( X3, Y3) ( Xo, Yo) W3 W1 ( X4, Y4) W2 W4 W ( X2, Y2) roghibin's blog

  24. ( Xo, Yo) ( X1, Y1) ( X3, Y3) W W3 W1 ( X4, Y4) W2 W4 ( X2, Y2) W.Xo = w1.x1+ w2.x2 + w3.x3 + w4.x4 W = w1 + w2 + w3 + w4 Xo = w1.x1+ w2.x2 + w3.x3 + w4.x4 w1 + w2 + w3 + w4 W.Yo = w1.y1+ w2.y2 + w3.y3 + w4.y4 Yo = w1.y1+ w2.y2 + w3.y3 + w4.y4 w1 + w2 + w3 + w4 roghibin's blog

  25. If we concern about AREA Xo = A1.x1+ A2.x2 + A3.x3 + A4.x4 A1 + A2 + A3 + A4 Yo = A1.y1+ A2.y2 + A3.y3 + A4.y4 A1 + A2 + A3 + A4 roghibin's blog

  26. If we concern about VOLUME Xo = V1.x1+ V2.x2 + V3.x3 + V4.x4 V1 + V2 + V3 + V4 Yo = V1.y1+ V2.y2 + V3.y3 + V4.y4 V1 + V2 + V3 + V4 roghibin's blog

  27. If we concern about LENGTH Xo = l1.x1+ l2.x2 + l3.x3 + l4.x4 l1 + l2 + l3 + l4 Yo = l1.y1+ l2.y2 + l3.y3 + l4.y4 l1 + l2 + l3 + l4 roghibin's blog

  28. Example : 1 Determine the coordinate of weight point, from following area object ! 10 3 2 6 roghibin's blog

  29. answer y 20 1 , 5 10 12 4 , 1,5 (1,5) (4,1 ½ ) A1.x1+ A2.x2 A1 + A2 Xo = 3 = 20.1+12.4 20+12 = 68 32 = 2,125 x 2 6 A1.y1+ A2.y2 A1 + A2 = 118 32 yo = = 20.5 + 12.1,5 20 + 12 = 100 + 18 32 = 3, 688 roghibin's blog

  30. Example : 2 Determine the coordinate of weight point, from following volume object ! 2R 2R 2R roghibin's blog

  31. Answer : y 2πR3 0, R 2R -2/3πR3 0, 3/8R 2/3πR3 2R 0, ½ R x 2R roghibin's blog

  32. Xo = V1.x1+ V2.x2 + V3.x3 V1 + V2 + V3 XO= 0 Yo = V1.y1+ V2.y2 + V3.y3 V1 + V2 + V3 roghibin's blog