roghibin's blog EQUILIBRIUM OF RIGID BODIES KESETIMBANG AN BENDA TEGAR

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

roghibin's blog Equilibrium Of POINT ( Kesetimbangan Titik) Point W T1T1 T2T2 Syarat Setimbang: 1.Σ F x = 0 2.Σ Fy = 0

roghibin's blog Equilibrium Of POINT ( Kesetimbangan Titik) Point T1T1 W T2T2 α βT 1 cos α T 1 sinα T 2 cos β T 2 sinβ Σ Fx = 0 T1 cos α- T2 cos β = 0 Σ Fy = 0 T1 sinα + T2 sinβ – W = 0

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

roghibin's blog Answer T2T2 T1 T3T3 30 o 60 o T2 cos 30 o T2 sin30 o T3 cos60 o T3 sin60 o T1 = W = m.g = 50 N Σ Fx = 0 T2 cos 30 o - T3 cos60 o = 0 T2 ½ = T3 ½ T3 = T2 Σ Fy = 0 T2 sin30 o + T3 sin60 o – W = 0 T2 ½ + T3 ½ = 50 T2 + T3 = 100 T 2 +T 2. 3=100 4 T 2 =100 T 2 = 25 N T 3 = 25 N

roghibin's blog Shortcut Formula T1 T2 T3 α1α1 α2α2 α3α3 Notes : Di kwadran 2 berlaku : Sin ( 180 – α ) = sin α

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

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

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

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

roghibin's blog

Equilibrium Of Rigid Bodies Prerequisite Of Equilibrium of Rigid Bodies

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

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

roghibin's blog 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. A B

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

roghibin's blog 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 A CB X =? o

roghibin's blog A CB X =? o N 1,5m1,2m 800N 300N200N N – 800 – 300 – 200 = 0 N = 1300 N Poros O 300.1, (1,2+x)-800.1,5 = x= x = x = 600 X = 600/200 X = 3 meter

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

roghibin's blog W4 W3 W2 W1 ( X1, Y1) ( X2, Y2) ( X3, Y3) ( X4, Y4) W ( Xo, Yo) W.Xo = w1.x1+ w2.x2 + w3.x3 + w4.x4 Xo = w1.x1+ w2.x2 + w3.x3 + w4.x4 w1 + w2 + w3 + w4 W = 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 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 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 If we concern about LENGTH Xo = l 1.x1+ l 2.x2 + l 3.x3 + l 4.x4 l1 + l2 + l3 + l4 Yo = l 1.y1+ l 2.y2 + l 3.y3 + l 4.y4 l1 + l2 + l3 + l4

roghibin's blog Example : 1 Determine the coordinate of weight point, from following area object !

roghibin's blog answer x y formA(x, y) I II (1,5) (4,1 ½ ) , 5 4, 1,5 A1.x1+ A2.x2 A1 + A2 X o = = = = 2,125 y o = A1.y1+ A2.y2 A1 + A2 = , = = = 3, 688

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

roghibin's blog Answer : 2R x y formvolume(x, y ) 2πR32πR3 -2/3πR 3 2/3πR 3 0, R 0, 3/8R 0, ½ R

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