Question

Chapter 3, Problem 3/064 One of the smooth vertical walls supporting end B of the 240-kg uniform shaft is turned through a 33° angle as shown here. End A is supported by the ball-and-socket connection in the horizontal x-y plane. Calculate the magnitudes of the forces P and R exerted on the ball end B of the shaft by the vertical walls C and D, respectively. 6.6 m 33 5.2 m 2.5 ñy Answers:

Answer 1

Given m= 240 kg W = mg-240×9.81 = 23.4.4 N. 6.6= √5.2² +2.5? th? - 3.205 m Coordinates: A (2.5,5.2,0)m B(0,0, 3.205) G (2.5, 5-2, 3.205) G (1-25, 2:6, 1.6025) on Position vector : AG = od od = (1.25-2.5) [+ (2.6-5.2) 5+ (1.6025_0JE =(-1.25 ? – 2.6] + 1.6025k) . AB - 0B0A = (-2.50 -5.2} + 3.205k) in Vector} P = p; R = R los 3s + Rsin 33 ) forms] W = – Wii –– 2354.6k N Consider moment about A: SMA=0 (AGxW) + AB X (B + R) = 0 S (-1.25? _ 2.69 + 1. Goz SR ) × (-23 544R) +(-2.5i - s.29 +3-2056)( PÎ + Rcos 3g în treino (-2943; +6121.432) 419 9 Â Hea -2.5 -5.2 3.2051 - 0 1 Roos33 (Ptasin 33 0 1 ang vitame

(-2943 ) + 6121461) + ? (-3-205 (P +R$ in 33) - Î (-3.205xRC0333°)+ ic (-2:56peSin339) +5.2Rko533) 5 -29459 46121. Guî – 3,205pt 1.1455pỉ + 2.688RÎ 1-0 -25PR - 10362 R K + 4.36 kk i (6121.44 -3.205P-1.14558) +î (-2945+2.688R) YAO +(-2.5P _ 1.362 R +436R) k) Equating I coefficients:- - 2943 +2.688R=0 R = 1094-866 N. Equating ħ coefficients: 2.50_1.362 R + 4.36 R=0 - 2.50 +2.998 R=0 P - 2.998x1094-866 2.5 TP-1372.96 N)