N_-_Rigid_Body_Equilibrium_2-D

# Solution 2 10 ft rbx b 12 ft 5000 lb 10 ft 5000 lb

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Unformatted text preview: ft 5000 lb 10 ft 5000 lb RAX A RAY Step 1: Draw a FBD and Show all known external forces and the support reactions SOLUTION 2 -Continued 10 ft RBX B 12 ft 5000 lb 5000 lb 10 ft RAX A RAY Step 2: Write the equilibrium equations Σ Fx = 0; RAX - RBX = 0 Σ Fy = 0; RAY – 5000 = 0 Σ MA = 0; 12 RBX – 10(5000) = 0 SOLUTION 2 -Continued 10 ft RBX B 12 ft 5000 lb 10 ft 5000 lb RAX A RAY Step 3: Solve the system equations RAY = 5000 lb RBX = 4167 lb (Left Direction) RAX = 4167 lb FBD Each ball weighs 2943 N FBD – A FBD – B FBD EXAMPLE Smooth wall Assume the ladder weight = W, Draw FBD of the system FBD SOLUTION Smooth wall RBX Wladder Ffriction RAY FBD Draw FBD FBD 10650 lb 12.48 ft RAX RAY RBY FBD 10 lb The idealized model A, B, and C are smooth contact points FC FC 10 lb FC The free body diagram FBD 900 N 3.33 m 1000 N RCY 2.5 m RAX RAY FBD FBD REX RAY REY FBD FBD RAX RD RAY FBD FBD MA RAX A RAY EXAMPLE 1 Determine all the support reactions at points A and E SOLUTION 1 Y A X RE X RAY REY Step 1: Draw FBD SOLUTION 1 - continued Y A X RA YR A Y Y R E Y – REX = 0 R E RX E REX Step 2: write the equilibrium equations Σ FX = 0; Σ FY = 0; Σ MA = 0; RAY – 12 – 14 – 18 + REY = 0 – 3(12) – 6(14) – 9(18) + 12(REY) = 0 SOLUTION 1 -Continued Y A X RA YR A Y Y R E Y R E RX E REX Step 3: Solve the three equilibrium equations REX = 0 kN REY = (3(12) + 6(14) + 9(18))/12 = 23.5 kN RAY = 12 + 14 + 18 – REY = 44 – 23.5 = 20.5 kN EXAMP...
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## This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.

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