Question

Determine the tension forces that developed in the cables AC, AD, and AB. If the hook supports a flowerpot that has a mass of 42 kg and the distance d = 3.8 m. Use the standard acceleration 9.81 m/s2 in your solution. 1-2 md-2 m- D 3 m 6 m B 6 m

Answer 1

Page: No. 1 -2m-*2m N & DE zm 90 90 TAD 90° ITAC 6m de 198 ) B2 А) T ། B TAR TAB om d=3.8m Yw=mg = 42x9.81 = 412:02 N mass of flower pot, m=42 kg Standard acceleration, g= 9.81m|s2 x= x= x2= tant tai 2.0) - tor) -156 2 6. 8= B 1 = 82 = tant = 18.43499 (13= 32-3474 &z= tant (3.8

Page, No. 2 Z TAD Sind ( Tacsinai ATABsings X TAD COS 92 COSB2 TAC Cosa cospl TAC Cosq, sinßi TADCO SA sing2 K T=W=412.02 NI TAB COSO х Using equilibrium equations, sty=0 TAD COSAZ sin 82 - TAC Cast, sinßi=0 - TAD= TAC Cosa, sinki Cosaa sinße TAG Cos 25.3769 sin 18 4349 Cos 25:3769 sin 18.4349 - TAD = TAC

Page. No:3 {Tic=0 + TAB COS A3 - Tac Cosar, cosßi - TAD Cosaz cos $2 = 0 from equationO, TAD=Tac > TAB cusaz - "Tac cosay cospi - Tac cos az.cosB2 = 0 TAB = TAC (cosa, cos Bi + cosas cos B2, cos93 TAC X 2 cosa casf COSKB = TACX 2 Cos 25.3769 cas 18.4349 COS 32:3474 2 - TAB 2:0232 Tác EF=0 + TAD Sinaz + TAC sing+TAB Sinaz - W=0 (from equation and @) ->) Tac sing + Tac sina +(2:0292) Tac sina 3 -412:02=0 → Tac (2 sind +2:0292 sinaz) = 412:02 TAC = (2sing +2:0292 sin 03) - TAC 412.02

Page No 4 - TAC = 412.02 (2 sin 25-3769 + 2:0292 sin 32:3474 TAC 212. 06 79 N Tension force developed in cable AC from equation o in cable AD from equation TAB = 2.0292 X 212,0679 developed TAB 430•3282 N Tension force in cable AB TAD =212.0679 N Tension force developed 2