N_-_Rigid_Body_Equilibrium_2-D

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online