N_-_Rigid_Body_Equilibrium_2-D

5 6003 3002 6rcy 0 ra x a solution 3 continued f3 b

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Unformatted text preview: RA X A SOLUTION 3 - continued F3 B F2 F1 RC Y RA X A RA Y Step 3: Solve the equilibrium equations RCY = (600(3) + 300(2) – 1000(2.5))/6 = – 16.7 N (downward) RAY = 600 + 300 – RCY = 916.7 N RAX = 1000 N EXAMPLE 4 600 lb Find the support reactions at points A and B if the weight of the boom is 125 lb, center of mass is at G and the load is 600 lb. SOLUTION 4 AY AX 1 ft 40° A 1 ft B FB 3 ft G 125 lb 600 lb 5 ft D FBD of the Boom 600 lb Note: CB is a two-force member, the number of unknowns at B is reduced from two to one. Equations of equilibrium yield: ∑ MA = -125*4-600∗ 9+(FBsin40°)(1)+(FBcos40°)(1)=0; ∑ FX = AX + 4188 cos40° = 0; ∑ FY = AY + 4188 sin40° – 125 – 600 = 0; F = 4188 lb A = –3210 lb A = –1970 lb EXAMPLE 5 The load on the bent rod is supported by a smooth inclined surface at point B and a collar at A. The collar is free to slide over the fixed inclined rod. Determine the support reactions at A and B. SOLUTION 5 MA 5 4 3 3 ft FBD of the Rod 100 lb 200 lb-ft 3 ft 5 13 12 NB 2 ft NA ∑ FX = (4/5) NA – (5/13) NB = 0 ∑ FY = (3/5) NA – (12/13) NB – 100 = 0 ∑ MA = MA – 100∗ 3 – 200 + (12/13) NB∗ 6 – (5/13) NB∗ 2 = 0 Solving th...
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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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