Phys 170 - 2019 Exam 2 Z 1. 1 m 2 m Q QD 1'm 2 m C(1,0,2) 12 p(-2,0,1) 1 A x B ·G ( 0, 0 , 0) y BCO,2 Ay -1 m 1 m Ax G (0, 1 -0.5 ) y V 80 kg . 9.8 ln = 784.8N X FBC = FBC : < 1 , - 2 , 2 > J(1)2+22+22 : ?Bc ·< ?’ ? , , 2) 3 BD = FBD : < - 2,-2, 1> Ja2 + 22 + 12 = FBD : < = 2 , = 2 , 1 > 3 b) Cartesian Equations SFx =0 Ax + ? BBC 2 BBD =0 EF4 = 0 Ay -2 F. y 3 FBC - 2 FRD =0 3 EF2 = 0 A2 + 2= FBC + 1_ FBD - 784 . 8N = 0) c) Moments ' Equations SMB = 0 -4002 + 200k + i J K 0 -2 0 + J K 0 -1 0 Ax Ay Az 0 0 - 789.5 (8MB)2 =D -400 - 2A, + 784,5 =0
Solve Equations in B, sub in what we know solve w / these two 1 FBC - 2 FBD 3 -? FBC - ? FBD = - A y 2 F 3 BC 3 + 1 FBD = 592.5SN FBC = 771.06N 771N FBD = 235.53 ~ 235NV Solve for Ay WIN 771.06 + 235.53 ) N = A, 1 7 671,06N = A = 671N
Review Relative Motion (2) P Block A ( the P force push it up & right y Block B 1 Por. Ff (a-b) NA ( equal & opposite reactions ) 0 WA NA Ef ( a- b ) 7 - LWB x (F) B NB b) Block A: EFx = max This should say aB + ~A / B COS Q P - NA sint - ( Fg ), cost = m (@ x ) EFy = may OA/B Sint - WA + No cost -(F) - sino = m Block B: EF x = max No sing + (Fc) A-B COSQ -(F) B = mBax SiFy = may - NA cost +(FF) A-B sint - WB + NB = mphy =0